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context is that of the limit of a sequence. Let (sn; n ≥ 1) be a sequence of real numbers. If there is a number s such Deﬁnition that \|sn − s\| may ultimately always be as small as we please, then s is said to be the limit of the sequence sn. Formally, we write if and only if for any ε > 0, there is a ﬁnite n0 such that \|sn − s\| < ε for all n > n0. lim n→∞ sn = s 22 0 Introduction Notice that sn need never actually take the value s, it must just get closer to it in the long run (e.g., let xn = n−1). Let (ar ; r ≥ 1) be a sequence of terms, with partial sums Inﬁnite Series sn = n r =1 ar, n ≥ 1. ∞ If sn has a ﬁnite limit s as n → ∞, then the sum s. Otherwise, it diverges. If convergent. ∞ r =1 \|ar \| converges, then ∞ r =1 ar is said to converge with sum r =1 ar is said to be absolutely For example, in the geometric sum in I above, if \|x\| < 1, then \|x\|n → 0 as n → ∞. Hence, ∞ r =x\| < 1, and the series is absolutely convergent for \|x\| < 1. In particular, we have the negative binomial theorem: ∞ r =1 − x)−n. This is true even when n is not an integer. For example, (1 − x)−1/0 2r r − 1) x r /! + · · · In particular, we often use the case n = 2: ∞ r =0 (r + 1)x r = (1 − x)−2. Also, by deﬁnition, for all x, where e is the base of natural logarithms, exp x = ex = ∞ r =0 x r r! Inﬁnite Series 23 and, for \|x\| < 1, − log(1 − x) = ∞ r =1 sr r. An important property of ex is the exponential limit theorem: as n → ∞, n 1 + x n → ex. This has a useful general
ization: let r (n, x) be any function such that nr (n, x) → 0 as n → ∞, then 1 + x n Finally, note that we occasionally use special identities such as → ex, as n → ∞. + r (n, x) n ∞ r =1 1 r 2 = π 2 6 and ∞ r =1 1 r 4 = π 4 90. 1 Probability And of all axioms this shall win the prize. ‘Tis better to be fortunate than wise. John Webster Men’s judgements are a parcel of their fortunes. W. Shakespeare, Antony and Cleopatra 1.1 Notation and Experiments In the course of everyday life, we become familiar with chance and probability in various contexts. We express our ideas and assessments in many ways, such as: It will very likely rain. It is almost impossible to hole this putt. That battery may work for a few more hours. Someone will win the lottery, but it is most unlikely to be one of us. It is about a 50-50 chance whether share prices will rise or fall today. You may care to amuse yourself by noting more such judgments of uncertainty in what you say and in the press. This large range of synonyms, similes, and modes of expression may be aesthetically pleasing in speech and literature, but we need to become much more precise in our thoughts and terms. To aid clarity, we make the following. (1) Deﬁnition Any well-deﬁned procedure or chain of circumstances is called an experiment. The end results or occurrences are called the outcomes of the experiment. The set of possible outcomes is called the sample space (or space of outcomes) and is denoted by the Greek letter. In cases of interest, we cannot predict with certainty how the experiment will turn out, rather we can only list the collection of possible outcomes. Thus, for example: Experiment (a) Roll a die (b) Flip a coin Possible outcomes One of the faces Head or tail 24 1.1 Notation and Experiments 25 (c) Buy a lottery ticket (d) Deal a bridge hand Win a prize, or not All possible arrangements of 52 cards into (e) Run a horse race Any ordering of the runners four equal parts Typically, probability statements do not refer to individual outcomes in the sample space; instead, they tend to embrace collections of outcomes or subsets of. Here
are some examples: Experiment (a) Deal a poker hand (b) Buy a share option (c) Telephone a call centre (d) Buy a car (e) Get married Set of outcomes of interest Have at least a pair Be in the money at the exercise date Get through to a human in less than 1 hour It runs without major defects for a whole year Stay married Clearly this is another list that you could extend without bound. The point is that in typical probability statements of the form which we also write as the probability of A is p, P(A) = p, the symbol A represents groups of outcomes of the kind exempliﬁed above. Furthermore, we concluded in Chapter 0 that the probability p should be a number lying between 0 and 1 inclusive. A glance at the Appendix to that chapter makes it clear that P(.) is in fact simply a function on these subsets of, which takes values in [0, 1]. We make all this formal; thus: (2) (3) Deﬁnition An event A is a subset of the sample space. Deﬁnition [0, 1]. The probability of the event A is denoted by P(A). Probability is a function, deﬁned on events in, that takes values in (4) Example: Two Dice Suppose the experiment in question is rolling two dice. (Note that in this book a “die” is a cube, conventionally numbered from 1 to 6, unless otherwise stated.) Here are some events: (a) A = Their sum is 7. (b) B = The ﬁrst die shows a larger number than the second. (c) C = They show the same. We may alternatively display these events as a list of their component outcomes, so A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, and so on, but this is often very tedious. 26 1 Probability Of course this experiment has the type of symmetry we discussed in Section 0.3, so we can also compute P(A) = P(C) = 1 6, and P(B) = 15 36 = 5 12. (5) Example: Darts A dart is thrown to hit a chess board at random. Here, “at random” clearly means that it is equally likely to hit any point of the board
. Possible events are: (a) A = It hits a white square. (b) B = It lands within one knight’s move of a corner square. This problem is also symmetrical in the sense of Section 0.3, and so we calculate P(A) = 1 2 and P(B) = 1 8. Of course, the brief deﬁnitions above raise more questions than they answer. How should the probability function behave when dealing with two or more events? How can it be extended to cope with changes in the conditions of the experiment? Many more such questions could be posed. It is clear that the brief summary above calls for much explanation and elaboration. In the next few sections, we provide a few simple rules (or axioms) that deﬁne the properties of events and their probabilities. This choice of rules is guided by our experience of real events and their likelihoods, but our experience and intuition cannot prove that these rules are true or say what probability “really” is. What we can say is that, starting with these rules, we can derive a theory that provides an elegant and accurate description of many random phenomena, ranging from the behaviour of queues in supermarkets to the behaviour of nuclear reactors. 1.2 Events Let us summarise our progress to date. Suppose we are considering some experiment such as tossing a coin. To say that the experiment is well deﬁned means that we can list all the possible outcomes. In the case of a tossed coin, the list reads: (head, tail). For a general (unspeciﬁed) experiment, any particular outcome is denoted by ω; the collection of all outcomes is called the sample space and is denoted by. Any speciﬁed collection of outcomes in is called an event. Upper case letters such as A, B, and C are used to denote events; these may have sufﬁces or other adornments such as Ai, ¯B, C ∗, and so on. If the outcome of the experiment is ω and ω ∈ A, then A is said to occur. The set of outcomes not in A is called the complement of A and is denoted by Ac. In particular, the event that contains all possible outcomes is the certain event and is denoted by. Also, the event containing no outcomes is the impossible event and is denoted by φ. Obviously, φ
= c. What we said in Chapter 0 makes it natural to insist that P() = 1 and that P(φ) = 0. It is also clear from our previous discussions that the whole point of probability is to say how likely the various outcomes are, either individually or, more usually, collectively in events. Here are some more examples to illustrate this. 1.2 Events 27 Suppose n people are picked at random and interrogated as to Example: Opinion Poll their opinion (like or dislike or do not care) about a brand of toothpaste. Here, the sample space is all collections of three integers (x, y, z) such that x + y + z = n, where x is the number that like it, y the number that dislike it, and z the number that do not care. Here an event of interest is A ≡ more like it than dislike it which comprises all the triples (x, y, z) with x > y. Another event that may worry the manufacturers is B ≡ the majority of people do not care, which comprises all triples (x, y, z), such that x + y < z. Example: Picking a Lottery Number In one conventional lottery, entries and the draw choose six numbers from 49. The sample space is therefore all sextuples {x1,..., x6}, where all the entries are between 1 and 49, and no two are equal. The principal event of interest is that this is the same as your choice. Example: Coins notation above, we usually write If a coin is tossed once, then = {head, tail}. In line with the = {H, T }. The event that the coin shows a head should strictly be denoted by {H }, but in common with most other writers we omit the braces in this case, and denote a head by H. Obviously, H c = T and T c = H. Likewise, if a coin is tossed twice, then = {H H, H T, T H, T T }, and so on. This experiment is performed even more often in probability textbooks than it is in real life. Because events are sets, we use the usual notation for combining them; thus: A ∩ B denotes outcomes in both A and B; their intersection. A ∪ B denotes outcomes in either A or B or both; their union. AB denotes outcomes in either A or B, but not both; their symmetric difference. A\B denotes outcomes in
A that are not in B; their difference. ∞ A j denotes outcomes that are in at least one of the countable collection (A j ; j ≥ 1); j=1 their countable union. [Countable sets are in one–one correspondence with a subset of the positive integers.] A ⊆ B denotes that every outcome in A is also in B; this is inclusion. A = {ω1, ω2, ω3,..., ωn} denotes that the event A consists of the outcomes ω1,..., ωn A × B denotes the product of A and B; that is, the set of all ordered pairs (ωa, ωb), where ωa ∈ A and ωb ∈ B. 28 1 Probability Figure 1.1 The interior of the rectangle represents the sample space, and the interior of the circle represents an event A. The point ω represents an outcome in the event Ac. The diagram clearly illustrates the identities Ac ∪ A = and \A = Ac. These methods of combining events give rise to many equivalent ways of denoting an event. Some of the more useful identities for any events A and B are: (1) (2) (3) (4) (5) (6) AB = (A ∩ Bc) ∪ (Ac ∩ B) A = (A ∩ B) ∪ (A ∩ Bc) A\B = A ∩ Bc Ac = \A A ∩ Ac = φ A ∪ Ac =. These identities are easily veriﬁed by checking that every element of the left-hand side is included in the right-hand side, and vice versa. You should do this. Such relationships are often conveniently represented by simple diagrams. We illustrate this by providing some basic examples in Figures 1.1 and 1.2. Similar relationships hold Figure 1.2 The interior of the smaller circle represents the event A; the interior of the larger circle represents the event B. The diagram illustrates numerous simple relationships; for example, the region common to both circles is A ∩ B ≡ (Ac ∪ Bc)c. For another example, observe that A B = Ac Bc. 1.2 Events 29 between combinations of three or more events and some of these are given in the problems at the end of this chapter. When A ∩ B = φ
we say that A and B are disjoint (or mutually exclusive). (7) Example A die is rolled. The outcome is one of the integers from 1 to 6. We may denote these by {ω1, ω2, ω3, ω4, ω5, ω6}, or more directly by {1, 2, 3, 4, 5, 6}, as we choose. Deﬁne: A the event that the outcome is even, B the event that the outcome is odd, C the event that the outcome is prime, D the event that the outcome is perfect (a perfect number is the sum of its prime factors). Then the above notation compactly expresses obvious statements about these events. For exampleω6} A ∪ B = C\A = B\{ω1} and so on. It is natural and often useful to consider the number of outcomes in an event A. This is denoted by \|A\|, and is called the size or cardinality of A. It is straightforward to see, by counting the elements on each side, that size has the following properties. If A and B are disjoint, then \|A ∪ B\| = \|A\| + \|B\|, and more generally, for any A and B \|A ∪ B\| + \|A ∩ B\| = \|A\| + \|B\|. If A ⊆ B, then For the product A × B, Finally, \|A\| ≤ \|B\|. \|A × B\| = \|A\|\|B\|. \|φ\| = 0. (8) (9) (10) (11) (12) (13) Example The Shelmikedmu are an elusive and nomadic tribe whose members are unusually heterogeneous in respect of hair and eye colour, and skull shape. A persistent anthropologist establishes the following facts: (i) 75% have dark hair, the rest have fair hair. (ii) 80% have brown eyes; the rest have blue eyes. 30 1 Probability (iii) No narrow-headed person has fair hair and blue eyes. (iv) The proportion of blue-eyed broad-headed tribespeople is the same as the proportion of blue-eyed narrow-headed tribespeople. (v) Those who are blue-eyed and broad-headed are fair-haired or dark-haired in equal proportion. (vi) Half the tribe is dark-haired and broad-headed.
(vii) The proportion who are brown-eyed, fair-haired, and broad-headed is equal to the proportion who are brown eyed, dark-haired, and narrow-headed. The anthropologist also ﬁnds n, the proportion of the tribe who are narrow-headed, but unfortunately this information is lost in a clash with a crocodile on the difﬁcult journey home. Is another research grant and ﬁeld trip required to ﬁnd n? Fortunately, not if the anthropologist uses set theory. Let B be the set of those with blue eyes C be the set of those with narrow heads D be the set of those with dark hair Then the division of the tribe into its heterogeneous sets can be represented by Figure 1.3. This type of representation of sets and their relationships is known as a Venn diagram. The proportion of the population in each set is denoted by the lower case letter in each compartment, so a = \|Bc ∩ C c ∩ Dc\|/\|\|, b = \|B ∩ C c ∩ Dc\|/\|\|, Figure 1.3 Here the interior of the large circle represents the entire tribe, and the interior of the small circle represents those with narrow heads. The part to the right of the vertical line represents those with dark hair, and the part above the horizontal line represents those with blue eyes. Thus, the shaded quadrant represents those with blue eyes, narrow heads, and fair hair; as it happens, this set is empty by (iii). That is to say B ∩ C ∩ Dc = φ, and so g = 0. 1.2 Events 31 and so on. The required proportion having narrow heads is n = \|C\|/\|\| = e + f + g + h and, of course. The information in (i)–(vii), which survived the crocodile, yields the following relationships.75 a + d + e + h = 0..5 a = e The anthropologist (who has a pretty competent knowledge of algebra) solves this set of equations to ﬁnd that.15 + 0.1 + 0.05 = 0.3 Thus, three-tenths of the tribe are narrow-headed. This section concludes with a technical note (which you may omit on a ﬁrst reading). We have noted that events are subsets of. A natural question is, which subsets of
are entitled to be called events? It seems obvious that if A and B are events, then A ∪ B, Ac, A ∩ B, and so on should also be entitled to be events. This is a bit vague; to be precise, we say that a subset A of can be an event if it belongs to a collection F of subsets of, obeying the following three rules: ∈ F; ifA ∈ F then Ac ∈ F; ifA j ∈ F for j ≥ 1, then ∞ j=1 A j ∈ F. (i) (ii) (iii) (iv) (v) (vi) (vii) (14) (15) (16) The collection F is called an event space or a σ -ﬁeld. Notice that using (1)–(6) shows that if A and B are in F, then so are A\B, AB and A ∩ B. (17) Example (7) Revisited It is easy for you to check that {φ, A, B, } is an event space, and {φ, A ∪ C, B\C, } is an event space. However, {φ, A, } and {φ, A, B, D, } are not event spaces. In general, if is ﬁnite, it is quite usual to take F to be the collection of all subsets of, which is clearly an event space. If is inﬁnite, then this collection is sometimes too big to be useful, and some smaller collection of subsets is required. (1) (2) (3) (4) (5) 32 1 Probability 1.3 The Addition Rules for Probability An event A has probability P(A). But any experiment may give rise to a great many events of interest, and we have seen in Section 1.2 that these can be combined in numerous ways. We need some rules that tell us how to deal with probabilities of complicated events. Naturally, we continue to require that for any event A 0 ≤ P(A) ≤ 1 and, in particular, that the certain event has probability 1, so P() = 1. It turns out that we require only one type of rule, the addition rules. The simplest form of this is as follows: The Basic Addition Rule If A and B are disjoint events, then
P(A ∪ B) = P(A) + P(B). This rule lies at the heart of probability. First, let us note that we need such a rule, because A ∪ B is an event when A and B are events, and we therefore need to know its probability. Second, note that it follows from (3) (by induction) that if A1, A2,..., An is any collection of disjoint events, then n P i=1 Ai = P(A1) + · · · + P(An). The proof is a simple exercise, using induction. Third, note that it is sometimes too restrictive to conﬁne ourselves to a ﬁnite collection of events (we have seen several sample spaces, with inﬁnitely many outcomes), and we therefore need an extended version of (4). Extended Addition Rule If A1, A2,... is a collection of disjoint events, then P(A1 ∪ A2 ∪ · · ·) = P(A1) + P(A2) + · · ·. Equation (5), together with (1) and (2), 0 ≤ P(A) ≤ 1 and P() = 1, are sometimes said to be the axioms of probability. They describe the behaviour of the probability function P deﬁned on subsets of. In fact, in everyday usage, P is not referred to as a probability function but as a probability distribution. Formally, we state the following. Let be a sample space, and suppose that P(·) is a probability function Deﬁnition on a family of subsets of satisfying (1), (2), and (5). Then P is called a probability distribution on. 1.3 The Addition Rules for Probability 33 The word distribution is used because it is natural to think of probability as something that is distributed over the outcomes in. The function P tells you just how it is distributed. In this respect, probability behaves like distributed mass, and indeed in many books authors do speak of a unit of probability mass being distributed over the sample space, and refer to P as a probability mass function. This metaphor can be a useful aid to intuition because, of course, mass obeys exactly the same addition rule. If two distinct objects A and B have respective masses m(A) and m(B), then the mass of
their union m(A ∪ B) satisﬁes m(A ∪ B) = m(A) + m(B). Of course, mass is also nonnegative, which reinforces the analogy. We conclude this section by showing how the addition rule is consistent with, and suggested by, our interpretations of probability as a proportion. First, consider an experiment with equally likely outcomes, for which we deﬁned prob- ability as the proportion P(A) = \|A\| \|\|. If A and B are disjoint then, trivially, \|A ∪ B\| = \|A\| + \|B\|. Hence, in this case, P(A ∪ B) = \|A ∪ B\| \|\| = \|A\| \|\| + \|B\| \|\| = P(A) + P(B). Second, consider the interpretation of probability as reﬂecting relative frequency in the long run. Suppose an experiment is repeated N times. At each repetition, events A and B may or may not occur. If they are disjoint, they cannot both occur at the same repetition. We argued in Section 0.4 that the relative frequency of any event should be not too far from its probability. Indeed, it is often the case that the relative frequency N (A)/N of an event A is the only available guide to its probability P(A). Now, clearly N (A ∪ B) = N (A) + N (B). Hence, dividing by N, there is a powerful suggestion that we should have P(A ∪ B) = P(A) + P(B). Third, consider probability as a measure of expected value. For this case, we resurrect the benevolent plutocrat who is determined to give away $1 at random. The events A and B are disjoint. If A occurs, you get $1 in your left hand; if B occurs you get $1 in your right hand. If (A ∪ B)c occurs, then Jack gets $1. The value of this offer to you is $P(A ∪ B); the value to your left hand is $P(A); and the value to your right hand is $P(B). Obviously, it does not matter in which hand you get the money, so P(A ∪ B) = P(A) + P(B). Finally, consider the case where we imagine a point is picked
at random anywhere in some plane region of area \|\|. If A ⊆, we deﬁned P(A) = \|A\| \|\|. 34 1 Probability Because area also satisﬁes the addition rule, we have immediately, when A ∩ B = φ, that P(A ∪ B) = P(A) + P(B). It is interesting and important to note that in this case the analogy with mass requires the unit probability mass to be distributed uniformly over the region. We can envisage this distribution as a lamina of uniform density \|\|−1 having total mass unity. This may seem a bizarre thing to imagine, but it turns out to be useful later. In conclusion, it seems that the addition rule is natural and compelling in every case where we have any insight into the behaviour of probability. Of course, it is a big step to say that it should apply to probability in every other case, but it seems inevitable. Doing so has led to remarkably elegant and accurate descriptions of the real world. This property (4) is known as ﬁnite additivity and (5) is countable additivity. Note that if A ⊆ B, then (6) (7) P(A) ≤ P(B) and ﬁnally, using A ∪ Ac =, we have P(φ) = 0. Once again, these statements are quite consistent with our intuition about likelihoods, as reinforced by experience. Historically, the theory of probability has its roots ﬁrmly based in observation of games of chance employing cards, dice, and lotteries. (8) Example Three dice are rolled and the numbers on the upper faces are added together. The outcomes 9 and 10 can each be obtained in six distinct ways; thus: 10 =. Some time before 1642, Galileo was asked to explain why, despite this, the outcome 10 is more likely that the outcome 9, as shown by repeated experiment. He observed that the sample space has 63 = 216 outcomes, being all possible triples of numbers from 1 to 6. Of these, 27 sum to 10, and 25 sum to 9, so P(10) = 27 216 and P(9) = 25 216. This provides an explanation for the preponderance of 10 over 9. It is just this kind of agreement between theory and experiment that justiﬁes our adoption of the rules above. We will see many more
examples of this. 1.4 Properties of Probability We have agreed that, given a space F of events A, the probability function P(·) satisﬁes the following rules (or axioms) that we display as a deﬁnition. 1.4 Properties of Probability 35 Deﬁnition The function P(·) : F → [0, 1] is a probability function if (1) (2) (3) (4) (5) (6) (7) (8) and P(A) ≥ 0 for all A ∈ F, P() = 1 ∞ j=1 P A j = ∞ j=1 P(A j ) whenever A1, A2,... are disjoint events (which is to say that Ai ∩ A j = φ whenever i = j). In passing, we note that (3) is known as the property of countable additivity. Obviously, it implies ﬁnite additivity so that, in particular, if A ∩ B = φ, then P(A ∪ B) = P(A) + P(B). From these three rules we can derive many important and useful relationships, for example, P(φ) = 0 P(Ac) = 1 − P(A), n P 1 Ai = P(A\B) = P(A) − P(A ∩ B), P(A ∪ B) = P(A) + P(B) − P(A ∩ B), n 1 P(Ai ) − i< j P(Ai ∩ A j ) + i< j<k + (−)n+1P(A1 ∩ A2 ∩... ∩ An). P(Ai ∩ A j ∩ Ak) + · · · The following examples begin to demonstrate the importance and utility of (1)–(8). (9) Example Let us prove (4), (5), and (7) above. First, by (2) and (3), for any A ∈ F, 1 = P() = P(A ∪ Ac) = P(A) + P(Ac) which proves (5). Now, setting A = establishes (4). Finally, using (3) repeatedly, we obtain (10) and P(B ) = P(B �
� A) + P(B ∩ Ac), P(A ∪ B) = P(A ∪ (B ∩ Ac)) = P(A) + P(B ∩ Ac) = P(A) + P(B) − P(B ∩ A) by (1), which proves (7). 36 1 Probability You should now prove (6) as an elementary exercise; the proof of (8) is part of Problem 12. (11) Example: Inequalities for P(·) (i) If A ⊆ B then B ∩ A = A, so from (10) P(B) = P(A) + P(B ∩ Ac) ≥ P(A) by (1). (ii) For any A, B, we have Boole’s inequalities (12) P(A) + P(B) ≥ P(A ∪ B) ≥ max{P(A), P(B)} ≥ P(A ∩ B) ≥ P(A) + P(B) − 1 by, (7). by part (i) by part (i) again by (7) again (13) Example: Lottery An urn contains 1000 lottery tickets numbered from 1 to 1000. One is selected at random. A fairground performer offers to pay $3 to anyone who has already paid him $2, if the number on the ticket is divisible by 2, 3, or 5. Would you pay him your $2 before the draw? (If the ticket number is not divisible by 2, 3, or 5 you lose your $2.) Solution Let Dk be the event that the number drawn is divisible by k. Then P(D2) = 500 1000 = 1 2 and so on. Also P(D2 ∩ D3) = P(D6) = 166 1000 and so forth. Using (8) with n = 3 and making several similar calculations, we have P(D2 ∪ D3 ∪ D5) = P(D2) + P(D3) + P(D5) + P(D2 ∩ D3 ∩ D5) − P(D2 ∩ D3) − P(D3 ∩ D5) − P(D2 ∩ D5) = 10−3(500 + 333 + 200 + 33 − 166 − 66 − 100) = 367
500. The odds on winning are thus better than 2:1, and you should accept his generous offer. 1.5 Sequences of Events This section is important, but may be omitted at a ﬁrst reading. Often, we are confronted by an inﬁnite sequence of events (An; n ≥ 1) such that A = limn→∞ An exists. In particular, if An ⊆ An+1 for all n, then (1) lim n→∞ An = ∞ j=1 A j = A, and A is an event by (1.2.16). It is of interest, and also often useful, to know P(A). The following theorem is therefore important as well as attractive. (2) Theorem If An ⊆ An+1 ∈ F for all n ≥ 1 and A = limn→∞ An, then 1.6 Remarks 37 P(A) = lim n→∞ P(An). Proof. Because An ⊆ An+1, we have (A j+1\A j ) ∩ (Ak+1\Ak) = φ, for k = j. Also, setting A0 = φ n 1 (A j \A j−1) = An. Furthermore, (3) P(An+1\An) = P(An+1) − P(An). Hence, because A0 = φ P(A) = P = P lim n→∞ An A j+1\A j ∞ j=0 ∞ = P(A j+1\A j ) by (1.4.3) j=0 = lim n→∞ P(An) by (3). From this result, it is a simple matter to deduce that if An ⊇ An+1 for all n, then (4) lim n→∞ P(An) = P(A). With a bit more work, one can show more generally that if limn→∞ An = A, then (4) is still true. Because of this, the function P(.) is said to be a continuous set function. 1.6 Remarks Simple problems in probability typically require the calculation of the probability P(E) of some event E, or at least the calculation of bounds for P(E). The underlying experiment may be implicit or explicit; in any case, the
ﬁrst step is always to choose the sample space. (Naturally, we choose the one that makes ﬁnding P(E) easiest.) Then you may set about ﬁnding P(E) by using the rules and results of Section 1.4, and the usual methods for manipulating sets. Useful aids at this simple level include Venn diagrams, and identities such as de Morgan’s laws, namely: (1) (2) and c Ai = n i=1 n i=1 Ac i c Ai = n i=1 n i=1. Ac i 38 1 Probability These are readily established directly or by induction, as follows. First, draw a Venn diagram to see that (A ∪ B)c = Ac ∩ Bc. Now if n j=1 A j = Bn, then n+1 j=1 c A j = (An+1 ∪ Bn)c = Ac n+1 ∩ Bc n. Hence, (1) follows by induction on n. The result (2) can also be proved directly, by induction, or by using (1). You can do this, and you should do it now. We end this section with a note for the more demanding reader. It has been claimed above that probability theory is of wide applicability, yet most of the examples in this chapter deal with the behaviour of coins, dice, lotteries, and urns. For most of us, weeks or even months can pass without any involvement with dice or urns. (The author has never even seen an urn, let alone removed a red ball from one.) The point is that such simple problems present their probabilistic features to the reader unencumbered by strained assumptions about implausible models of reality. The penalty for simplicity is that popular problems may become hackneyed through overuse in textbooks. We take this risk, but reassure the reader that more realistic problems feature largely in later chapters. In addition, when considering some die or urn, students may be pleased to know that they are treading in the footsteps of many eminent mathematicians as they perform these calculations. Euler or Laplace may have pondered over exactly the same difﬁculty as you, although perhaps not for so long. 1.7 Review and Checklist for Chapter 1 We began by introducing the idea of experiments and their outcomes, with events and their probabilities. Then we used
our experience of chance events, and intuitive ideas about probability as a proportion, to formulate a set of rules (or axioms) for probability. Of these, the addition rule is the most important, and we derived several useful corollaries that will be in constant use. In the course of this, we also introduced some standard notation. For convenience, the key rules and results are summarized here. Notation sample space of outcomes ≡ certain event A, B, C possible events in φ P(.) P(A) A ∪ B A ∩ B Ac A\B impossible event the probability function the probability that A occurs either A or B occurs, or both occur (union) both A and B occur (intersection) A does not occur (complement) difference: A occurs and B does not 1.7 Review and Checklist for Chapter 1 39 either A or B, but not both if A occurs, then B occurs is a partition of if the Ai are disjoint with union AB A ⊆ B {Ai } A ∩ B = φ A and B are disjoint F \|A\| event space (or σ -ﬁeld) cardinality (or size) of A Axioms and Rules Any subset A of is an event if it belongs to a collection F of subsets of such that ∈ F if A ∈ F, then Ac ∈ F if Ai ∈ F for i ≥ 1, then Ai ∈ F. ∞ i=1 A function P(.) deﬁned on events is a probability function or probability distribution if P(A) ≥ 0 for all A in F P() = 1 ∞ P 1 Ai = i P(Ai ) for any disjoint events {Ai }. n Ai. 1 The function P(.) obeys the following rules: Range: 0 ≤ P(A) ≤ 1 Impossible event: P(φ) = 0 Inclusion-exclusion: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) Complement: P(A) = 1 − P(Ac) Difference: P(A\B) = P(A) − P(B) when B ⊆ A General inclusion-exclusion: n P Ai = P(Ai ) − P(Ai ∩ A j ) + · · · − (−1)nP
1 i i< j n P 1 Ai ≤ i P(Ai ). Checklist of Terms for Chapter 1 Boole’s inequalities: 1.1 experiment outcome sample space event probability Venn diagram 1.2 certain event impossible event 1 Probability 40 event space σ -ﬁeld 1.3 addition rules probability distribution countable additivity 1.4 axioms of probability Boole’s inequalities 1.5 continuous set function 1.6 de Morgan’s Laws.8 Example: Dice You roll two dice. What is the probability of the events: (a) They show the same? (b) Their sum is seven or eleven? (c) They have no common factor greater than unity? Solution First, we must choose the sample space. A natural representation is as ordered pairs of numbers (i, j), where each number refers to the face shown by one of the dice. We require the dice to be distinguishable (one red and one green, say) so that 1 ≤ i ≤ 6 and 1 ≤ j ≤ 6. Because of the symmetry of a perfect die, we assume that these 36 outcomes are equally likely. (a) Of these 36 outcomes, just six are of the form (i, i), so using (1.3.1) the required probability is \|A\| \|\| = 6 36 = 1 6. (b) There are six outcomes of the form (i, 7 − i) whose sum is 7, so the probability that the sum is 7 is 6 36 = 1 6. There are two outcomes whose sum is 11—namely, (5, 6) and (6, 5)—so the prob- ability that the sum is 11 is 2 36 = 1 18. Hence, using (1.4.3), the required probability is 1 6 + 1 18 = 2 9. (c) It is routine to list the outcomes that do have a common factor greater than unity. They are 13 in number, namely: {(i, i); i ≥ 2}, (2, 4), (4, 2), (2, 6), (6, 2), (3, 6), (6, 3), (4, 6), (6, 4). This is the complementary event, so by (1.4.5) the required probability is 1 − 13 36 Doing it this way gives a slightly quicker enumeration than the direct approach. = 23 36. Worked Examples and Ex
ercises 41 (1) (2) (3) (4) (5) Exercise What is the probability that the sum of the numbers is 2, 3, or 12? Exercise What is the probability that: (a) The sum is odd? (b) The difference is odd? (c) The product is odd? Exercise What is the probability that one number divides the other? Exercise What is the probability that the ﬁrst die shows a smaller number than the second? Exercise What is the probability that different numbers are shown and the smaller of the two numbers is r, 1 ≤ r ≤ 6? Remark It was important to distinguish the two dice. Had we not done so, the sample space would have been {(i, j); 1 ≤ i ≤ j ≤ 6}, and these 21 outcomes are not equally likely, either intuitively or empirically. Note that the dice need not be different colours in fact, it is enough for us to be able to suppose that they are. 1.9 Example: Urn An urn contains n heliotrope and n tangerine balls. Two balls are removed from the urn together, at random. (a) What is the sample space? (b) What is the probability of drawing two balls of different colours? (c) Find the probability pn that the balls are the same colour, and evaluate lim n→∞ pn. Solution drawn together, a natural sample space is = {HH, H T, T T }. (a) As balls of the same colour are otherwise indistinguishable, and they are (b) The outcomes in exhibit no symmetry. Taking our cue from the previous example, we choose to distinguish the balls by numbering them from 1 to 2n, and also suppose that they are drawn successively. Then the sample space is the collection of ordered pairs of the form (i, j), where 1 ≤ i, j ≤ 2n, and i = j, because we cannot pick the same ball twice. These 2n(2n − 1) outcomes are equally likely, by symmetry. In n2 of them, we draw heliotrope followed by tangerine, and in n2, we draw tangerine followed by heliotrope. Hence, P(H T ) = n2 + n2 2n(2n − 1) = n 2n − 1. (c) By (1.4.5) pn = 1 − P(
H T ) = n − 1 2n − 1 → 1 2 as n → ∞. Find P(HH) when the sample space is taken to be all unordered pairs of distinguishable (1) (2) (3) Exercise balls. Exercise tangerine? Exercise removed. What is the probability that it is tangerine? What is the probability that (a) the ﬁrst ball is tangerine? (b) the second ball is Half the balls are removed and placed in a box. One of those remaining in the urn is 42 1 Probability (4) A fair die with n sides is rolled. If the r th face is shown, r balls are removed from Exercise the urn and placed in a bag. What is the probability that a ball removed at random from the bag is tangerine? 1.10 Example: Cups and Saucers A tea set has four cups and saucers with two cups and saucers in each of two different colours. If the cups are placed at random on the saucers, what is the probability that no cup is on a saucer of the same colour? Solution I Call the colours azure and blue; let A be the event that an azure cup is on an azure saucer, and B the event that a blue cup is on a blue saucer. Because there are only two places for blue cups not to be on blue saucers, we see that A occurs if and only if B occurs, so A = B and P((A ∪ B)c) = P(Ac) = 1 − P(A). Now P(A) = P(A1 ∪ A2), where A1 and A2 denote the events that the ﬁrst and second azure cups, respectively, are on azure saucers. There are 24 equally probable ways of putting the four cups on the four saucers. In 12 of them, A1 occurs; in 12 of them, A2 occurs; and in 4 of them, A1 ∩ A2 occurs, by enumeration. Hence, by (1.4.7), P(A) = 12 24 + 12 24 − 4 24 = 5 6, and the required probability is 1 6. Solution II Alternatively, instead of considering all the ways of placing cups on saucers, we may consider only the distinct ways of arranging the cups by colour with the saucers �
�xed. There are only six of these, namely: aabb; abba; abab; baab; baba; bbaa; and by symmetry they are equally likely. By inspection, in only one of these arrangements is no cup on a saucer of the same colour, so the required probability is 1 6. Remark In this example, considering a smaller sample space makes the problem easier. This is in contrast to our solutions to Examples 1.7 and 1.8, where we used larger sample spaces to simplify things. (1) (2) Exercise What is the probability that exactly (a) One cup is on a saucer of the same colour? (b) Two cups are on saucers of the same colour? Exercise What is the probability that no cup is on a saucer of the same colour if the set comprises four cups and saucers in four distinct colours? Worked Examples and Exercises 43 1.11 Example: Sixes Three players, Achilles, Briseis, and Chryseis, take turns to roll a die in the order ABC, ABC, A.... Each player drops out of the game immediately upon throwing a six. (a) What is the sample space for this experiment? (b) Suppose the game stops when two players have rolled a six. What is the sample space for this experiment? (c) What is the probability that Achilles is the second player to roll a six? (d) Let Dn be the event that the third player to roll a six does so on the nth roll. Describe the event E given by E = ∞ c Dn. n=1 Solution that (a) Let U be the collection of all sequences x1,..., xn for all n ≥ 1, such x j ∈ {1, 2, 3, 4, 5} xn = 6. for 1 ≤ j < n, Then each player’s rolls generate such a sequence, and the sample space consists of all selections of the triple (u1, u2, u3), where ui ∈ U for 1 ≤ i ≤ 3. We may denote this by U × U × U or even U 3, if we want. (b) Let V be the collection of all sequences x1, x2,..., xn, for n ≥ 1, such that x j ∈ {1, 2, 3, 4, 5
} for 1 ≤ j ≤ n. Then the sample space consists of two selections u1, u2 from U, corresponding to the players who roll sixes, and one selection v from V corresponding to the player who does not. The length of v equals the longer of u1 and u2 if this turn in the round comes before the second player to get a six, or it is one less than the longer of u1 and u2 if this turn in the round is later than the second player to get a six. (c) Despite the answers to (a) and (b), we use a different sample space to answer this question. Suppose the player who is ﬁrst to roll a six continues to roll the die when the turn comes round, these rolls being ignored by the others. This does not affect the respective chances of the other two to be the next player (of these two) to roll a six. We therefore let be the sample space consisting of all sequences of length 3r + 1, for r ≥ 0, using the integers 1, 2, 3, 4, 5, or 6. This represents 3r + 1 rolls of the die, and by the assumed symmetry the 63r +1 possible outcomes are all equally likely for each r. Suppose Achilles is the second player to roll a six on the 3r + 1th roll. Then his r + 1 rolls include no six except his last roll; this can occur in 5r ways. If Briseis was ﬁrst to roll a six, then her r rolls include at least one six; this may be accomplished in 6r − 5r ways. In this case, Chryseis rolled no six in r attempts; this can be done in 5r ways. Hence, Achilles is second to Briseis in 5r.5r.(6r − 5r ) outcomes. Likewise, he is second to Chryseis in 5r.5r.(6r − 5r ) outcomes. Hence, the probability that he is second to roll a six on the 3r + 1th roll is pr = 2(6r − 5r )52r 63r +1, for r ≥ 1. 44 1 Probability By (1.4.3), therefore, the total probability that Achilles is second to roll a six is the sum of these, namely ∞ r =1 pr = 300 1001. ∞ (d) The event n=1 D
n is the event that the game stops at the nth roll for some n ≥ 1. Therefore, E is the event that they never stop. For each player, ﬁnd the probability that he or she is the ﬁrst to roll a six. Show that P(E) = 0. Find the probability that the Achilles rolls a six before Briseis rolls a six. (Hint: use (1) (2) (3) (4) Exercise Exercise Exercise a smaller sample space.) Exercise that he is more likely to be last than to be second? Show that the probability that Achilles is last to throw a six is 305 1001. Are you surprised The interesting thing about the solution to (c) is that the sample space Remark includes outcomes that are not in the original experiment, whose sample space is described in (a). The point is that the event in question has the same probability in the original experiment and in the modiﬁed experiment, but the required probability is obtained rather more easily in the second case because the outcomes are equally likely. This idea of augmenting the sample space was ﬁrst used by Pascal and Fermat in the seventeenth century. In fact, we ﬁnd easier methods for evaluating this probability in Chapter 2, using new concepts. 1.12 Example: Family Planning A woman planning her family considers the following schemes on the assumption that boys and girls are equally likely at each delivery: (a) Have three children. (b) Bear children until the ﬁrst girl is born or until three are born, whichever is sooner, and then stop. (c) Bear children until there is one of each sex or until there are three, whichever is sooner, and then stop. Let Bi denote the event that i boys are born, and let C denote the event that more girls are born than boys. Find P(B1) and P(C) in each of the cases (a) and (b). Solution (a) If we do not consider order, there are four distinct possible families: BBB, GGG, GGB, and BBG, but these are not equally likely. With order included, there are eight possible families in this larger sample space: (1) {B B B; B BG; BG B; G B B; GG B; G BG; BGG; GGG} = and by symmetry they are equally likely. Now, by (1.3
.1), P(B1) = 3 fact that P(C) = 1 2 is also clear by symmetry. 8 and P(C) = 1 2. The Worked Examples and Exercises 45 Now consider (b). There are four possible families: F1 = G, F2 = BG, F3 = B BG, and F4 = B B B. Once again, these outcomes are not equally likely, but as we have now done several times we can use a different sample space. One way is to use the sample space in (1), remembering that if we do this then some of the later births are ﬁctitious. The advantage is that outcomes are equally likely by symmetry. With this choice, F2 corresponds to {BGG ∪ BG B} and so. Likewise, F1 = {GGG ∪ GG B ∪ G BG ∪ G B B} and so P(C) = 1. P(B1) = P(F2) = 1 2 4 (2) (3) (4) Exercise Exercise Exercise Find P(E) in all three cases. Find P(B1) and P(C) in case (c). Find P(B2) and P(B3) in all three cases. Let E be the event that the completed family contains equal numbers of boys and girls. 1.13 Example: Craps You roll two fair dice. If the sum of the numbers shown is 7 or 11, you win; if it is 2, 3, or 12, you lose. If it is any other number j, you continue to roll two dice until the sum is j or 7, whichever is sooner. If it is 7, you lose; if it is j, you win. What is the probability p that you win? Solution Suppose that you roll the dice n times. That experiment is equivalent to rolling 2n fair dice, with the sample space 2n being all possible sequences of length 2n, of the numbers 1, 2, 3, 4, 5, 6, for any n ≥ 1. By symmetry, these 62n outcomes are equally likely, and whether you win or you lose at or before the nth roll of the pair of dice is determined by looking at the sum of successive pairs of numbers in these outcomes. The sample space for the roll of a pair of dice (2) has 36 equally likely outcomes. Let n j denote the number of outcomes in which the sum
of the numbers shown is j, 2 ≤ j ≤ 12. Now let Ak be the event that you win by rolling a pair with sum k, and consider the 11 distinct cases: (a) P(A2) = P(A3) = P(A12) = 0, because you always lose with these. (b) For A7 to occur, you must get 7 on the ﬁrst roll. Because n7 = 6 P(A7) = n7 \|2\| = 6 36 = 1 6. (c) Likewise, = 2 36 (d) For A4 to occur, you must get 4 on the ﬁrst roll and on the nth roll, for some n ≥ 2, 4 (36 − n4 − n7)n−2 with no 4 or 7 in the intervening n − 2 rolls. You can do this in n2 P(A11) = n11 36 since n11 = 2. = 1 18, 46 1 Probability ways and, therefore, P(A4) = ∞ n=2 = (e) Likewise, 4(36 − n4 − n7)n−2 n2 62n = 1 36 n2 4 36(n4 + n7) by (1.4.3) because n4 = 3. P(A5) = n2 5 36(n5 + n7) = 2 45 = P(A9) because n5 = 4 because n9 = 4. P(A6) = P(A8) = 25 396 because n6 = n8 = 5 Finally, and P(A10) = P(A4) because n10 = n4 = 3. Therefore, the probability that you win is + 1 P(A7) + P(A11) + 2P(A4) + 2P(A5) + 2P(A6) = 1 6 18 0.493. + 1 18 + 4 45 + 25 198 (1) (2) (3) (4) Exercise What is the probability that you win on or before the second roll? Exercise What is the probability that you win on or before the third roll? Exercise What is the probability that you win if, on the ﬁrst roll, (a) The ﬁrst die shows 2? (b) The ﬁrst die shows 6? Exercise number would you choose?
If you could ﬁx the number to be shown by one die of the two on the ﬁrst roll, what 1.14 Example: Murphy’s Law A fair coin is tossed repeatedly. Let s denote any ﬁxed sequence of heads and tails of length r. Show that with probability one the sequence s will eventually appear in r consecutive tosses of the coin. (The usual statement of Murphy’s law says that anything that can go wrong, will go wrong). If a fair coin is tossed r times, there are 2r distinct equally likely outcomes Solution and one of them is s. We consider a fair die with 2r faces; each face corresponds to one Problems 47 of the 2r outcomes of tossing the coin r times and one of them is face s. Now roll the die repeatedly. Let Ak be the event that face s appears for the ﬁrst time on the kth roll. There are 2r k distinct outcomes of k rolls, and by symmetry they are equally likely. In (2r − 1)k−1 of them, Ak occurs, so by (1.3.1), P(Ak) = (2r − 1)k−1 Because Ak ∩ A j = φ for k = j, we have by (1.4.3) that 2r k. m m (1) P Ak = P(Ak) = 1 − 1 1 m, 2r − 1 2r which is the probability that face s appears at all in m rolls. Now consider n tosses of the coin, and let m = [ n r ] (where [x] is the integer part of x). The n tosses can thus be divided into m sequences of length r with a remainder n − mr. Let Bn be the event that none of these m sequences is s, and let Cn be the event that the sequence s does not occur anywhere in the n tosses. Then Cn ⊆ Bn = c, m 1 Ak because rolling the die m times and tossing the coin mr times yield the same sample space of equally likely outcomes. Hence, by Example 1.4.11(i) and (1.4.5) and (1), P(Cn) ≤ P(Bn) = 2r − 1 2r m → 0 as n → ∞. Now the event that s eventually occurs is lim n→∞ (C
c n), so by (1.4.5) and (1.5.4), P lim n→∞ C c n = lim n→∞ P = 1 − lim n→∞ P(Cn) = 1. C c n (2) (3) (4) If the coin is tossed an unbounded number of times, show that the probability that a If the coin is tossed n times, show that the probability that it shows heads on an odd Exercise number of tosses (and tails on the rest) is 1 2 Exercise head is ﬁrst shown on an odd numbered toss is 2 3 Exercise If Malone tosses his coin m times and Watt tosses his coin n times, show that the probability that they get the same number of heads each is equal to the probability that Beckett gets m heads in m + n tosses of his coin.B. Unless otherwise stated, coins are fair, dice are regular cubes and packs of cards are well shufﬂed with four suits of 13 cards. You are given a conventional pack of cards. What is the probability that the top card is an ace? You count a pack of cards (face down) and ﬁnd it defective (having only 49 cards!). What is the probability that the top card is an ace? 1 2 48 1 Probability 3 4 5 6 7 8 9 10 11 12 A class contains seven boys and eight girls. (a) If two are selected at random to leave the room, what is the probability that they are of different sexes? (b) On two separate occasions, a child is selected at random to leave the room. What is the probability that the two choices result in children of different sexes? An urn contains 100 balls numbered from 1 to 100. Four are removed at random without being replaced. Find the probability that the number on the last ball is smaller than the number on the ﬁrst ball. Let F be an event space. Show that the total number of events in F cannot be exactly six. What integers can be the number of events in a ﬁnite event space? To start playing a game of chance with a die, it is necessary ﬁrst to throw a six. (a) What is the probability that you throw your ﬁrst six at your third attempt? (b) What is the probability that you require more than three attempts? (c) What is the most likely
number of attempts until you ﬁrst throw a six? (d) After how many throws would your probability of having thrown a six be at least 0.95? Let A, B, and C be events. Write down expressions for the events where (a) At least two of A, B, and C occur. (b) Exactly two of A, B, and C occur. (c) At most two of A, B, and C occur. (d) Exactly one of A, B, and C occurs. A die is loaded in such a way that the probability that a 6 is thrown is ﬁve times that of any other number, each of them being equally probable. (a) By what factor is the probability of a total of 24 from four throws greater than that for an unloaded die? (b) Show that for the loaded die, the probability of obtaining a total of six from four throws is two and half times that of obtaining ﬁve, and compare the probability of obtaining 23 with that of obtaining 24 from four throws. A fair coin is tossed four times. What is the probability of (a) At least three heads? (b) Exactly three heads? (c) A run of three or more consecutive heads? (d) A run of exactly three consecutive heads? Find the probability that in 24 throws of two dice, double six fails to appear. Two dice are rolled and their scores are denoted by S1 and S2. What is the probability that the quadratic x 2 + x S1 + S2 = 0 has real roots? (a) If P(A) is the probability that an event A occurs, prove that n i=1 P Ai = n P(Ai ) − P(Ai ∩ A j ) + P(Ai ∩ A j ∩ Ak) + · · · i=1 + (−1)n+1P(A1 ∩ A2 ∩... ∩ An), i< j≤n i< j<k≤n where A1, A2,..., An are events. (b) A tea set consists of six cups and saucers with two cups and saucers in each of three different colours. The cups are placed randomly on the saucers. What is the probability that no cup is on a saucer of the same colour? 13 An urn contains three tickets numbered 1,
2, and 3, and they are drawn successively without replacement. What is the probability that there will be at least one value of r (r = 1, 2, 3) such that on the r th drawing a ticket numbered r will be drawn? 14 15 16 17 18 19 20 21 Problems 49 Four red balls and two blue balls are placed at random into two urns so that each urn contains three balls. What is the probability of getting a blue ball if (a) You select a ball at random from the ﬁrst urn? (b) You select an urn at random and then select a ball from it at random? (c) You discard two balls from the second urn and select the last ball? Four fair dice are rolled and the four numbers shown are multiplied together. What is the probability that this product (a) Is divisible by 5? (b) Has last digit 5? Suppose that n fair dice are rolled, and let Mn be the product of the numbers shown. (a) Show that the probability that the last digit of Mn is 5 is a nonincreasing function of n. (b) Show that the probability that Mn is divisible by 5 is a non-decreasing function of n. (c) Find the limits of the probabilities in (a) and (b) and interpret this. The consecutive integers 1, 2,..., n are inscribed on n balls in an urn. Let Dr be the event that the number on a ball drawn at random is divisible by r. (a) What are P(D3), P(D4), P(D3 ∪ D4), and P(D3 ∩ D4)? (b) Find the limits of these probabilities as n → ∞. (c) What would your answers be if the n consecutive numbers began at a number a = 1? Show that if A and B are events, then P(A ∩ B) − P(A)P(B) = P(A)P(Bc) − P(A ∩ Bc) = P(Ac)P(B) − P(Ac ∩ B) = P((A ∪ B)c) − P(Ac)P(Bc) Show that (a) min {1, P(A) + P(B)} ≥ P(A ∪ B) ≥ max {P(A), P(B)}
. (b) min {P(A), P(B)} ≥ P(A ∩ B) ≥ max {0, P(A) + P(B) − 1}. n n ≥ Ai P(Ai ) − (n − 1). (c) P The function d(x, y) is deﬁned on the event space by d(A, B) = P(AB). (a) Show that for any events A, B, and C, i=1 1 d(A, B) + d(B, C) − d(A, C) = 2(P(A ∩ Bc ∩ C) + P(Ac ∩ B ∩ C c)). (b) When is d(A, B) zero? (c) Let A1, A2,... be a monotone sequence of events such that Ai ⊆ A j for i ≤ j. Show that for i ≤ j ≤ k, d(Ai, Ak) = d(Ai, A j ) + d(A j, Ak). An urn contains x ≥ 2 xanthic balls and y ≥ 1 yellow balls. Two balls are drawn at random without replacement; let p be the probability that both are xanthic. (a) If p = 1 2 (b) If p = 1 8 (c) If p = r −2, where r is an integer, show that r ≥ 6, and ﬁnd values of x and y that yield p = 1 36, ﬁnd the smallest possible value of x in the two cases when y is odd or even., ﬁnd the smallest possible value of x.. 22 When are the following true? (a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (b) A ∩ (B ∩ C) = (A ∩ B) ∩ C (c) A ∪ (B ∪ C) = A\(B\C) (d) (A\B)\C = A\(B\C) 50 1 Probability (e) A (B C) = (A B) C (f) A\(B ∩ C) = (A\B) ∪ (A\C) (g) A\(B ∪ C) = (A
\B) ∩ (A\C). Birthdays If m students born in 1985 are attending a lecture, show that the probability that at least two of them share a birthday is p = 1 − (365)! (365 − m)!(365)m Show that if m ≥ 23, then p > 1 2 Let (An; n > 1) be a collection of events. Show that the event that inﬁnitely many of the An occur is given by. What difference would it make if they were born in 1988? ∞ Am. n≥1 m=n 23 24 25 Boole’s Inequality Show that n P 1 Ai ≤ n i=1 P(Ai ). 2 Conditional Probability and Independence Now and then there is a person born who is so unlucky that he runs into accidents which started to happen to somebody else. Don Marquis 2.1 Conditional Probability Suppose you have a well-shufﬂed conventional pack of cards. Obviously (by symmetry), the probability P(T ) of the event T that the top card is an ace is P(T ) = 4 52 However, suppose you notice that the bottom card is the ace of spades SA. What now is the probability that the top card is an ace? There are 51 possibilities and three of them are aces, so by symmetry again the required probability is 3 51. To distinguish this from the original probability, we denote it by P(T \|SA) and call it the conditional probability of T given that the bottom card is the ace of spades. = 1 13. Similarly, had you observed that the bottom card was the king of spades SK, you would conclude that the probability that the top card is an ace is Here is a less trivial example. P(T \|SK ) = 4 51. Example: Poker r ) denotes the number of ways of choosing r cards from n cards. If you are unfamiliar with this notation, omit this example at a ﬁrst reading.] [Note: In this example the symbol ( n Suppose you are playing poker. As the hand is dealt, you calculate the chance of being dealt a royal ﬂush R, assuming that all hands of ﬁve cards are equally likely. (A royal ﬂush comprises 10, J, Q, K, A in a single suit.) Just as you get the answer P(R) = 4 −1 52 5 = 1 6497
40, the dealer deals your last card face up. It is the ace of spades, SA. If you accept the card, what now is your chance of picking up a royal ﬂush? 51 52 2 Conditional Probability and Independence Intuitively, it seems unlikely still to be P(R) above, as the conditions for getting one have changed. Now you need your ﬁrst four cards to be the ten to king of spades precisely. (Also, had your last card been the two of spades, S2, your chance of a royal ﬂush would deﬁnitely be zero.) As above, to distinguish this new probability, we call it the conditional probability of R given SA and denote it by P(R\|SA). Is it larger or smaller than P(R)? At least you do have an ace, which is a start, so it might be greater. But you cannot now get a ﬂush in any suit but spades, so it might be smaller. To resolve the uncertainty, you assume that any set of four cards from the remaining 51 cards is equally likely to complete your hand and calculate that P(R\|SA) = −1 51 4 = 13 5 P(R). Your chances of a royal ﬂush have more than doubled. Let us investigate these ideas in a more general setting. As usual we are given a sample space, an event space F, and a probability function P(.). We suppose that some event B ∈ F deﬁnitely occurs, and denote the conditional probability of any event A, given B, by P(A\|B). As we did for P(.), we observe that P(.\|B) is a function deﬁned on F, which takes values in [0, 1]. But what function is it? Clearly, P(A) and P(A\|B) are not equal in general, because even when P(Bc) = 0 we always have P(Bc\|B) = 0. Second, we note that given the occurrence of B, the event A can occur if and only if A ∩ B occurs. This makes it natural to require that Finally, and trivially, P(A\|B) ∝ P(A ∩ B). P(B\|B) = 1. After a moment’s thought about these three observations, it appears that an attractive candidate to play
the role of P(A\|B) is P(A ∩ B)/P(B). We make these intuitive reﬂections formal as follows. Deﬁnition ditional probability that A occurs is denoted by P(A\|B) and deﬁned by Let A and B be events with P(B) > 0. Given that B occurs, the con- (1) P(A\|B) = P(A ∩ B) P(B). When P(B) = 0, the conditional probability P(A\|B) is not deﬁned by (1). However, to avoid an endless stream of tiresome reservations about special cases, it is convenient to adopt the convention that, even when P(B) = 0, we may still write P(A ∩ B) = P(A\|B)P(B), both sides having the value zero. Thus, whether P(B) > 0 or not, it is true 2.1 Conditional Probability 53 that P(A ∩ B) = P(A\|B)P(B). Likewise, P(A ∩ Bc) = P(A\|Bc)P(Bc) and hence, for any events A and B, we have proved the following partition rule: (2) Theorem P(A) = P(A ∩ B) + P(A ∩ Bc) = P(A\|B)P(B) + P(A\|Bc)P(Bc). The reader will come to realize the crucial importance of (1) and (2) as he or she discovers more about probability. We begin with a trivial example. Example: Poker Revisited formal discussion earlier in this section. By (1) Let us check that Deﬁnition 1 is consistent with our in- P(R\|SA) = P(R ∩ SA)/P(SA) = 1 52 5 = 51 4 52 5 −1. 51 4 Here is a more complicated example. Example: Lemons An industrial conglomerate manufactures a certain type of car in three towns called Farad, Gilbert, and Henry. Of 1000 made in Farad, 20% are defective; of 2000 made in Gilbert, 10% are defective, and of 3000 made in Henry, 5% are defective. You buy a car from a distant dealer. Let D be the event that
it is defective, F the event that it was made in Farad and so on. Find: (a) P(F\|H c); (b) P(D\|H c); (c) P(D); (d) P(F\|D). Assume that you are equally likely to have bought any one of the 6000 cars produced. Solution (a) (b) P(F\|H c) = P(F ∩ H c) P(H c) by (1), = P(F) P(H c) because F ⊆ H c, = 1000 6000 3000 6000 = 1 3. P(D\|H c) = P(D ∩ H c) by (1) P(H c) = P(D ∩ (F ∪ G)) P(H c) = P(D ∩ F) + P(D ∩ G) P(H c) because H c = F ∪ G, because F ∩ G = φ 54 (c) (d) 2 Conditional Probability and Independence = P(D\|F)P(F) + P(D\|G)P(G) P(H c) · 1 3 + 1 10 · 1 6 by (1) 1 5 = on using the data in the question, 1 2 = 2 15. P(D) = P(D\|H )P(H ) + P(D\|H c)P(H c) by (2) + 2 15 · 1 2 on using the data and (b) · 1 2 = 1 20 = 11 120. by (1) P(F\|D) = P(F ∩ D) P(D) = P(D\|F)P(F) P(D) 11 120 · 1 6 = 1 5 = 4 11 by (1) on using the data and (c). We often have occasion to use the following elementary generalization of Theorem 2. (3) Theorem We have P(A) = i P(A\|Bi )P(Bi ) whenever A ⊆ i Bi and Bi ∩ B j = φ for i = j; the extended partition rule. Proof This is immediate from (1.4.3) and (1). For example, with the notation of (3), we may write P(B j \|A) = P(B j ∩ A)/
P(A) = P(A\|B j )P(B j ) P(A), and expanding the denominator using (3), we have proved the following celebrated result; also known as Bayes’s Rule: 2.1 Conditional Probability 55 Bayes’s Theorem If A ⊆ n 1 Bi, and Bi ∩ B j = φ for i = j, then P(B j \|A) = P(A\|B j )P(B j ) n P(A\|Bi )P(Bi ) 1 ; P(A) > 0. The following is a typical example of how (4) is applied in practice. Example: False Positives You have a blood test for some rare disease that occurs by chance in 1 in every 100 000 people. The test is fairly reliable; if you have the disease, it will correctly say so with probability 0.95; if you do not have the disease, the test will wrongly say you do with probability 0.005. If the test says you do have the disease, what is the probability that this is a correct diagnosis? Solution says you do. Then, we require P(D\|T ), which is given by Let D be the event that you have the disease and T the event that the test P(D\|T ) = P(T \|D)P(D) P(T \|D)P(D) + P(T \|Dc)P(Dc) by (4) = (0.95)(0.00001) (0.95)(0.00001) + (0.99999)(0.005) 0.002. Despite appearing to be a pretty good test, for a disease as rare as this the test is almost useless. It is important to note that conditional probability is a probability function in the sense deﬁned in Section 1.4. Thus, P(\|B) = 1 and, if Ai ∩ A j = φ for i = j, we have i P Ai \|B = i P(Ai \|B). From these, we may deduce various useful identities (as we did in Section 1.4); for example: P(A ∩ B ∩ C) = P(A\|B ∩ C)P(B\|C)P(C), n P 1 Ai = P A1\| Ai P A2\| n 3
n 2 Ai... P(An) P(A\|B) = 1 − P(Ac\|B), P(A ∪ B\|C) = P(A\|C) + P(B\|C) − P(A ∩ B\|C), and so on. (4) (5) (6) (7) (8) (9) 56 2 Conditional Probability and Independence (10) Example P i Let us prove (5), (6), (7), (8), and (9). First, Ai ∩ B P(B) by (1). Ai \|Ai ∩ B) P(B) i P(Ai ∩ B)/P(B) by (1.4.3), because the Ai are disjoint, P(Ai \|B) by (1) again, and we have proved (5). Second, by repeated use of (1), P(A\|B ∩ C)P(B\|C)P(C) = P(A ∩ B ∩ C) P(B ∩ C) · P(B ∩ C) P(C) ·P(C) = P(A ∩ B ∩ C), if the denominator is not zero. If the denominator is zero, then (6) still holds by convention, both sides taking the value zero. The relation (7) follows by induction using (6); and (8) and (9) are trivial consequences of (5). (11) Example: Repellent and Attractive Events The event A is said to be attracted to B if P(A\|B) > P(A). If P(A\|B) < P(A), then A is repelled by B and A is indifferent to B if (12) P(A\|B) = P(A). (a) Show that if B attracts A, then A attracts B, and Bc repels A. (b) A ﬂimsy slip of paper is in one of n bulging box ﬁles. The event that it is in the jth box ﬁle is B j, where P(B j ) = b j > 0. The event that a cursory search of the jth box ﬁle fails to discover the slip is Fj, where P(Fj \|B j ) = φ j < 1.
Show that B j and Fj are mutually repellent, but Fj attracts Bi, for i = j. Solution ing by P(A), we have P(B\|A) > P(B). Furthermore, by Theorem 2, (a) Because B attracts A, by (1), P(A ∩ B) > P(A)P(B), whence, on divid- P(A\|Bc)P(Bc) = P(A) − P(A\|B)P(B) < P(A)(1 − P(B)), because B attracts A, = P(A)P(Bc). So Bc repels A (on dividing through by P(Bc) = 0). (b) By Bayes’ theorem (4), P(B j \|Fj ) = P(Fj \|B j )P(B j ) n P(Fj \|Bi )P(Bi ) i=.2 Independence 57 because, obviously, for i = j, P(Fj \|Bi ) = 1. Hence, P(B j ) − P(B j \|Fj ) = b j (1 − b j )(. Therefore, B j is repelled by Fj. Also, for i = j, P(Bi \|Fj ) − P(Bi ) = bi 1 − b j + φ j b j − bi = bi b j ( so Fj attracts Bi, for i = j. Notice that this agrees with our intuition. We believe quite strongly that if we look in a ﬁle for a slip and fail to ﬁnd it, then it is more likely (than before the search) to be elsewhere. (Try to think about the consequences if the opposite were true.) This conclusion of Example 11 was not incorporated in our axioms, but follows from them. It therefore lends a small but valuable boost to their credibility. Finally, we consider sequences of conditional probabilities. Because conditional probability is a probability function [see (5)], we expect it to be continuous in the sense of Section 1.5. Thus if (as n → ∞) An → A and Bn → B, then by Theorem 1.5.2 we have and (13) lim n→∞ P(An\|B) = P(A\|B) lim n→∞ P(A\|Bn) = P(
A\|B). 2.2 Independence It may happen that the conditional probability P(A\|B) is the same as the unconditional probability P(A), so that P(A) = P(A\|B) = P(A ∩ B) P(B). This idea leads to the following: (1) Deﬁnition (a) Events A and B are independent when P(A ∩ B) = P(A)P(B). (b) A collection of events (Ai ; i ≥ 1) is independent when P i∈F Ai = i∈F P(Ai ) for any ﬁnite set F of indices. (c) Events A and B are conditionally independent, given C, when P(A ∩ B\|C) = P(A\|C)P(B\|C). 58 2 Conditional Probability and Independence This does not imply independence unless C =. (d) A collection of events (Ai ; i ≥ 1) is pairwise independent if P(Ai ∩ A j ) = P(Ai )P(A j ) for i = j. This does not imply independence in general. It is easy to see that independence is equivalent to the idea of indifference deﬁned in (2.1.12), but the term “indifference” is not in general use. It is usually, but not always, clear when two events are independent, as the next two examples illustrate. (2) Example: Sport Prior to a game of football, you toss a coin for the kick-off. Let C be the event that you win the toss, and let M be the event that you win the match. (a) Show that the outcome of the match is independent of whether you win the toss if and only if, for some p and p, with 0 < p, p < 1, P(C ∩ M) = pp, P(C ∩ M c) = p(1 − p), P(C c ∩ M) = (1 − p) p, and P(C c ∩ M c) = (1 − p)(1 − p). (b) Let B be the event that you win both or lose both, so B = {(C ∩ M) ∪ (C c ∩ M c)}. Suppose that C and M are indeed independent. Show that
C and B are independent if and only if p = 1 2. Solution deﬁnition P(C ∩ M) = pp and so on. (a) If C and M are independent, and P(C) = p and P(M) = p, then by Conversely, for the given probabilities P(C) = P(C ∩ M) + P(C ∩ M c) = pp + p(1 − p) = p and similarly we have P(M) = p. Hence, P(C)P(M) = pp = P(C ∩ M). This, together with three similar identities (exercises for you), demonstrates the independence. (b) Trivially, P(C ∩ B) = P(C ∩ M). Hence, C and B are independent if pp = P(C ∩ M) = P(C)P(B) = p( pp + (1 − p)(1 − p)). That is, if (1 − p)(1 − 2 p) = 0. Because p = 1, it follows that p = 1 trivial. 2. The converse is (3) Example: Flowers A plant gets two independent genes for ﬂower colour, one from each parent plant. If the genes are identical, then the ﬂowers are uniformly of that colour; 2.2 Independence 59 if they are different, then the ﬂowers are striped in those two colours. The genes for the colours pink, crimson, and red occur in the population in the proportions p:q:r, where p + q + r = 1. A given plant’s parents are selected at random; let A be the event that its ﬂowers are at least partly pink, and let B be the event that its ﬂowers are striped. (a) Find P(A) and P(B). (b) Show that A and B are independent if p = 2 (c) Are these the only values of p, q, and r such that A and B are independent? 3 and r = q = 1 6. Solution have (a) With an obvious notation (P for pink, C for crimson, and R for red), we P(PP) = P(P)P(P), = p2, by parents independence, because P occurs with probability p. Likewise, P(PR) = P(R)
P(P) = r p = P(RP). Hence, P(A) = P(P P ∪ P R ∪ P ∪ PC ∪ C P) by (1.4.3), = p2 + 2 pr + 2 pq = 1 − (1 − p)2, because p + q + r = 1. (Can you see how to get this last expression directly?) Similarly, P(B) = P(PC ∪ P R ∪ RC) = 2( pq + qr + r p). (b) The events A and B are independent, if and only if, P(A)P(B) = P(A ∩ B) = P(PC ∪ P R) = 2( pq + pr ). From part (a), this is equivalent to (4) (1 − (1 − p)2)( pq + qr + pr ) = p(q + r ), and this is satisﬁed by the given values of p, q, and r. (c) No. Rearranging (4), we see that A and B are independent for any values of q and r lying on the curve rq = 2rq(, in the r − q plane. You may care to amuse yourself by showing that this is a loop from the origin. Outside the loop, A and B are attractive; inside the loop, A and B are repellent. (5) Example 1.13 Revisited: Craps Let us reconsider this game using conditional probability and independence. Recall that Ak is the event that you win by rolling a pair with sum k. Let Sk be the event that any given roll yields sum k. Now, for example, A4 occurs only if S4 occurs at the ﬁrst roll and S4 occurs before S7 in later rolls. However, all the rolls after the ﬁrst until the ﬁrst occurrence of S4 or S7 are irrelevant, and rolls are independent. 60 Hence, 2 Conditional Probability and Independence P(A4) = P(S4)P(S4\|S4 ∪ S7) = (P(S4))2 P(S4 ∪ S7) = 3 36 2 3 36 + 6 36 = 1 36. Now, performing a similar calculation for A5, A6, A8, A9, and A10
yields the solution to Example 1.13. (6) Example Suppose A and B are independent, and B and C are independent. (a) Are A and C independent in general? (b) Is B independent of A ∪ C? (c) Is B independent of A ∩ C? Solution (a) No. Take A ≡ C. [Then do Problem 4(a).] (b) No. Consider Example 2(b), with p = p = 1 2, and let A = M. Then, B is indepen- dent of M and C, but and P(B ∩ (C ∪ M)) = P(C ∩ M) = 1 4 P(B)P(C ∪ M) = ( pp + (1 − p)(1 − p))(1 − (1 − p)(1 − p)) = 3 8 = 1 4. (c) No. With the same notation as in (b), following Example 2(b) again, we have P(B ∩ C ∩ M) = P(C ∩ M) = 1 4, and P(B)P(C ∩ M) = ( pp + (1 − p)(1 − p)) pp = 1 8 = 1 4. 2.3 Recurrence and Difference Equations Many problems in probability have a structure in which the repetition of some procedure is essential. At a trivial level, one may repeatedly roll dice, catch ﬁsh, have children, and so on; more important problems involve the same idea. In Chapter 1, it is necessary to suppose that all the probabilities on the sample space were given or that all outcomes were equally likely. Conditional probability provides a more natural way of deﬁning such problems; conditional on the procedure having reached some stage, it supplies the probabilities of the consequent events. By emphasizing this recurrent aspect of some experiment, conditional probability enables us to tackle problems by deriving recurrence relations. These often turn out to be simple difference equations. Many of the worked examples illustrate these ideas; the following results are useful. 2.3 Recurrence and Difference Equations 61 (1) Theorem Let (ak; k ≥ 0) be a sequence of real numbers. (i) If the sequence (uk; k ≥ 0) satisﬁes then (ii) If (uk; k ≥ 0) satisﬁes uk+1 − akuk =
0, uk = u0 k−1 0 a j. uk+1 − uk = cαk, where α and c are constants, then (iii) If (uk; k ≥ 0) satisﬁes uk − u0 = c αk − 1 α − 1. uk+1 − uk = c, for some constant c, then uk − u0 = kc. (iv) If for some constants a, α, c1, and c2, we have then uk+1 − auk = c1 + c2αk, a = α, uk = u0ak + c1(1 − ak) 1 − α + c2(αk − ak) α − a. Proof These results are veriﬁed simply by substitution. (2) Example: Fly A room has four walls, a ﬂoor, and a ceiling. A ﬂy moves between these surfaces. If it leaves the ﬂoor or ceiling, then it is equally likely to alight on any one of the four walls or the surface it has just left. If it leaves a wall, then it is equally likely to alight on any of the other three walls, the ﬂoor, or the ceiling. Initially, it is on the ceiling. Let Fk denote the event that it is on the ﬂoor after k moves. What is fk = P(Fk)? Solution Let Ck denote the event that it is on the ceiling after k moves, and Nk, Ek, Wk, Sk denote the corresponding event for the four walls. Set ck = P(Ck), and so on. Then by Theorem 2.1.3, (3) P(Fk) = P(Fk\|Fk−1)P(Fk−1) + P(Fk\|Ck−1)P(Ck−1) + P(Fk\|Nk−1)P(Nk−1) + P(Fk\|Ek−1)P(Ek−1) + P(Fk\|Wk−1)P(Wk−1) + P(Fk\|Sk−1)P(Sk−1) = 1 5 fk−1 + 0 + 4 5 wk−1, (4) (5) (6
) (7) (8) 62 2 Conditional Probability and Independence where we have used the fact that, by symmetry, nk = ek = wk = sk. Likewise, fk−1 + 1 5 wk−1 + 1 5 wk−1 + 1 5 ck−1 = 3 5 (1 − 4wk−1), wk = 3 5 on using the fact that which follows from the observation that the ﬂy has to be somewhere in the room. Hence, fk + ck + 4wk = 1, wk = − 1 5 wk−1 + 1 5. Because the ﬂy starts on the ceiling, w0 = 0, and so by Theorem 1(iv) wk = 1 6 1 − − 1 5 k. Substituting into (3) gives fk = 1 5 fk−1 + 2 15 1 − − 1 5 k−1. Hence, for some constant A, fk =. The arbitrary constant A is determined by recalling that the ﬂy starts on the ceiling. Thus, f0 = 0, and fk = −. Alternatively, we may substitute (5) into (3) to get directly: fk = 1 25 fk−2 + 4 25. It is a simple matter to check that, for any constants A and B, fk = satisﬁes (8). Because f0 = f1 = 0, the solution (7) is recovered immediately. Notice that as k → ∞, (6) and (7) yield fk → 1 6. In the long run, the ﬂy is equally likely to be on any surface. 6. It follows from (4) that ck → 1 6 and wk → 1 2.4 Remarks Independence and conditioning greatly add to your armoury of weapons for attacking problems. If an event requires the occurrence of a number of independent events, then calculations are simpliﬁed by using Deﬁnition 2.2.1. 2.4 Remarks 63 Where independence fails, it may be possible to ﬁnd a family of disjoint events Bi whose union includes the event A of interest, and for which P(A\|Bi ) is easily calculated. The required probability is then calculated using Theorem 2.1.3. Such events Bi can also be found to yield P(A) as the solution
of some recurrence relation, as exempliﬁed in Section 2.3. Note that you are warned to avoid the painful student error that asserts that A and B are independent when A ∩ B = φ. This is wrong, except possibly when one of P(A) or P(B) is zero, which is not a case of any great interest in general. Finally, we take this opportunity to stress that although conditional probability is extremely useful and powerful, it also greatly increases the possibilities for making egregious errors. The celebrated Monty Hall problem is a recent classic example, which we discuss in Example 2.13. But cautionary examples had been in existence for many years before Marilyn vos Savant made that one famous. Here are two classics from the nineteenth century. Galton’s Paradox (1894) Suppose you ﬂip three fair coins. At least two are alike and it is an evens chance whether the third is a head or a tail, so the chance that all three are the same is 1 2. Solution In fact, P(all same) = P(T T T ) + P(. What is wrong? Resolution As is often the case, this paradox arises from fudging the sample space. This “third” coin is not identiﬁed initially in, it is determined by the others. The chance whether the “third” is a head or a tail is a conditional probability, not an unconditional probability. Easy calculations show that P(3rd is H \|H H ) = 1 4 P(3rd is T \|H H ) = 3 4 P(3rd is T \|T T ) = 1 4 P(3rd is H \|T T ) = 3 4 H H denotes the event that there are at least two heads. T T denotes the event that there are at least two tails. In no circumstances, therefore, is it true that it is an evens chance whether the “third” is a head or a tail; the argument collapses. Bertrand’s Box Paradox (1889) There are three boxes. One contains two black counters, one contains two white counters, and one contains a black and a white counter. Pick a box at random and remove a counter without looking at it; it is equally likely to be black or white. The other counter is equally likely to be black or white. Therefore, the chance that your box contains identical counters is 1 2.
However, this is clearly false: the correct answer is 2 3. 64 2 Conditional Probability and Independence Resolution This is similar to Galton’s paradox. Having picked a box and counter, the probability that the other counter is the same is a conditional probability, not an unconditional probability. Thus, easy calculations give (with an obvious notation) (1) P(both black\|B) = 2 3 = P(both white\|W ); in neither case is it true that the other counter is equally likely to be black or white. 2.5 Review and Checklist for Chapter 2 In practice, we may have partial knowledge about the outcome of an experiment, or the conditions of an experiment may change. We therefore used intuitive ideas about probability to deﬁne conditional probability and deduce the key result known as Bayes’s rule (or Bayes’s theorem). It may also be the case that occurrence of an event has no effect on the probability of another. This led us to deﬁne the concept of independence. These ideas are particularly useful when experiments have (or can be reformulated to have) a sequential structure. Use of conditional probability and independence often gives rise to recurrence relations and difference equations in these cases. Notation: P(A\|B) The conditional probability of A given B. RULES: Conditioning Rule: P(A\|B) = P(A ∩ B)/P(B) Addition Rule: P(A ∪ B\|C) = P(A\|C) + P(B\|C), when A ∩ C and B ∩ C are disjoint. Multiplication Rule: P(A ∩ B ∩ C) = P(A\|B ∩ C)P(B\|C)P(C) Partition Rule: P(A) = P(A\|Bi )P(Bi ), when (Bi : i ≥ 1) are disjoint events and A ⊆ ∪i Bi. Bayes’s Rule (or Theorem): P(Bi \|A) = P(A\|Bi )P(Bi )/P(A) = P(A\|Bi )P(Bi ) j P(A\|B j )P(B j ) Extended addition Rule: P(∪i Ai \|C) = i P(Ai \|C) when {Ai } is a partition; which is to
say Ai ∩ A j Extended multiplication Rule: = φ, and C ⊆ ∪i Ai. P n i=1 Ai = P(A1\|A2 ∩... ∩ An)... P(An−1\|An)P(An) Independence Rule: A and B are independent if and only if P(A ∩ B) = P(A)P(B); this is equivalent to P(A\|B) = P(A) and to P(B\|A) = P(B). More generally, (Ai ; 1 ≤ i ≤ n) are independent if P( Ai ) = i∈I P(Ai ) for any i∈I choice of index set I ⊆ {1,..., n}. Worked Examples and Exercises 65 Conditional independence Rule: A and B are conditionally independent given C when P(A ∩ B\|C) = P(A\|C)P(B\|C). Note that independence does not imply conditional independence, nor is it implied by it. Pairwise independence Rule: (Ai ; 1 ≤ i ≤ n) are pairwise independent if P(Ai ∩ A j ) = P(Ai )P(A j ), i = j. Checklist of Terms for Chapter 2 2.1 conditional probability partition rules Bayes’s theorem (or rule) 2.2 independence conditional independence pairwise independence 2.3 difference equation 2.4 paradoxes.6 Example: Sudden Death Two golfers (Atropos and Belladonna) play a series of holes. Atropos wins each hole with probability p, Belladonna wins each hole with probability q, and holes are halved with probability r. Holes are independent, and the game stops on the ﬁrst occasion when a hole is not halved. What is the probability that Atropos wins? We give two methods of solution. Let An be the event that Atropos wins the match at the nth hole, Hk the Solution I event that the kth hole is halved, and Wk the event that Atropos wins the kth hole. Then, (1) P(An) = P(H1 ∩ H2 ∩... ∩ Hn−1 ∩ Wn) = n−1 1 P(Hk)P(Wn
) by independence = r n−1 p. Hence, the probability that Atropos wins the match is ∞ P An = ∞ P(An) (2) 1 by (1.4.3) because Ak ∩ A j = φ for k = j. Now ∞ ∞ 1 P(An) = 1 n=1 r n−. 66 2 Conditional Probability and Independence Solution II Let V be the event that Atropos wins the match. Then by Theorem 2.1.3, (3) P(V ) = P(V \|W1)P(W1) + P(V \|H1)P(H1) + P(V \|W c 1 ∩ H c 1 )P(W c 1 1 ). ∩ H c Now, P(V \|W1) = 1, and Also, because holes are independent P(V \|. Hence, substituting into (3), P(V \|H1) = P(V ). P(V ) = p + P(V )r, so P(V ) = p/(1 − r ), in agreement with Solution I. Remark The ﬁrst solution harks back to Chapter 1, by dividing up the sample space into disjoint events and using (1.4.3). The second solution exploits the power of conditional probability by conditioning on the outcome of the ﬁrst hole. You will use this second idea frequently in tackling problems in probability. (4) (5) (6) (7) Show that the probability un that Atropos wins at or before the nth hole is p(1 − Exercise r n)/(1 − r ). Exercise Given that Atropos wins at or before the nth hole, show that: (a) The probability that the ﬁrst hole was halved is r (1 − r n−1)/(1 − r n), (b) The probability that the ﬁrst hole was won is (1 − r )/(1 − r n). Exercise hole? Exercise What is the probability that Atropos wins, given that exactly n holes have been played when the match is won? Use this to solve the example by a third method. Given that Atropos wins, what is the probability that she has won before the third 2.7 Example: Polya’s Urn An urn contains b blue
balls and c cyan balls. A ball is drawn at random, its colour is noted, and it is returned to the urn together with d further balls of the same colour. This procedure is repeated indeﬁnitely. What is the probability that: (a) The second ball drawn is cyan? (b) The ﬁrst ball drawn is cyan given that the second ball drawn is cyan? Solution Let Cn denote the event that the nth drawn ball is cyan. Then (a) Now given C1, the urn contains c + d cyan balls on the second drawing, so P(C2) = P(C2\|C1)P(C1) + P P. C2\|C c 1 C c 1 P(C2\|C1. Worked Examples and Exercises 67 Likewise, given C c 1 the urn contains c cyan balls on the second drawing, so C2\|. Hence, (b) P(C2) = c + d = P(C1). b + c + d P(C1\|C2) = P(C1 ∩ C2)P(C2) = P(C2\|C1)P(C1)/P(C21) (2) (3) (4) (5) (1) (2) using the results of (a). Find the probability that the ﬁrst drawn ball is cyan given that the following n drawn Show that P(Cn) = P(C1) for all n ≥ 1. Find the probability that the ﬁrst drawn ball is cyan given that the nth drawn ball is Exercise Exercise cyan. Exercise balls are all cyan. What is the limit of this probability as n → ∞? Show that for any j, k, P(Ck\|C j ) = P(C j \|Ck). Exercise Show that in m + n drawings, the probability that m cyan balls are followed by n blue Exercise balls is the same as the probability that n blue balls are followed by m cyan balls. Generalize this result. 2.8 Example: Complacency In a factory, if the most recent accident occurred exactly k days before today, then the probability that an accident occurs today is pk; there is no accident with probability 1 − pk. During the n successive days immediately after an accident, what is the probability that (a) There are no
accidents? (b) There is exactly one accident? Solution accidents, n ≥ 1. We are given that (a) Let An be the event that the n days following an accident are free of P(An\|An−1) = 1 − pn, and P(A1) = 1 − p1. The crucial observation is that n ≥ 2 P(An) = P(An\|An−1)P(An−1) = (1 − pn)P(An−1). This is almost completely obvious, but we labour the point by giving two detailed veriﬁcations. I Notice that An ⊆ An−1. Hence, P(An) = P(An ∩ An−1) = P(An\|An−1)P(An−1) by (2.1.1). II Alternatively, by Theorem 2.1.2, P(An) = P(An\|An−1)P(An−1) + P because P An\|Ac n−1 = 0. An\|Ac n−1 P Ac n−1 = P(An\|An−1)P(An−1) 68 2 Conditional Probability and Independence Returning to (1), we iterate this relation to get P(An) = n 1 (1 − p j ) = πn (say). (b) Let Bk be the event that the ﬁrst accident in the n day period occurs on the kth day. Then, P(Bk) = P Ac k ∩ Ak−1 = P \|Ak−1 P(Ak−1) by (2.1.1) Ac k k−1 1 = pk (1 − p j ) = pkπk−1 Now, given an accident on the kth day, the event that there are no accidents in the succeeding n − k days has the same probability as An−k. Hence, the probability of exactly one accident is n n P(Bk)P(An−k) = πk−1 pkπn−k = sn (say). k=1 k=1 (3) (4) (5) (6) Show that if pn is nondecreasing in n (and pn > 0), then an accident is certain to occur Evaluate πn and sn when pn = p. Exercise sometime. Exercise Exercise
What is the probability that in the n days following an accident (a) There is at least one accident? (b) There are exactly two accidents? (c) Evaluate these probabilities when pn = p. Exercise Show that if ci is a collection of numbers satisfying 0 ≤ ci ≤ 1, i ≥ 1, then ∞ i−1 ci c1 + (1 − c j ) + ∞ (1 − ci ) = 1. i=2 j=1 i=1 (7) Exercise What condition on ( pn; n ≥ 1) would allow a nonzero chance of no second accident? 2.9 Example: Dogﬁght Algy, Biggles, and the Commodore are ﬁghting an air battle. In the ﬁrst round, each ﬁres one shot in alphabetical order, and each may ﬁre at any unhit opponent. Anyone hit drops out of the battle immediately. Survivors play successive rounds with the same rules as the ﬁrst round until one winner remains. On any shot aimed at an opponent, Algy hits with probability α, the Commodore hits with probability γ, and Biggles never misses. Show that if shots are independent and, then Algy should ﬁre his ﬁrst shot into the ground. Solution Suppose that Algy were to ﬁre at Biggles and hit him. Algy would then be the Commodore’s target, and the battle would continue with shots alternating between these two until a hit is scored. Let C A be the event that the Commodore wins this two-man battle, and consider the following three events: A1 ≡ the Commodore scores a hit with his ﬁrst shot. Worked Examples and Exercises 69 A2 ≡ the Commodore misses and Algy scores a hit with his ﬁrst returned shot. A3 ≡ the Commodore and Algy both miss their ﬁrst shots. If A3 occurs, then the next round begins under the same conditions; hence, P(C A\|A3) = P(C A). Also, P(C A\|A1) = 1, and P(C A\|A2) = 0. Because Ai ∩ A j = φ for i = j and ∪3 we may use Theorem 2.1.3 to give 1 Ai =, yielding P(
C A) = γ + 0 + (1 − γ )(1 − α)P(C A), P(C A) = γ 1 − (1 − γ )(1 − α). However, if Algy misses, then Biggles will certainly ﬁre at the Commodore because α < γ, and hit him. Then Algy can win only if his second round shot at Biggles hits; otherwise, Biggles surely hits him with his second round shot. Thus, in this case, Algy wins with probability α. Hence, missing Biggles gives Algy a better chance if 1 − P(C A) = α(1 − γ ) 1 − (1 − γ )(1 − α) > α that is if α > 1 − γ /(1 − γ ). (1) (2) (3) Exercise If Algy does ﬁre his ﬁrst shot at Biggles, what is the probability that he wins the battle? Exercise Algy is not a competent probabilist, and decides whether to shoot at Biggles by tossing a coin (heads he does, tails he doesn’t). Given that the battle ends with the fourth shot, what is the probability that Algy aimed to miss? Exercise is β. If α = 0.5, β = 0.875, and γ = 0.75, where should Algy aim his ﬁrst shot? Suppose that Biggles is not infallible; in fact, the probability that any shot of his hits 2.10 Example: Smears In a population of women, a proportion p have abnormal cells on the cervix. The Pap test entails taking a sample of cells from the surface of the cervix and examining the sample to detect any abnormality. (i) In a case where abnormal cells are present, the sample will fail to include any with probability µ. (ii) In a sample including abnormal cells, examination fails to observe them with proba- bility ν. (iii) In a sample free of abnormal cells, normal cells are wrongly classiﬁed as abnormal with probability π. All sampling and identiﬁcation errors are independent. If a randomly selected woman has such a test: (a) What is the probability that the result is wrong? (b) If an abnormality is reported, what is the probability
that no abnormal cells are present? 70 2 Conditional Probability and Independence (a) Let E denote the event that the result is in error, A denote the event that Solution abnormal cells are present, and S denote the event that the sample fails to include abnormal cells when they are present. Using Theorem 2.1.2, we have P(E) = P(E\|A)P(A) + P(E\|Ac)P(Ac) = P(E\|A) p + P(E\|Ac)(1 − p). By (iii), P(E\|Ac) = π. Also, by (2.1.5), P(E\|A) = P(E ∩ S\|A) + P(E ∩ Sc\|A). By (iii) and (i), and by (ii) and (i), Hence, P(E ∩ S\|A) = (1 − π )µ P(E ∩ Sc\|A) = ν(1 − π ). P(E) = p((1 − π)µ + ν(1 − µ)) + (1 − p)π. (b) Let D denote the event that an abnormality is reported. By (2.1.1), P(Ac\|D) = P(Ac ∩ D) P(D). Now, by Theorem 2.1.2, P(D) = P(D\|A)P(A) + P(D\|Ac)P(Ac) = P(D ∩ S\|A)P(A) + P(D ∩ Sc\|A)P(A) + π (1 − p) = πµp + (1 − ν)(1 − µ) p + π(1 − p) by (2.1.5) and Hence, P(Ac ∩ D) = π(1 − p). P(Ac\|D) = π(1 − p) πµp + π(1 − p) + (1 − ν)(1 − µ) p. Notice that this is an example of Bayes’ Theorem (2.1.4). (1) (2) (3) Evaluate P(E) and P(Ac\|D) when Exercise (a) p = 10% and µ =
ν = π = 10−1, and when (b) p = 50% and µ = 10−1 and ν = π = 10−2. Exercise What is the probability that the result is wrong if no abnormality is reported? Evaluate this in the above two cases, and compare P(Ac\|D) and P(A\|Dc). Exercise Whatever the result of the test, it is recorded wrongly in the letter to the patient with probability ρ independently of other errors. Let L be the event that the letter is wrong, and let M be the event that the letter reports abnormalities to the patient. Find P(L), P(Ac\|M), and P(A\|M c). 2.11 Example: Gambler’s Ruin You enter a casino with $k, and on each spin of a roulette wheel you bet $1 at evens on the event R that the result is red. The wheel is not fair, so P(R) = p < 1 2. If you lose all $k, Worked Examples and Exercises 71 you leave; and if you ever possess $K ≥ $k, you choose to leave immediately. What is the probability that you leave with nothing? (Assume spins of the wheel are independent. Note that this is an idealized casino.) Let pk be the probability that you leave with nothing. If the ﬁrst spin results Solution in red, you gain $1 and are in the same position as if you had just entered with $k + 1. Thus, conditional on R, your chance of leaving with nothing is pk+1. Similarly, if the ﬁrst spin results in black (or zero), you have $k − 1 and your chance of leaving with nothing is pk−1. Hence, (1) pk = ppk+1 + (1 − p) pk−1, 0 < k < K. If k = 0, then you certainly leave with nothing; if k = K, you leave before betting. Hence, p0 = 1 and pK = 0. Writing (1) as pk+1 − pk = 1 − p p ( pk − pk−1) for p > 0, gives (on using p0 = 1) pk+1 − pk = k 1 − p p ( p1 − 1), by Theorem 2.3.1(a). Hence
, pk = 1 + ( p1 − 1, by Theorem 2.3.1(b). Because pK = 0, setting k = K in this gives 0 = 1 + ( p1 − 1 and now eliminating p1 gives ﬁnally pk = Show that as K → ∞ in (2), pk → 1. Find pk when p = 1 2. Show that with probability one you do not remain in the casino forever. Given that you leave with nothing, what is the probability that you never possessed Exercise Exercise Exercise Exercise more than your initial $k? Let n(k, K, r ) be the number of sequences of red and black that result in your leavExercise ing the casino with $K on the rth spin of the wheel. Show that the probability of this event is n(k, K, r ) p(r +K −k)/2(1 − p)(r −K +k)/2. (2) (3) (4) (5) (6) (7) 72 2 Conditional Probability and Independence (8) (9) Let K = 2k. Show that the probability that you leave the casino on the r th spin given Exercise that you leave with nothing is the same as the probability that you leave the casino on the r th spin given that you leave with $K. Exercise this tell you about most gamblers? Show that doubling the stakes increases your chance of leaving with $K. What does 2.12 Example: Accidents and Insurance In any given year, the probability that a given male driver has a mishap entailing a claim from his insurance company is µ, independently of other years. The equivalent probability in female drivers is λ. Assume there are equal numbers of male and female drivers insured with the Acme Assurance Association, which selects one of them at random. (a) What is the probability that the selected driver makes a claim this year? (b) What is the probability that the selected driver makes a claim in two consecutive years? (c) If the insurance company picks a claimant at random, what is the probability that this claimant makes another claim in the following year? (a) Let A1 and A2 be the events that a randomly chosen driver makes a claim Solution in each of the ﬁrst and second years. Then conditioning on the sex of the driver (M or F) yields P(A1) = P(
A1\|M)P(M) + P(A1\|F)P(F) = 1 2 (µ + λ) because P(F) = P(M) = 1 2. (b) Likewise, P(A1 ∩ A2) = P(A1 ∩ A2\|M)P(M) + P(A1 ∩ A2\|F)P(F) = 1 2 (µ2 + λ2). (c) By deﬁnition, P(A2\|A1) = P(A2 ∩ A1)/P(A1) = µ2 + λ2 µ + λ (1) (2) (3) (4) (5) Note that A1 and A2 are conditionally independent given the sex of the driver. Are Exercise they ever independent? Exercise Exercise (a) Find the probability that a driver makes a claim in a third year given that the driver has claimed Show that P(A2\|A1) ≥ P(A2). When does equality hold? in each of the two preceding years. (b) Find the probability that a driver claims in year n, given that this driver has claimed in all of the preceding n years. (c) Find the limit in (b) as n → ∞. Exercise (a) Male (b) Female. Exercise Find the probability that a claimant is Find the probability that a driver claiming in n successive years is male. Worked Examples and Exercises 73 (6) Exercise and female drivers are insured with the AAA. Now rework the example and exercises without assuming that equal numbers of male Remark The fact that a claimant is more likely to have a second accident, even though accidents were assumed independent for a given driver, is an example of what is sometimes called a sampling paradox (though it is not a paradox). It is the reason why insurance companies offer no-claims discounts (or at least, one of the reasons). It is the case in practice that µ > λ. 2.13 Example: Protocols Part A: Boys and Girls Consider the following question: “Tom has exactly one sibling. What is the probability that it is a girl?” (a) Do you think this question has a well-deﬁned answer? (b) If so, write down your answer, and then consider the following arguments: (i) There
are three family possibilities; two girls, two boys, or one of each. Two girls is impossible, which leaves equal chances that the sibling is a boy or a girl. The answer is 1 2. (ii) Families with a child called Tom arise in four equally likely ways: T B, BT, T G, GT. So Tom has a brother as often as he has a sister. The answer is 1 2. (iii) There are four cases: B B, BG, G B, GG. The last is impossible, and in two of the remaining three cases the sibling is a girl. The answer is 2 3. (iv) Assuming that the sex of siblings is independent, the other sibling is equally likely to be a girl or a boy. The answer is 1 2. Are any of these correct? Is yours correct? (You may assume that any given birth gives rise to one girl or one boy with equal probability.) Solution (a) The question is ill-posed; there is no correct answer because the sample space is not deﬁned. This is the same as saying that the underlying experiment (selecting Tom) is not described. (b) We may consider some well-posed questions. I A woman has two children that are independently equally likely to be a boy or a girl. One of them at least (Tom) is male. Now the sample space has four equally likely outcomes: (1) = {BG, G B, B B, GG}; the event of interest is A = {BG, G B}, and we are given that B occurs, where B = {BG, G B, B B}. Hence, P(A\|B) = P(A ∩ B) P(B) = P(A) P(B) = 2 3. II A woman has two children that are independently equally likely to be a boy or a girl. Her ﬁrst son is called Tom with probability p1 < 1. If she has two sons, and the oldest is 74 2 Conditional Probability and Independence not Tom, the second son is called Tom with probability p2. The sample space is (2) = {B B, BT, T B, T G, BG, GT, G B, GG} where, for example, P(BT ) = 1 orem 2.1.4, 4 (1 − p1) p2. Then the required probability is, using The- P(GT ) +
P(T G) P(T B) + P(BT ) + P(T G) + P(GT ) = p1 + p2 2 p1 + 2 p2 − p1 p2. Notice that this is equal to 1 and we deﬁne p2 to be zero. In any case, 2 if either p1 = 0 or p2 = 0, but not both. It is also 1 2 if p1 = 1 ≤ 1 2 p1 + p2 2 p1 + 2 p2 − p1 p2 ≤ 2 3. Notice that we have assumed that families and names are independent; that is, that women are not more (or less) likely to have boys because they want to call them Tom (or not), and that having a girl does not change the chance that a boy is called Tom. III A boy is selected at random from a number of boys who have one sibling. This sample space has four equally likely outcomes = {B∗ B, B B∗, B∗G, G B∗} where the star denotes the boy (Tom) who was picked at random. (The experiment amounts to picking one of the B-symbols in (1) with equal chance of picking any.) Hence, the event of interest is A = {B∗G, G B∗} and the required probability is 1 2. IV A chance acquaintance is introduced as Tom who has just one sibling. What is the chance that it is a sister? The sample space is the set of your chance acquaintances. This is too vague to allow further progress. Remark The arguments of (b) (i), (ii), (iii), and (iv) appeared in letters to The Guardian in June 1989. An answer can only be deﬁned when the exact procedure (also known as a protocol) for selecting Tom is decided. If, for example, you meet Tom at a club for identical twins, the problem is different again. Notice also that parts of this example are getting rather distant from the type of experiment used to justify our axioms. You may well see no particular reason to suppose that our theory of probability is relevant in, say, Case IV, or even in Case II. (3) (4) (5) In the framework of Case II, consider the following two procedures: Exercise (a) Select one of her two children at random (b) Select one of her sons (if
any) at random. In each case, ﬁnd the probability that the child is the elder given that his name is Tom. Exercise likely outcomes? Exercise Suppose a woman has three children and each is independently equally likely to be male or female. Show that the event “they are either all girls or all boys” is independent of the event “at least two children are boys.” In the framework of Case II, can it be the case that T G, T B, BT, and GT are equally Worked Examples and Exercises 75 Part B: Goats and Cars: The Monty Hall Problem Suppose yourself to be participating in the following bizarre contest. You have a choice of three doors. Behind one door is a costly automobile, behind the other two doors are cheap goats. You choose the ﬁrst door, whereupon the master of ceremonies opens the third door to reveal a goat; he then offers you the opportunity to change your choice of door. Can you calculate the probability that the car lies behind the second door? (You are given the object behind the door you open.) Solution No, you cannot. To see this, let Ci be the event that the car lies behind the ith door, and let G be the event that a goat is revealed to you behind the third door. You require P(C2\|G), which we can write as (6) P(C2 ∩ G)/P(G) = P(G\|C2)P(C2) P(G\|C1)P(C1) + P(G\|C2)P(C2) = P(G\|C2) P(G\|C1) + P(G\|C2), on the reasonable assumption that the car is equally likely to be behind any door, so that P(C2) = P(C1). Now observe that all three terms in the denominator and numerator of (6) depend on the decisions of the master of ceremonies. His rules for making his decision once again form a protocol. If you do not know his protocol for the contest, you cannot calculate P(C2\|G). Remark This problem was presented in Parade magazine (1990, 1991, distributed in the USA) and generated an extensive correspondence in that and several other periodicals. Almost all participants assumed (wrongly) that the problem as stated has one solution, and chieﬂy disputed as to
whether the answer should be 1 2 or 2 3. (7) (8) (9) Show that if you have paid the master of ceremonies enough to ensure that you win, Exercise then P(C2\|G) = 1. Exercise Show that if the master of ceremonies has decided that (i) whatever you choose, he will show you a goat; and (ii) if he has a choice of two goats, he will pick one at random, then P(C2\|G) = 2 3. Exercise Show that if the master of ceremonies has decided that (i) whatever you choose, he will open a different door; and (ii) he will pick it at random, then P(C2\|G) = 1 2. Show that if the master of ceremonies has decided that (i) if a goat is behind the ﬁrst door, he will open it for you; and (ii) if a car lies behind the ﬁrst door, he will open another door, then P(C2\|G) = 0. (10) Exercise Remark There are many famous problems equivalent to these two, to all of which the correct answer is, there is no unique answer (e.g., the ‘Prisoners’ paradox’, ‘Red Ace’). There seems to be no way of preventing the futile, acrimonious, and incorrect discussions accompanying their regular appearance in the popular press. The so-called “Doomsday Argument” provides a slightly different but equally fallacious example. 2.14 Example: Eddington’s Controversy Four men each tell the truth independently with probability 1 3. D makes a statement that C reports to B, and B then reports C’s statement to A. If A asserts that B denies that C claims that D is a liar, what is the probability that D spoke the truth? 76 2 Conditional Probability and Independence Let SA be the event that A makes the given statement. Further, let A denote Solution the event that A tells the truth, Ac denote the event that A lies, and so on. Then, obviously, D ∩ C ∩ B ∩ A ⊆ SA, for if they all tell the truth then A makes the given statement. Also, by the independence, P(D ∩ C ∩ B ∩ A) = 1 81 Likewise, by following through the chain of assertions, we see that
D ∩ C c ∩ Bc ∩ A, D ∩ C c ∩ B ∩ Ac and D ∩ C ∩ Bc ∩ Ac are included in SA, each having probability 4 81.. When D lies, Dc ∩ C c ∩ Bc ∩ Ac ⊆ SA for A also makes the given statement if they are all liars. Here, P(Dc ∩ C c ∩ Bc ∩ Ac) = 16 81 and likewise Dc ∩ C c ∩ B ∩ A, Dc ∩ C ∩ Bc ∩ A and Dc ∩ C ∩ B ∩ Ac are included in SA, each having probability 4 81. These mutually exclusive outcomes exhaust the possibilities, so by conditional probability, P(D\|SA) = P(D ∩ SA) P(D ∩ SA) + P(Dc ∩ SA) = 13 81 + 28 81 13 81 = 13 41. (1) (2) (3) (4) 1 Exercise What is the probability that C did claim that D is a liar, given SA? Exercise What is the probability that both C and D lied, given SA? Exercise rather than by listing outcomes. Exercise you reconstruct the argument that led him to this answer? Eddington himself gave the answer to this problem as 25 Prove the result of the example more laboriously by using conditional probability, 71. This is the controversy! Can Remark This example is similar to Example 2.13, in that it is entertaining but of no real practical value. Our theory of probability does not pretend to include this type of problem, and nothing can be said about the credibility of real reports by these methods. Despite this, the ﬁrst attempt to do so was made in the seventeenth century, and such attempts have been repeated sporadically ever since. P R O B L E M S The probability that an archer hits the target when it is windy is 0.4; when it is not windy, her probability of hitting the target is 0.7. On any shot, the probability of a gust of wind is 0.3. Find the probability that: (a) On a given shot, there is a gust of wind and she hits the target. (b) She hits the target with her ﬁrst shot. (c) She hits the target exactly once in two shots.
(d) There was no gust of wind on an occasion when she missed. 2 3 4 5 6 7 8 9 Problems 77 Let A, B be two events with P(B) > 0. Show that (a) If B ⊂ A, then P(A\|B) = 1, (b) If A ⊂ B, then P(A\|B) = P(A)/P(B). Three biased coins C1, C2, C3 lie on a table. Their respective probabilities of falling heads when, 2 tossed are 1 3, and 1. A coin is picked at random, tossed, and observed to fall heads. Calculate the 3 probability that it is Ck for each k = 1, 2, 3. Given that a coin has been tossed once and observed to fall heads, calculate the probability that a second throw of the same coin will also produce heads. The experiment is begun again with the same three coins. This time the coin selected is tossed twice and observed to fall heads both times. Calculate the probability that it is Ck for each k = 1, 2, 3. Given that a coin has been tossed twice and observed to fall heads both times, calculate the probability that a third throw of the same coin will also produce heads. (a) An event E is independent of itself. Show that it has probability either 0 or 1. (b) Events A and B are disjoint. Can you say whether they are dependent or independent? (c) Prove that if events A and B are independent then so are the events Ac and B, and the events Ac and Bc. Candidates are allowed at most three attempts at a given test. Given j − 1 previous failures, the probability that a candidate fails at his jth attempt is p j. If p1 = 0.6, p2 = 0.4, and p3 = 0.75, ﬁnd the probability that a candidate: (a) Passes at the second attempt: (b) Passes at the third attempt: (c) Passes given that he failed at the ﬁrst attempt; (d) Passes at the second attempt given that he passes. Dick throws a die once. If the upper face shows j, he then throws it a further j − 1 times and adds all j scores shown. If this sum is 3, what is the probability that he only threw the die (a) Once altogether? (
b) Twice altogether? A man has ﬁve coins in his pocket. Two are double-headed, one is double-tailed, and two are normal. They can be distinguished only by looking at them. (a) The man shuts his eyes, chooses a coin at random, and tosses it. What is the probability that the lower face of the coin is a head? (b) He opens his eyes and sees that the upper face is a head. What is the probability that the lower face is a head? (c) He shuts his eyes again, picks up the coin, and tosses it again. What is the probability that the lower face is a head? (d) He opens his eyes and sees that the upper face is a head. What is the probability that the lower face is a head? An urn contains four dice, one red, one green, and two blue. (a) One is selected at random; what is the probability that it is blue? (b) The ﬁrst is not replaced, and a second die is removed. What is the chance that it is: (i) blue? or (ii) red? (c) The two dice are thrown. What is the probability that they show the same numbers and are the same colour? (d) Now the two remaining in the urn are tossed. What is the probability that they show the same number and are the same colour, given that the ﬁrst two did not show the same number and colour? A 12-sided die A has 9 green faces and 3 white faces, whereas another 12-sided die B has 3 green faces and 9 white faces. A fair coin is tossed once. If it falls heads, a series of throws is made with die A alone; if it falls tails then only the die B is used. 10 11 12 13 14 15 16 17 78 2 Conditional Probability and Independence (a) Show that the probability that green turns up at the ﬁrst throw is 1 2. (b) If green turns up at the ﬁrst throw, what is the probability that die A is being used? (c) Given that green turns up at the ﬁrst two throws, what is the probability that green turns up at the third throw? Suppose that any child is male with probability p or female with probability 1 − p, independently of other children. In a family with four children, let A be
the event that there is at most one girl, and B the event that there are children of both sexes. Show that there is a value of p, with 0 < p < 1 2, such that A and B are independent. Suppose that parents are equally likely to have (in total) one, two, or three offspring. A girl is selected at random; what is the probability that the family includes no older girl? (Assume that children are independent and equally likely to be male or female.) Two roads join Ayton to Beaton, and two further roads join Beaton to the City. Ayton is directly connected to the City by a railway. All four roads and the railway are each independently blocked by snow with probability p. I am at Ayton. (a) Find the probability that I can drive to the City. (b) Find the probability that I can travel to the City. (c) Given that I can travel to the City, what is the probability that the railway is blocked? An urn contains b blue and r red balls, which may be withdrawn at random according to one of the following three schemes. (a) The balls are removed at random one at a time until all those remaining are of the same colour. (b) The balls are removed until a ball differs in colour from its predecessor. This ﬁrst different ball is replaced in the urn; this process is then continued until the remaining balls are all the same colour. (c) The balls are removed one by one and inspected. The ﬁrst is discarded. Each succeeding ball that is the same colour as its predecessor is replaced, the others are discarded, until the remaining balls are all the same colour. In each case, ﬁnd the probability that the remaining balls are all red. Let A1, A2,..., An be independent events. Show that the probability that none of the events A1,..., An occur is less than exp (− Let A and B be independent events. Show that n 1 P(Ai )). max{P((A ∪ B)c), P(A ∩ B), P(A B)} ≥ 4 9. A coin is tossed repeatedly; on each toss, a head is shown with probability p or a tail with probability 1 − p. All tosses are mutually independent. Let E denote the event that the ﬁrst run of r successive heads occurs earlier than the �
�rst run of s successive tails. Let A denote the outcome of the ﬁrst toss. Show that P(E\|A = head) = pr −1 + (1 − pr −1)P(E\|A = tail). Find a similar expression for P(E\|A = tail) and hence ﬁnd P(E). After marking the papers of a certain student, the examiners are unable to decide whether he really understands the subject or is just blufﬁng. They reckon that the probability that he is a bluffer is p, 0 < p < 1, and the probability that he understands is q = (1 − p). They therefore give him a viva voce consisting of n independent questions, each of which has a probability u of being answered by someone who understands the subject. Unfortunately, there is also a probability b, 0 < b < 1, that the answer can be guessed by someone who does not understand. Show that the probability that the student understands given that he manages to answer k questions correctly is given by r, where r = quk(1 − u)n−k quk(1 − u)n−k + pbk(1 − b)n−k. 18 19 20 21 22 23 24 25 Problems 79 Show that if the student gets every single question right and u > b, then as n increases the probability that the student really understands tends to 1. How many questions must the student get right to convince the examiners that it is more likely that he understands the subject than that he is blufﬁng? A team of three students Amy, Bella, and Carol answer questions in a quiz. A question is answered, or 1 by Amy, Bella, or Carol with probability 1 6, respectively. The probability of Amy, Bella, 2, or 3 or Carol answering a question correctly is 4 5, respectively. What is the probability that the 5 team answers a question correctly? Find the probability that Carol answered the question given that the team answered incorrectly., 1 3, 3 5 The team starts the contest with one point and gains (loses) one point for each correct (incorrect) answer. The contest ends when the team’s score reaches zero points or 10 points. Find the probability that the team will win the contest by scoring 10 points, and show that this is approximately 4 7. A and B play a sequence of games. in each of which A has a
probability p of winning and B has a probability q (= 1 − p) of winning. The sequence is won by the ﬁrst player to achieve a lead of two games. By considering what may happen in the ﬁrst two games, or otherwise, show that the probability that A wins the sequence is p2/(1 − 2 pq). If the rules are changed so that the sequence is won by the player who ﬁrst wins two consecutive games, show that the probability that A wins the sequence becomes p2(1 + q)/(1 − pq). Which set of rules gives that weaker player the better chance of winning the sequence? You toss a coin. If it shows a tail, you roll one die and your score is the number it shows. If the coin shows a head, you toss ﬁve more coins and your score is the total number of heads shown (including the ﬁrst coin). If you tell me only that your score is two, what is the probability that you rolled a die? Three fair dice labelled A, B, and C are rolled on to a sheet of paper. If a pair show the same number a straight line is drawn joining them. Show that the event that the line AB is drawn is independent of the event that BC is drawn. What is the probability that a complete triangle is drawn? (The dice are not colinear.) You roll a fair die n times. What is the probability that (a) You have rolled an odd number of sixes? (b) You have not rolled a six on two successive rolls? (c) You rolled a one before you rolled a six, given that you have rolled at least one of each? Irena throws at a target. After each throw she moves further away so that the probability of a hit is two-thirds of the probability of a hit on the previous throw. The probability of a hit on the ﬁrst throw is 1 4. Find the probability of a hit on the nth throw. Deduce that the probability of never hitting the target is greater than 1 4. A fair coin is tossed three times. What is the probability that it lands “heads” at least once? In a coin-tossing game, a player tosses ﬁve fair coins. If he is content with the result, he stops. If not, he picks up one or more of the coins and tosses
them a second time. If he is still dissatisﬁed, he may for one last time pick up and throw again one or more of the coins. Show that if the player’s aim is to ﬁnish with ﬁve heads showing, and if he uses the best strategy, then the probability that he will succeed is ( 7 A second player plays the same game but aims to ﬁnish with either all heads or all tails showing. 8 )5. What is the probability of his succeeding? Alf and Bert play a game that each wins with probability 1 2. The winner then plays Charlie whose probability of winning is always θ. The three continue in turn, the winner of each game always playing the next game against the third player, until the tournament is won by the ﬁrst player to win two successive games, Let p A, pB, pC be the probabilities that Alf, Bert, and Charlie, respectively, win the tournament. Show that pC = 2θ 2/(2 − θ + θ 2). Find p A and pB, and ﬁnd the value of θ for which p A, pB, pC are all equal. (Games are independent.) If Alf wins the tournament, what is the probability that he also won the ﬁrst game? 26 Box A contains three red balls and two white balls; box B contains two red balls and two white balls. A fair die is thrown. If the upper face of the die shows 1 or 2, a ball is drawn at random from 80 2 Conditional Probability and Independence box A and put in box B and then a ball is drawn at random from box B. If the upper face of the die shows 3, 4, 5 or 6, a ball is drawn at random from box B and put in box A, and then a ball is drawn at random from box A. What are the probabilities (a) That the second ball drawn is white? (b) That both balls drawn are red? (c) That the upper face of the red die showed 3, given that one ball drawn is white and the other red? A fair six-sided die, with faces numbered from 1 to 6, is thrown repeatedly on to a ﬂat surface until it ﬁrst lands with the 6 face uppermost. Find the probability that this requires: (a) n throws. (b) An even
number of throws. (c) Show that the probability that the 5 face appears at least once before the ﬁrst 6 is 1 2, and ﬁnd the probability that all the faces 1 to 5 appear before the ﬁrst 6. Suppose that n water lily leaves are placed so that the base of each leaf lies on a circle. A frog is initially on leaf L 1; she hops clockwise to the adjacent leaf L 2 with probability p, or anticlockwise to leaf L n with probability q. Succeeding hops are independent, and go to the nearest leaf clockwise with probability p or the nearest leaf anticlockwise with probability q. Find the probability that: (a) The frog returns to L 1 before visiting all n leaves. (b) The ﬁrst hop on to L 1 has the same orientation as the ﬁrst hop off L 1. (c) What is the probability that the ﬁrst hop on to L 1 is clockwise? Anselm and Bill toss a fair coin repeatedly. Initially, Anselm has m marks, where 1 ≤ m ≤ n − 1. If the coin shows a head, then Anselm gains a mark from Bill; otherwise, he forfeits a mark to Bill. Whenever Anselm has n marks, he must immediately give one to Bill. Let pk m be the probability that Anselm has n marks on k occasions before the ﬁrst moment m, with appropriate boundary at which he has no marks. Write down a difference equation for pk conditions, and deduce that for k ≥ 1, pk m = m n2 1 − 1 n k−1. Explain how you could have shown this without solving the equation for pk is certain to lose all his marks eventually. A and B each have $60. They play a sequence of independent games at each of which A wins $x from B with probability p, or loses $x to B with probability q, where p + q = 1. The stake x is determined by rolling a fair die once, and setting x as the number shown by the die; 1 ≤ x ≤ 6. (a) What is the probability that A wins his opponent’s fortune before losing his own? (b) If A could choose the stake to be an integer x such that 1 ≤ x ≤ 6, and p < q, what value m. Show that Anselm
should he choose for x? 27 28 29 30 31 A document is equally likely to be in any of three boxﬁles. A search of the ith box will discover the document (if it is indeed there) with probability pi. What is the probability that the document is in the ﬁrst box: (a) Given that I have searched the ﬁrst box once and not found it? (b) Given that I have searched the ﬁrst box twice and not found it? (c) Given that I have searched all three boxes once and not found it? Assume searches are independent. A network forming the edges of a cube is constructed using 12 wires, each 1 metre long. An ant is placed on one corner and walks around the network, leaving a trail of scent as it does so. It never turns around in the middle of an edge, and when it reaches a corner: 32 Problems 81 (i) If it has previously walked along both the other edges, it returns along the edge on which it has just come. (ii) If it has previously walked along just one of the other edges, it continues along the edge along which it has not previously walked. (iii) Otherwise, it chooses one of the other edges arbitrarily. Show that the probability that the ant passes through the corner opposite where it started after walking along just three edges is 1 2, but that it is possible that it never reaches the opposite corner. In the latter case, determine the probability of this occurring. What is the greatest distance that the ant has to walk before an outside observer (who knows the rules) will know whether the ant will ever reach the corner opposite where it started? Show that the rules may be modiﬁed to guarantee that the ant (whose only sense is smell) will be able to reach the corner opposite the corner where it started by walking not more than a certain maximum distance that should be determined. The ant can count. Pooling to be the contaminated one, and you need to identify it. Luckily, you have an infallible test. (a) If you test the jars one at a time, ﬁnd the probability that you require t tests to identify the You have (n!)2 jars of ﬂuid, one of which is contaminated. Any jar is equally likely contaminated jar. (b) Alternatively, you may arrange the jars in j groups of size k, where jk = (n!)2. A
sample from each of the jars in a group is pooled in one jar, and this pooled sample is tested. On ﬁnding the contaminated pooled sample, each jar of this group is tested separately. Find the probability that you require t tests to ﬁnd the contaminated jar. (c) What is the best choice for j and k? Simpson’s Paradox Two drugs are being tested. Of 200 patients given drug A, 60 are cured; and of 1100 given drug B, 170 are cured. If we assume a homogeneous group of patients, ﬁnd the probabilities of successful treatment with A or B. Now closer investigation reveals that the 200 patients given drug A were in fact 100 men, of whom 50 were cured, and 100 women of whom 10 were cured. Further, of the 1100 given drug B, 100 were men of whom 60 were cured, and 1000 were women of whom 110 were cured. Calculate the probability of cure for men and women receiving each drug; note that B now seems better than A. (Results of this kind indicate how much care is needed in the design of experiments. Note that the paradox was described by Yule in 1903, and is also called the Yule-Simpson paradox.) In Problem 34, given that a randomly chosen patient is cured, ﬁnd: (a) The probability that the patient is male. (b) The probability that the patient is female. Prisoners’ Paradox Three prisoners are informed by their warder that one of them is to be released and the other two shipped to Devil’s Island, but the warder cannot inform any prisoner of that prisoner’s fate. Prisoner A thus knows his chance of release to be 1 3. He asks the warder to name some one of the other two who is destined for Devil’s Island, and the warder names B. Can A now calculate the conditional probability of his release? Let A and B be events. Show that P(A ∩ B\|A ∪ B) ≤ P(A ∩ B\|A). When does equality hold? Explain the following “paradox” posed by Lewis Carroll. We are provided with a supply of balls that are independently equally likely to be black or white. Proposition If an urn contains two such balls, then one is black and the other white. Proof Initially, P(B B) = P(BW ) = P(W B) = P(W W )
= 1 4. Add a black ball, so that now P(B B B) = P(B BW ) = P(BW B) = P(BW W ) = 1 4. 33 34 35 36 37 38 82 2 Conditional Probability and Independence Now pick a ball at random. By (2.1.3), P(black ball drawn) = 1. 1 4 + 2 3 But if I pick a ball at random from three, with probability 2 3 of drawing a black ball, then two are black and one is white. Hence, before adding the black ball the urn contained one white ball and one black ball. 39 40 (*) 41 Let M1, M2,..., Mn be a sequence of men such that M j reports to M j+1 on a statement made by M j−1. Let Rn be the event that Mn reports that Mn−1 reports that... that M2 reports that M1 is a liar. If each reporter lies independently with probability p, ﬁnd pn, the probability that M1 told the truth given Rn. Show that as n → ∞, pn → 1 − p. Suppose that for events S, A, and B, P(S\|A) ≥ P(S) P(A\|S ∩ B) ≥ P(A\|S) P(A\|Sc) ≥ P(A\|Sc ∩ B). (a) Show that, except in trivial cases, P(S\|A ∩ B) ≥ P(S\|B). (b) Show that P(S\|A) ≥ P(A). (c) Show that if (∗) is replaced by P(S\|B) ≥ P(S), then P(S\|A ∩ B) ≥ P(S\|A). You have to play Alekhine, Botvinnik, and Capablanca once each. You win each game with respective probabilities pa, pb, and pc, where pa > pb > pc. You win the tournament if you win two consecutive games, otherwise you lose, but you can choose in which order to play the three games. 42 43 Weather Show that to maximize your chance of winning you should play Alekhine second. Show that the events A and B are mutually attractive if and only if P(B\|A) > P(B\|Ac). Days can be sunny or cloudy. The weather tomorrow
is the same as the weather today with probability p, or it is different with probability q, where p + q = 1. If it is sunny today, show that the probability sn that it will be sunny n days from today satisﬁes where s0 = 1. Deduce that sn = ( p − q)sn−1 + q; n ≥ 1, sn = 1 2 Flats, Sharps, and Craps Dice can be crooked (or weighted) in various ways. One way is to shorten the distance between one pair of opposite faces, thus making them more likely; these are called ﬂats (also known as broads or doctors). (1 + ( p − q)n); n ≥ 1. Another way is to taper all four sides of the die, creating a truncated pyramid, or to insert a weight in the base. The top face becomes more likely and the base less likely; these are called sharps. You have three pairs of dice: (i) A fair pair for which p1 = p2 = p3 = p4 = p5 = p6 = 1 6. (ii) A pair of 1–6 ﬂats for which p1 = p6 = 1 4 and p2 = p3 = p4 = p5 = 1 8. (iii) A pair of 5–2 sharps for which p5 = 1 12, and p1 = p3 = p4 = p6 = 1 6. 4 With which pair would you prefer to play craps? (See Example 1.12 for the rules.) The Monty Hall Problem: Example 2.13B Suppose that the presenter’s protocol requires him to show you a goat when he opens another door. With a choice of two goats (called Bill and Nan, say), he shows you Bill with probability b. Show that the conditional probability that the third door conceals the car, given that you are shown Bill, is, p2 = 1 1 1+b. 44 45 3 Counting What I say is, patience, and shufﬂe the cards. Cervantes This chapter deals with a special subject and may be omitted on a ﬁrst reading. Its contents are important and useful, but are not a prerequisite for most of the following chapters. 3.1 First Principles We have seen that many interesting problems in probability can be solved by counting the number of outcomes in an event. Such counting often
turns out to also be useful in more general contexts. This chapter sets out some simple methods of dealing with the commonest counting problems. The basic principles are pleasingly easy and are perfectly illustrated in the following examples. (1) Principle If I have m garden forks and n ﬁsh forks, then I have m + n forks altogether. (2) Principle If I have m different knives and n different forks, then there are mn distinct ways of taking a knife and fork. These principles can be rephrased in general terms involving objects, operations, or symbols and their properties, but the idea is already obvious. The important points are that in (1), the two sets in question are disjoint; that is a fork cannot be both a garden fork and a ﬁsh fork. In (2), my choice of knife in no way alters my freedom to choose any fork (and vice versa). Real problems involve, for example, catching different varieties of ﬁsh, drawing various balls from a number of urns, and dealing hands at numerous types of card games. In the standard terminology for such problems, we say that a number n (say) of objects or things are to be divided or distributed into r classes or groups. The number of ways in which this distribution can take place depends on whether (i) The objects can be distinguished or not. (ii) The classes can be distinguished or not. (iii) The order of objects in a class is relevant or not. 83 84 3 Counting (iv) The order of classes is relevant or not. (v) The objects can be used more than once or not at all. (vi) Empty classes are allowed or not. We generally consider only the cases having applications in probability problems. Other aspects are explored in books devoted to combinatorial theory. (3) Example (a) Six dice are rolled. What is the probability that they all show different faces? (b) What is the probability that ﬁve dice show different faces when rolled? Solution the assumed symmetry of the dice, (a) Let A be the event that they all show different faces. Then, because of P(A) = \|A\| \|\|. Now the upper face of each die may be freely chosen in six different ways, so by Principle 2 \|\| = 66 = 46656. However, for outcomes in A, the upper faces are required to be different. Thus, when the upper face of one die is freely
chosen in six ways, the upper face of the next can be freely chosen in ﬁve ways (different from the ﬁrst choice). The next may be freely chosen in four ways, and so on. Hence, by Principle 2, \|A\| = 6! and = 5 324 P(A) = 6! 66 (b) Let ˆA be the event that the ﬁve dice show different faces. By the same argument as. above, \|\| = 65. For outcomes in ˆA we may ﬁrst make a free choice of which different ﬁve faces are to be shown; this is the same as choosing one face not to be shown, which we can do in six ways. Then the ﬁrst face is freely chosen in ﬁve ways, the second in four ways, and so on. Hence, \| ˆA\| = 6! and P( ˆA) = 6! 65 = 5 54. 3.2 Permutations: Ordered Selection Suppose that several objects are placed randomly in a row; playing cards or lottery numbers provide trite examples (but important if you hold a ticket). The number of ways in which this arrangement may occur depends on how many objects there are, whether they are all distinct, whether they may be repeated, and so on. Such arrangements are called permutations. (1) Theorem Given n distinct symbols, the number of distinct permutations (without repetition) of length r ≤ n is n(n − 1)... (n − r + 1) = n! (n − r )!. 3.2 Permutations: Ordered Selection 85 (2) Theorem Given n distinct symbols which may be repeated any number of times, the number of permutations of length r is nr. Proof Theorems (1) and (2) are easily proved by induction. You do it. (3) r Theorem i=1 ni symbols of r distinct types, where ni are of type i and are otherwise indistinguishable, the number of permutations (without repetition) of all n symbols is Given n = Mn(n1,..., nr ) = n! r ni! i=1. Proof Suppose that the symbols of each type are numbered so that they are all distinguishable. Then in each originally unnumbered permutation, the symbols of type 1 can be permuted in n1!
ways, the symbols of type 2 in n2! ways, and so on. Thus, the total number of permutations is Mn(n1,..., nr )n1!n2!... nr!. However, we already know from (1) that the number of permutations of n objects is n!. Hence, which proves (3). Mn(n1,..., nr )n1!... nr! = n!, The number Mn is known as a multinomial coefﬁcient. A particularly important case that arises frequently is when r = 2. This is a binomial coefﬁcient, and it has its own special notation: (4) Mn(k, n − k) = n k = nCk = n! k!(n − k)! in most books. in some older books.. (5) Example You are playing bridge. When you pick up your hand, you notice that the suits are already grouped; that is, the clubs are all adjacent to each other, the hearts likewise, and so on. Given that your hand contains four spades, four hearts, three diamonds, and two clubs, what is the probability P(G) of this event G? Solution There are 13! permutations of your hand, which we assume are equally likely by symmetry. Now there are 4! permutations in which the spades are adjacent in any given position, 4! where the hearts are adjacent, and so on. Furthermore, there are 4! permutations of the order in which the respective suits may be placed in their adjacent blocks. Hence, the number of permutations in which G occurs is 4!4!4!3!2! and (6) P(G) = (4!)33!2! 13! = 4! M13(4, 4, 3, 2). 86 3 Counting Alternatively, you may observe from (3) that there are M13(4, 4, 3, 2) permutations of your hand where cards of the same suit are regarded as indistinguishable. For each order of suits, only one of these is in G. Because there are 4! permutations of the suit order, we immediately recover (6) again. Finally, we remark that the deﬁnition of the symbol (n other than that when n and r are integers with 0 ≤ r ≤ n. r ) is
sometimes extended to cases (7) Deﬁnition For real x and nonnegative integer r, x r = x(x − 1)... (x − r + 1) r! This deﬁnition can occasionally provide more compact expressions, for example, we have −x r = (−)r x + r − 1 r. 3.3 Combinations: Unordered Selection In a bridge hand or an election, the order in which you get your cards or the politician his votes is irrelevant. In problems of this type we do not arrange, we choose; a choice of objects or symbols is also called a combination. (1) Theorem distinct symbols without repetition is The number of ways of choosing a set of r symbols from a set of n n! r!(n − r )! =. n r Proof This is just a special case of Theorem 3.2.3. (2) Theorem sizes n1, n2,..., nr, where The number of ways of dividing n distinct objects into r distinct groups of r i=1 ni!. i=1 ni = n, is n!/ r Proof This is also a simple corollary of Theorem 3.2.3. r 1 ni symbols (where the ith distinct set contains ni indistin- Out of n = (3) Theorem guishable symbols), we can select i=1(ni + 1) − 1 combinations. r Proof Note that we can select any number of symbols from zero to ni, from each of the r sets, but we cannot take zero from all of them. (4) Theorem Given a set of n distinct symbols that may be repeated any number of times, the number of ways of choosing a set of size r is ( n+r −1 ). r Proof A proof of Theorem (4) may be found in Example 3.12 or Theorem 3.7.5. 3.4 Inclusion–Exclusion 87 (5) Example: Ark The wyvern is an endangered species in the wild. You want to form a captive breeding colony, and you estimate that a viable colony should initially contain r males and r females. You therefore trap a sequence of animals, each of which is independently male with probability p or female with probability q = 1 − p, where p = q (the females are more wary). Find the probability pn that it is necessary to capture n animals to create your
viable colony of r males and r females. Let An be the event that you ﬁrst possess r of each sex with the nth capture, Solution and let M be the event that the nth animal is male. Then, of the previous n − 1 captured animals, r − 1 are male and n − r are female. For any ﬁxed order of these sexes, the probability of being captured in that order is pr q n−r. The number of ways of ordering r males and n − r females, with a male last, is just the same as the number of ways of choosing r − 1 of the ﬁrst n − 1 captures to be male. By (1), this is ( n−1 r −1 ). Hence, for n ≥ 2r, P(An ∩ M) = n − 1 r − 1 pr q n−r. Likewise, when the last animal is female, P(An ∩ M c) = n − 1 r − 1 qr pn−r, and so n − 1 r − 1 pn = qr pr ( pn−2r + q n−2r ), n ≥ 2r. 3.4 Inclusion–Exclusion If a group of N men contains N (b1) who are bald, N (b2) who are bearded, and N (b1, b2) who are bald and bearded, how many altogether are bald, bearded, or both? The answer is N1 = N (b1) + N (b2) − N (b1, b2), because anyone who is both is counted once in all three terms on the right-hand side, and so contributes just one to the total N1 in (1). More generally, if a group of N objects may each have up to r distinct properties b1,..., br, then the number possessing at least one is N (bi ) − N1 = N (bi, b j ) + · · · + (−)r −1 N (b1,..., br ). bi bi <b j (1) (2) This is proved either by induction or by noting that any object having exactly k of the properties is counted ( k t ) times in the tth term, and it is the case that k t=1 (−)t+1 = 1. k t 88 3 Counting [See Problem 3.28(
a).] Hence, this object contributes just one to the total N1, as required. Notice that N0, the number of objects possessing none of the r properties, is just (3) N0 = N − N1. (4) Example: Derangements A permutation of the ﬁrst n integers is called a derangement if no integer is in its natural position. Thus, (3, 2, 1) is not a derangement of (1, 2, 3), but (2, 3, 1) is a derangement. Suppose one of the n! permutations of (1,..., n) is picked at random. Find pn, the probability that it is a derangement, and show that as n → ∞, pn → e−1. Let bk be the property that the integer k is in its natural position (the kth place). Solution Then, the number of derangements is n! − N1, where N1 is given by (2). Now, N (bk) is the number of permutations of (1,..., n) with k ﬁxed, namely, (n − 1)!. Likewise, N (bi, b j ) is the number of permutations of (1, 2,..., n) with i and j ﬁxed, namely, (n − 2)!, and so on. Hence, N1 = n j=1 (n − 1)! − = n(n − 1)! − (n − 2)! + · · · + (−)n−1 1≤i< j≤n n 2 (n − 2)! + · · · + (−)n−1. Now, because all permutations were supposed equally likely, the required probability is + 1 2! (n! − N1) = 1 − 1 1! + · · · + (−)n 1 n! pn = 1 n! − 1 3! → e−1 as n → ∞. 3.5 Recurrence Relations The answer a to a counting problem usually depends on some given parameters; such as the original number n of objects, the number r which are red (say), or the number s selected. We can make this explicit by writing, for example, a = a(n, r, s). It is often possible to ﬁnd relationships between
two or more of a(n, r, s), and then the problem is reduced to solving a recurrence relation. Some of these can indeed be solved. (1) Theorem If (an; n ≥ 0) satisﬁes then an = c(n)an−1, n ≥ 1, an = n k=1 c(k)a0. Proof Trivial, by induction. (2) Theorem If (an; n ≥ 0) satisﬁes an+2 + 2ban+1 + can = 0, n ≥ 0, and x = α and x = β are two distinct solutions of x 2 + 2bx + c = 0, then 3.5 Recurrence Relations an = 1 α − β ((a1 − βa0)αn − (a1 − αa0)βn). Proof Straightforward by induction. 89 Higher-order linear difference equations can be solved similarly; we omit the details. The recurrence can be in more than one variable, as the following example shows. (3) Example Show that Let a(n, k) be the number of ways of choosing k objects from n objects. a(n, k) =. n k Suppose we add an additional (n + 1)th object, and choose k of these n + 1. Solution This may be done in a(n + 1, k) ways. But also we may consider whether the additional object was among those chosen. The number of choices when it is not is a(n, k). If it is, then it is necessary to choose k − 1 further objects from the remaining n, which we may do in a(n, k − 1) ways. Then, a(n, k) satisﬁes the difference equation 0 ≤ k ≤ n, a(n + 1, k) = a(n, k) + a(n, k − 1), (4) with solution by inspection a(n, k) = ( n k ). The array generated by a(n, k); 0 ≤ k ≤ n, as n increases, is called Pascal’s triangle; it has many curious and interesting properties, which we do not explore here. Part of it is displayed in Figure 3.1. Figure 3.1 Pascal’s triangle written as an array of binomial coefﬁcients and as the array of their values. Obs
erve that by (4), each term is obtained by adding its two neighbours in the row above [except for ( 0 0 ) which is deﬁned to be 1]. 90 3 Counting 3.6 Generating Functions Even quite straightforward counting problems can lead to laborious and lengthy calculations. These are often greatly simpliﬁed by using generating functions (introduced by de Moivre and Euler in the early eighteenth century). Later examples will show the utility of generating functions; in this section, we give a fairly bald list of basic deﬁnitions and properties, for ease of reference. (1) Deﬁnition Given a collection of numbers (ai ; i ≥ 0), the function ga(x) = ∞ i=0 ai x i ai x i converges is called the generating function of (ai ). (It is, of course, necessary that somewhere if ga is deﬁned as a function of x. If we regard ga as an element of a ring of polynomials, such convergence is not necessary.) (2) Deﬁnition Given (ai ; i ≥ 0), the function ha(x) = ∞ i=0 ai x i i! is the exponential generating function of (ai ). (3) Deﬁnition Given a collection of functions ( fn(y); n ≥ 0); the function ∞ g(x, y) = x n fn(y) is a bivariate generating function. n=0 The following crucial result is a corollary of Taylor’s Theorem. We omit the proof. (4) Theorem (Uniqueness) If for some x0 and x1 we have ga(x) = gb(x) < ∞ for x0 < x < x1, then ai = bi for all i. Generating functions help to tackle difference equations; the following result is typical. (5) Theorem If (an; n ≥ 0) satisﬁes a recurrence relation an+2 + 2ban+1 + can = dn; n ≥ 0, then ga(x) = x 2gd (x) + a0 + a1x + 2ba0x 1 + 2bx + cx 2 with a corresponding result for higher-order equations., Proof To prove (5), multiply each side of the recurrence by x n+2 and sum
over n. 3.6 Generating Functions 91 The next four theorems are proved by rearranging the summation on the right-hand side in each case. You should do at least one as an exercise. (6) Theorem (Convolution) If (an) can be written as a convolution of the sequences (bn) and (cn), so then (7) Theorem (Tails) If then (8) Theorem If then an = n i=0 ci bn−i, n ≥ 0, ga(x) = gc(x)gb(x). bn = ∞ i=1 an+i, n ≥ 0, gb(x) = ga(1) − ga(x) 1 − x. cn = n i=0 ai, gc(x) = ga(x) 1 − x. (9) Theorem (Exponential Function) Let the function e(x) be deﬁned by e(x) = ∞ k=0 x k k!, then e(x + y) = e(x)e(y) = ex+y = ex ey. Finally, we have the celebrated (10) Binomial Theorems For integral n, n (1 + x)n = k=0 n k x k, n ≥ 0 (11) (12) and (1 − x)−n = ∞ k=, \|x\| < 1. 92 3 Counting (13) Example Let us prove the binomial theorems. Proof of (11) Considering the product (1 + x)n = (1 + x)(1 + x)... (1 + x), we see that a term x k is obtained by taking x from any choice of k of the brackets and taking 1 from the rest. Because there are ( n k ) times in the expansion. Because this is true for any k, the result follows. k ) ways of choosing k brackets, the term x k occurs ( n Alternatively, you can prove (11) by induction on n, using the easily veriﬁed identity. Proof of (12) Multiplication veriﬁes that 1 − x m+1 = (1 − x)(). Hence, for \|x\| < 1, we let m → ∞ to ﬁnd that (12) is true for n = 1. Now
we can prove (12) for arbitrary n in a number of ways; one possibility is by induction on n. Alternatively, we observe that a term x k is obtained in the product (1 + x + x 2 + · · ·)n if we choose a set of k xs from the n brackets, where we may take any number of xs from each bracket. But by Theorem 3.3.4, this may be done in just ( n+k−1 ) ways. The negative binomial theorem (12) follows. k Here is a classic example. (14) Example: The Coupon Collector’s Problem Each packet of an injurious product is equally likely to contain any one of n different types of coupon. If you buy r packets, what is the probability p(n, r ) that you obtain at least one of each type of coupon? Solution later, but the exponential generating function offers a particularly elegant answer. This famous problem can be approached in many different ways, as we see First, recall from Theorem 3.2.3 that the probability of getting r = n 1 ti coupons of n distinct types, where ti are of type i, is n−r Mr (t1,..., tn) = r! nr t1!... tn!. Then, p(n, r ) is the sum of all such expressions in which ti ≥ 1 for all i (so we have at least one coupon of each type). But now expanding ( t=1(s/n)t /t!)n by the multinomial theorem shows that the coef- ∞ ﬁcient of sr in this expansion is just p(n, r )/r!. Hence, exp − 1 s n n = ∞ r =n srp(n, r )/r!, and we have obtained the exponential generating function of the p(n, r ). Suppose that you are a more demanding collector who requires two complete sets of coupons, let p2(n, r ) be the probability that r packets yield two complete sets of n coupons. 3.7 Techniques 93 Then exactly similar arguments show that exp 2n sr r! p2(n, r ), and so on for more sets. In conclusion, it is worth remarking that multivariate generating functions are often useful, although they will not appear much at this early stage. We give one example. (15) Multinomial
Theorem Recall the multinomial coefﬁcients deﬁned in Theorem 3.2.3, We have Mn(n1,..., nr ) = n! r ni! i=1 ; r i=1 ni = n. (x1 + x2 + · · · + xr )n = Mn(n1,..., nr )x n1 1 x n2 2... x nr r, where the sum is over all (n1,..., nr ) such that ni = n. Proof This is immediate from the deﬁnition of Mn as the number of ways of permuting n symbols of which ni are of type i. In this case xi is of type i, of course. Corollary Setting x1 = x2 = · · · = xr = 1 gives Mn(n1,..., nr ) = r n. 3.7 Techniques When evaluating any probability by counting, it is ﬁrst essential to be clear what the sample space is, and exactly which outcomes are in the event of interest. Neglect of this obvious but essential step has led many a student into lengthy but nugatory calculations. Second, it is even more important than usual to be ﬂexible and imaginative in your approach. As the following examples show, a simple reinterpretation or reformulation can turn a hard problem into a trivial one. The main mechanical methods are: (i) Using the theorems giving the numbers of ordered and unordered selections, recalling in particular that the number of ordered selections of r objects is the number of unordered selections multiplied by r! (ii) Use of the inclusion–exclusion principle. (iii) Setting up recurrence relations. (iv) Use of generating functions. Third, we remark that the implicit assumption of this chapter can be turned on its head. That is, we have developed counting techniques to solve probability problems, but the solution of probability problems can just as well be used to prove combinatorial identities. 94 (1) Example (2) 3 Counting Prove the following remarkable identity: n + k k n k=0 2−(n+k) = 1. Hint: Consider an ant walking on a square lattice. Solution An ant walks on the square lattice of points with nonnegative integer coordinates (i, j), i ≥ 0
, j ≥ 0. It starts at (0, 0). If it is at (x, y), it proceeds next either to (x + 1, y) or to (x, y + 1) with equal probability 1 2. Therefore, at some transition (certainly less than 2n + 1 transitions), it leaves the square (0 ≤ x ≤ n, 0 ≤ y ≤ n). It does so either with an x-step from (n, y) to (n + 1, y), where 0 ≤ y ≤ n or with a y-step from (x, n) to (x, n + 1), where 0 ≤ x ≤ n. Let these 2n + 2 events be Sy(0 ≤ y ≤ n) and Sx (0 ≤ x ≤ n), respectively. Then by symmetry n n P(Sx ) = P(Sy) = 1 2. x=0 y=0 However, Sy occurs if the ant has taken exactly y vertical steps before its (n + 1)th horizontal step. There are ( n+y+1 ) choices for the y vertical steps and each route to (n + 1, y) has probability 2−(n+y+1). Hence, y P(Sy) = Substituting (4) into (3) yields (2). n + y + 1 y 2−(n+y+1). (3) (4) Finally, we remind the reader of our remark in 1.6 that many counting problems are traditionally formulated as “urn models.” They are used in this context for two reasons; the ﬁrst is utility. Using urns (instead of some realistic model) allows the student to see the probabilistic features without prejudice from false intuition. The second reason is traditional; urns have been used since at least the seventeenth century for lotteries and voting. (Indeed the French expression “aller aux urnes” means “to vote”.) It was therefore natural for probabilists to use them in constructing theoretical models of chance events. A typical example arises when we look at the distribution of n accidents among days of the week, and seek the probability of at least one accident every day or the probability of accident-free days. The problem is equivalent to placing n balls in 7 urns, and seeking the number of ways in which the outcome of interest occurs. The following result is central here.
(5) Theorem Let (x1,..., xn) be a collection of integers, such that x1 + · · · + xn = r. The number of distinct ways of choosing them is: (a) (b if xi ≥ 0 for all i. if xi > 0 for all i. 3.8 Review and Checklist for Chapter 3 95 Note that (a) is equivalent to the number of ways of placing r indistinguishable balls in n numbered urns, and (b) adds the additional requirement that there must be at least one ball in each urn. Also, (a) is equivalent to Theorem 3.3.4. Proof Using the urn model formulation, we simply imagine the r balls placed in a line. Then allocate them to urns by placing r − 1 dividers in the n − 1 gaps between the balls. Each such choice supplies a distinct set (x1,..., xn) with xi > 0, and there are ( r −1 n−1 ) such choices, by (3.3.1). This proves (b). For (a), we add n balls to the group, allocate them in ( n+r −1 n−1 ) ways by (b), and then remove one ball from each urn. All such allocations form a one–one correspondence with an allocation with xi ≥ 0 and 1 xi = r. This proves (a). n 3.8 Review and Checklist for Chapter 3 Many chance experiments have equally likely outcomes so that probabilities may be evaluated by counting outcomes in events of interest. Counting can be useful in other contexts also. In this chapter, we introduce a number of techniques for counting, foremost among which is the use of generating functions. We also introduced inclusion and exclusion, among other methods for counting the outcomes in some event of interest, and many examples to show how these techniques are applied to probability problems. In particular, we have that: The number of possible sequences of length r using elements from a set of size n is nr (with repetition permitted). The number of permutations of length r using elements from a set of size n ≥ r is n(n − 1)... (n − r + 1) (with repetition not permitted). The number of choices (combinations) of r elements from a set of size n ≥ r is n r = n(n − 1)... (n −
r + 1) r (r − 1)... 2.1. The number of subsets of a set of size n is 2n. The number of permutations of n symbols, of r types, with ni of the ith type, (n1 + · · · + nr = n), is Mn(n1,..., nr ) = n! r ni! i=1. Given a collection of numbers (ai ; i ≥ 0) their generating function is g(x) = and their exponential generating function is h(x) = The binomial theorems assert that for positive integer n, ∞ i=0 ai x i i!. ∞ i=0 ai x i, (1 + x)n = n k=0 n k x k, 96 and 3 Counting ∞ (1 − x)−n = x k. n + k − 1 k k=0 r The multinomial theorem asserts that if integers, i=1 ni = n, where all quantities are positive (x1 + x2 + · · · + xr )n = x ni 1 · · · x nr r. n! r ni! The number of derangements of a set of size n is i=1 n! 1 − 1 1! + 1 2! − 1 3! + · · · + (−1)n 1 n!. Finally, we take this opportunity to state two useful approximations to n!: √ formula n! ∼ 2πnnne−n, which is Stirling’s √ to say that as n increases 2πnnne−n/n! → 1. Robbins improved formula* says that − 1 12n √ < exp 2πnnne−n/n! < exp. −1 12n + 1 Checklist of Terms for Chapter 3 3.2 permutation multinomial coefﬁcient binomial coefﬁcient 3.3 combinations 3.4 inclusion–exclusion derangement 3.5 recurrence relation Pascal’s triangle 3.6 generating function exponential generating function convolution binomial theorems coupon collecting multinomial theorem *Named for H.E. Robbins (1915–2001), famous for the book What Is Mathematics?, co-authored with R. Courant. Worked Wxamples and Exercises 97.9 Example: Railway Trains Suppose
you work in a marshalling yard. Given n similar wagons and r similar locomotives, how many distinct ways are there of making up r trains if empty trains are allowed, and all the wagons are used, and either: (a) The wagons and locomotives bear distinct numbers? or (b) The wagons are anonymous, but the locomotives are numbered? or (c) Wagons and locomotives are anonymous? What are the corresponding numbers if each train must contain at least m wagons? Solution (a) First consider how many ways the trains may be sent out of the yard. This is just the number of permutations of n + r objects with the one constraint that the ﬁrst object has to be a locomotive! The number is r (n + r − 1)!. As regards just making up trains, we do not require an ordering of the locomotives, so we divide by r! to get the answer r (n + r − 1)! r! = (n + r − 1)! (r − 1)! (b) If the wagons are not numbered, then we cannot distinguish the n! permutations of wagons, so the answer is (n + r − 1)! n!(r − 1)! = n + r − 1 r − 1, which could alternatively have been obtained by observing that we can make up the trains by choosing r places to intersperse the locomotives with the wagons, given that one locomotive must be at the front. (c) If neither locomotives nor wagons bear numbers, then the number of ways of making up r trains is the number of distinct partitions of n into at most r nonnegative integers, denoted by pr (n). For example, the number 5 has altogether seven integral partitions so p∞(5) = 7; of these, four are of three or fewer integers, so p3(5) = 4. There are no simple expressions for pr (n), but we use the following result, which we state without proof. Theorem of partitions of n in which no part is greater than k. The number of partitions of n with at most k parts is the same as the number Now, if we deﬁne the generating function gr (x) = ∞ n=1 pr (n)x n, 98 we can see that 3 Counting 1 + gr (x by ﬁrst expanding
the right-hand side (by the negative binomial theorem 3.6.12), and then observing that the term x n arises just as often as n can be written as the sum of positive integers, no one of which is greater than r. (b) If each train must contain at least m wagons then we require mr ≥ n. In this case, ﬁrst suppose the wagons do not have numbers. We attach m to each train in essentially one way, leaving n − mr to be distributed in any way. Using (b), we get the answer ( n−r m+r −1 ) ways. (a) If the wagons have numbers then any of the n! permutations is distinct, giving r −1 n!( n−r m+r −1 r −1 ) ways. (c) If neither the locomotives nor wagons have numbers and each train must contain m wagons at least, then we require the number of partitions of m into at most r integers, all of which are not less than m. This is the same as the number of partitions of n − mr into at most r integers, that is, pr (n − mr ), which is the coefﬁcient of x n−mr in gr (x). Remark The results about partitions of n are not used elsewhere. They are included as an example of the power of generating functions, which are used extensively throughout probability and related subjects. How many ways of making up the r trains are there if you do not have to use all the (1) (2) (3) Exercise wagons? Exercise ways are there on the more realistic assumption that boys are indistinguishable? Exercise no section has less than 3 Tb? (Use only integral multiples of 1 Tb.) The hard disk on your PC stores 10 Tb. In how many ways can you divide it up so that In how many ways can ﬁve oranges be distributed among seven boys? How many 3.10 Example: Genoese Lottery Suppose that n cards in an urn each bear one of n consecutive integers; all the cards have a different number. Five cards are randomly selected from the urn without replacement. What is the probability that their numbers can be arranged as a run of three consecutive numbers and a nonadjacent run of two consecutive numbers? (For example, 34578 or 23789.) There are ( n Solution 5 ) ways of choosing a
set of ﬁve numbers, and these are assumed to be equally likely. We must count the number of sets providing the two required runs. If the run of three is ﬁrst and it starts at k, then the run of two may start at n − k − 4 places (1 ≤ k ≤ n − 5). Hence, the runs may occur in this order in n−5 k=1 n − k − 4 = n−5 j=1 j = 1 2 (n − 5)(n − 4) ways. Worked Examples and Exercises 99 The run of two is ﬁrst in another 1 5 ). (n − 5)(n − 4)/( n 2 (n − 5)(n − 4) cases, so the required probability is Remark This lottery is called Genoese because it was introduced by a member of the senate of Genoa named Benedetto Gentile in 1623. It was used to raise money in 1757 by the treasury of Louis XV of France on the advice of Casanova, who had a licence to sell tickets at 6% commission. The treasury also consulted d’Alembert; this was shortly after he had asserted that the chance of HH or TT in two spins of a fair coin is 2 3. When Frederick II of Prussia used this lottery to raise money for his treasury, he asked Euler to calculate the odds again. (6) As above, n cards bear n consecutive integers. Find the probability that: Exercise (a) If three cards are drawn, their numbers can form a run of length three. (b) If three cards are drawn, their numbers include a run of length exactly two. (c) If ﬁve cards are drawn, their numbers include two runs of length exactly two. 3.11 Example: Ringing Birds A wood contains n birds, none of which is ringed. Each day one bird is caught, ringed (if it does not already have a ring), and released. Each bird is equally likely to be caught on each day. This procedure is repeated on r successive days, r ≥ n. Show that the probability of ringing all the birds is p(r, n) where p(r, n) = n (−) j j=0 n j r 1 − j n Solution I The total number of outcomes (birds being distinguishable), is nr. Let N (r, n) be the number of outcomes in which every
bird is caught at least once. Because birds are distinct, some bird may be called the ﬁrst bird. Let Nk(r, n) be the number of outcomes in which the ﬁrst bird is caught exactly k times, and all the birds are caught at least once. There are ( r k ) ways of choosing the k days on which the ﬁrst bird is caught, and N (r − k, n − 1) ways in which the remaining n − 1 birds may be caught on the other r − k occasions. Hence, by Principle 3.1.2, Nk(r, n) = r k N (r − k, n − 1), and, by Principle 3.1.1, Of course, N (r, n) = r k=1 Nk(r, n). N (r, 1) = 1. (1) (2) (3) 100 3 Counting Substituting (1) into (2) yields a rather complicated recurrence relation. It can be solved by using the exponential generating function: (4) (5) By (3), we have Gn(s) = ∞ r =1 sr r! N (r, n); n > 1. G1(s) = es − 1. Now, multiply (2) by sr /r! and sum over r [using (1)], to give ∞ r Gn(s) = sr −ksk (r − k)!k! = r =1 ∞ Gn−1(s) k=1 sk k! = (es − 1)Gn−1(s) = (es − 1)n k=1 n = ens (−e−s) j n j = j=0 n ∞ r =0 j=0 Hence, (n − j)r r! (−1) j N (r − k, n − 1) by the convolution theorem 3.6.6, on iterating and using (5), by the binomial theorem on expanding e(n− j)s. sr n j N (r, n) = n j=0 (−1) j n j (n − j)r, using (4) and Theorem 3.6.4. Dividing by nr yields the required result. Solution II We may alternatively use the principle of inclusion and exclusion. In the terminology of Section 3.
4, “objects” are outcomes and an “object with the kth property” is an outcome in which k birds remain unringed. Now, the number of ways of selecting k, and the number of outcomes with k given birds unringed birds to remain unringed is is (n − k)r. Hence, by (3.4.1) and (3.4.3), n k N (r, n) = nr − n 1 (n − 1)r + · · · + (−1)n−1 n n − 1 (n − (n − 1))r, which yields the required result. (6) (7) (8) (9) Do the example again using (1.4.8). Exercise Exercise What is the probability that no bird is caught more than twice? Exercise What is the probability that every bird is caught at least twice? Exercise As r → ∞, show that the probability of ringing all the birds converges to one. Worked Wxamples and Exercises 101 3.12 Example: Lottery In each draw of a lottery, an integer is picked independently at random from the ﬁrst n integers 1, 2,..., n. What is the probability that in a sample of r successive draws the numbers are drawn in a nondecreasing sequence? There are nr possible sequences in all. We may take any nondecreasing Solution I sample sequence of r numbers, together with the ﬁrst n integers, and arrange these n + r integers as a nondecreasing sequence. Now, place a bar between adjacent different numbers in this sequence, and place a star between adjacent equal numbers. The number to the right of each star is a member of the original sample sequence; there are r stars and n + r − 1 places to choose to put them. Furthermore, each such choice corresponds to just one possible sample sequence. By Theorem 3.3.1, there are therefore ( n+r −1 ) nondecreasing sample sequences, so the required probability is ( n+r −1 )n−r. r r Take any sample sequence s = {s1, s2,..., sr } and add j − 1 to s j to Solution II get a unique new sequence t = {s1, s2 + 1, s3 + 2,..., sr + r − 1
}. The sequence t is a selection of r numbers without replacement from {1, 2, 3}, and subtracting j − 1 from t j yields a unique sequence s, which is selected with replacement from {1, 2,..., n}. ) ways of choosing the sequence t, this is also the number of Because there are ( n+r −1 r ways of choosing s. Remark Observe that this solution includes a proof of Theorem 3.3.4. (1) (2) (3) Exercise What is the probability that the r numbers are drawn in a strictly increasing sequence? Exercise Show that the probability that no two drawn numbers are consecutive (i.e., differ by unity) is r!( n−r +1 Exercise been drawn at least once. What is the probability that m draws are required to achieve this? r Integers are picked at random from {1, 2,..., n} until every integer k, 1 ≤ k ≤ n, has )n−r. 3.13 Example: The M´enages Problem Suppose that n married couples are seated randomly around a circular table so that men and women alternate. (a) Find the number of such seatings in which, for a given set of k couples, the husband and wife are sitting in adjacent seats (some other couples may be adjacent also). (b) Hence, deduce the probability that no couple is seated next to each other. Solution There are two ways of choosing in which seats the women will sit, and n! ways in which they can be arranged in them. The men can be seated in the remaining seats (alternating with the women) in n! ways. Thus, by Principle 3.1.2, there are 2(n!)2 possible seating arrangements, which are equally likely by hypothesis. (a) To count the number of arrangements in which a given set of k couples are adjacent, it is ﬁrst necessary to count the number of ways of choosing k pairs of adjacent seats for them to occupy. First, consider 2n − k chairs in a row. Now, choose k of these [which we may do in ( 2n−k ) ways], and place an extra chair by each of the k chosen chairs. This provides k 102 3 Counting a choice of k distinct nonoverlapping pairs of adjacent chairs in a row of 2n. Conversely, for any choice of k disjoint pairs of adjacent chairs from 2
n, we may discard a chair from each pair to give a choice of k from 2n − k chairs. This one–one correspondence shows that the number of ways of choosing k disjoint pairs of chairs from 2n is just ( 2n−k k Now, number the seats round the circular table from 1 to 2n. By Principle 3.1.1, the number of choices of k disjoint pairs of adjacent seats is the sum of the choices in which (1, 2n) is such a pair [which number ( 2n−2−(k−1) ) by the above result], and the choices in which (1, 2n) is not a pair [which number( 2n−k )]. k k−1 ). Hence, the disjoint pairs of seats can be chosen in 2n − k − 1 k − 1 2n − k k Mk = + = 2n 2n − k 2n − k k (1) (2) (3) (4) (5) ways. The k couples to occupy these seats can be chosen in ( n k ) ways, the women’s seats chosen in two ways, the k chosen couples arranged in k! ways, and the remaining men and women arranged in ((n − k)!)2 ways. Hence, using Principle 3.1.2, the number of seatings for which the k couples are in adjacent seats is Sk = 2n 2n − k. 2n − k k.2k!((n − k)!)2 = 4n(2n − k − 1)! ((n − k)!)2. (2n − 2k)! (b) Recalling the principle of inclusion and exclusion, we may interpret “an object with the kth property” as “a seating with the kth couple adjacent.” Hence, by using (3.4.3) and (3.4.2), the probability that no couple is adjacent is n k=0 (−)k n k Sk 2(n!)2 = n 0 (−)k 2n 2n − k 2n − k k (n − k)! n!. Exercise Why does it not matter whether we assume the table has a principal seat (head) or not? Suppose that n pairs of twins are seated randomly at a round table. What is the Exercise probability that no pair of twins sit next to each other? What is the limit
of this probability as n → ∞? Exercise What is the limit of (2) as n → ∞? The problem was ﬁrst discussed by E. Lucas in 1891 (m´enage is French Remark for household). This method of solution is due to K. Bogart and P. Doyle, American Mathematical Monthly, 1986. Show that 3.14 Example: Identity r j= Solution I Suppose we place r balls randomly in m urns. This is essentially the same as arranging r balls and m − 1 stars in a row, where the balls between successive stars are placed in successive urns. The positions for the stars may be chosen in ( m+r −1 r −1 ) Worked Examples and Exercises 103 ways. Now, in how many of these arrangements do we ﬁnd b balls in the ﬁrst urn? The answer is, in just as many ways as the other r − b balls can be put in the other m − 1 urns, that is in ( m+r −b−2 ) ways. Hence, by Principle 3.1.1, m− Setting m − 1 = k, r − b = j gives the required result (1). =0 =. Solution II we have Multiply each side of (1) by x r and sum from r = 0. By Theorem 3.6.8, 1 1 − x ∞ r =0 x r, r + k k and by (3.6.10), both sides equal (1 − x)−(k+1). The identity (1) follows by Theorem 3.6.4. Solution III Obviously, 1 (1 − x)k Equating the coefﬁcient of x r on each side of this identity yields (1). 1 (1 − x)k−1 1 1 − x =. Remark Up to a point, these three methods are really the same, but Solution III does illustrate how useful generating functions can be if you happen to hit on an appropriate identity when you need it. Show that for any number x and integer j, ( x j−+1 j ), and hence prove (1) (2) (3) (4) Exercise by a fourth method. Exercise Show that Exercise Show that n n j=0(−1) j ( x j=0( x− j j ) = (−1)n(
x−1 n ). r +1 ) − ( x−n r +1 ). ) = ( x+1 r 3.15 Example: Runs A fair coin is tossed repeatedly. A run of heads is all the heads shown between one tail and the next; if the ﬁrst toss is a head, there is an opening run up to the ﬁrst tail; and likewise, there may be a closing run after the last tail. A zero run is no run. (a) If the coin is tossed n times, show that the most likely number of runs (of heads, including the opening and closing runs) is [ n 4 ] + 1 when n is large. (b) Also, prove the identity m i= =. 2k + m m (a) There are 2n possible outcomes. We can choose any outcome with k head Solution runs in the following way. Visualize the n coins in a row. They provide n + 1 intervening spaces (including that before the ﬁrst and that after the last coin). Now, place 2k stars 104 3 Counting in the spaces, and let the coins between the (2r + 1)th and (2r + 2)th stars be heads (r = 0, 1,..., k − 1); the other coins are tails. There are ( n+1 2k ) ways to do this, so fk = P(k head runs) = 2−n Now considering fk/ fk+1, we ﬁnd n + 1 2k. fk fk+1 ≷ 1 according as k ≷ n2 + n − 2 4n. Therefore, the most likely number of runs is the integer next after nn +n−2 for large n. 4n, which is n 4 + 1 (b) Suppose that the n tosses result in a heads and b tails. First, we divide the a heads into k nonempty groups that form the runs. Imagining the heads in a row, there are a − 1 places to put k − 1 dividing lines, so the number of ways of getting the k runs of heads is ( a−1 k−1 ). The head runs alternate with runs of tails, so the b tails are divided into k + 1 groups of which the ﬁrst and last may be empty (providing an opening and/or closing run). If we add an auxiliary tail to
each, the same argument as used for head runs shows that the number of ways of arranging the tails is. Hence, the number of ways of getting k head runs in n tosses showing a heads is rk = The total number of ways of getting k head runs is therefore n−k+1 n−2k+1 a=k i=0 m rk = = = = =0 n + 1 2k 2k + m m by (a) as required. ; m = n − 2k + 1, Note: In the exercises, n tosses yield a heads and n − a tails. (1) (2) (3) (4) Exercise What is the probability that the ﬁrst run is a head run of length k? Exercise What is the probability that the last run is a head run of length k? Exercise What is the probability that the second run is a head run of length k? Exercise What is the probability that the ﬁrst run is of length k? Worked Examples and Exercises 105 3.16 Example: Fish A lake contains b bream and c chub. Any ﬁsh of each species is equally likely to take a hook. (a) If you catch n ﬁsh and don’t throw any back, what is the probability that you have caught x bream? (b) You then return all the ﬁsh to the lake (alive) and start ﬁshing again. You now catch m ﬁsh. What is the probability that exactly k bream are caught twice? Solution (a) The question clearly intends us to assume that all possible selections of n ﬁsh are equally likely to occur. Then, the number of ways of selecting n ﬁsh without repetition is ( b+c n ). The number of ways of catching x bream (and hence also n − x chub) is ( b n−x ), so the required probability is x )( c px = (1, max{0, n − c} ≤ x ≤ min{b, n}. (b) We assume that ﬁsh do not learn from experience, so that all ( b+c m ) selections of m ﬁsh are still equally likely. If x bream were in the ﬁrst catch, where k ≤ x ≤ b, then the number of ways of selecting m ﬁ
sh, of which k are bream being caught for the second time, is ( x m−k ). Therefore, the required conditional probability of catching k bream twice, given a ﬁrst catch of x bream, is k )( b+c−x pk\| by the same argument as in (a). Hence, by Theorem 2.1.3, the required unconditional probability is pk = x pk\|x px = b x=2) Exercise What is the probability of catching x bream if you catch n ﬁsh and (a) You throw bream back but not chub? (b) You throw both species back? Part (a) considers sampling with partial replacement, and part (b) considers sampling with replacement. Discuss the difference in your answers. (3) (4) Exercise Show that (1 + x)c = (1 + x)b+c.] Exercise the limit of the probabilities in (1) and (2) as b and c → ∞. Discuss. n ). [You may want k )( c ( b n−k ) = ( b+c min{b,n} k=0 Suppose that as b and c approach ∞, b/(b + c) → p and c/(b + c) → 1 − p. Find to recall that (1 + x)b 106 3 Counting 3.17 Example: Colouring Let K (b, c) be the number of different ways in which b indistinguishable balls may be coloured with c different colours. Show that K (b, c) = K (b − 1, c) + K (b, c − 1) and deduce that ∞ b=0 x b K (b, c) = (1 − x)−c. Use this to show K (b, c) = ( b+c−1 c−1 ). Solution Pick any colour and call it grurple. The number of colourings is the number of ways of colouring the balls which do not colour any grurple, plus the number of ways of colouring which do use grurple. Hence, K (b, c) = K (b − 1, c) + K (b, c − 1). Also, K (1, c) = c, and K (0, c) = 1 because there are c colours for one ball and only one way of colouring no balls
. Now let gc(x) = ∞ b=0 x b K (b, c). Multiply (1) by x b and sum from b = 0 to get gc(x) = xgc(x) + gc−1(x). Now using Theorem 3.5.1, we solve (2) to ﬁnd gc(x) = (1 − x)−c. Furthermore, we may write (1 − x)−c = (1 + x + x 2 + · · ·)(1 + x + x 2 + · · ·) · · · (1 + x + x 2 + · · ·) Where the right side is the product of c brackets. We get K (b, c) by picking a term from each bracket, and we can say that picking x k from the ith bracket is like picking k objects of type i. The coefﬁcient K (b, c) of x b is thus obtained by choosing b objects from c different types of objects with repetition. By Theorem 3.3.4, we have c + b − 1 b Let C(n, k) be the number of ways of choosing a set of k objects from n distinct K (b, c) =. n k=0 C(n, k)x k = (1 + x)n. Exercise objects. Show that Exercise 20 have? [For example, x1 = 0, x2 = 4, x3 = 16.] Exercise and x3? [For example, x1 = 5, x2 = 6, x3 = 9.] Exercise Show that K (b, c) = ( b+c−1 How many nonnegative integer valued solutions for x1, x2, and x3 does x1 + x2 + x3 = How many positive integer valued solutions does x1 + x2 + x3 = 20 have for x1, x2, c−1 ) by a method different from that in the above solution. (1) (2) (3) (4) (5) (6) (7) (1) (2) (3) Worked Examples and Exercises 107 3.18 Example: Matching (Rencontres) Suppose n different letters are typed with their corresponding envelopes. If the letters are placed at random in the envelopes, show that the probability that exactly r letters match their envelop
es is p(n, r ) = 1 r! n−r k=0 (−1)k k!. [This problem ﬁrst surfaced in France during the eighteenth century as a question about coincidences when turning over cards from packs (a kind of French snap).] Using (3.4.4) gives (1) easily; we display a different method for the sake of variety. Solution We can suppose that the order of the envelopes is ﬁxed. Let the number of permutations of the letters in which r out of the n letters match their envelopes be a(n, r ). Then, p(n, r ) = a(n, r ) n!. Suppose we have another letter sealed in its correct envelope. Consider the number A of arrangements that there are of this letter and n letters of which r match their envelopes. We can get this number A in two ways. Either: (i) We place the sealed letter in any one of n + 1 positions among the n letters to get (n + 1)a(n, r ) arrangements: or: (ii) We permute n + 1 unsealed letters of which r + 1 match and then choose one of the r + 1 matching letters to seal, giving (r + 1)a(n + 1, r + 1) arrangements. The two numbers must both be equal to A, (n + 1)a(n, r ) and hence, dividing by (n + 1)! we obtain so (r + 1)a(n + 1, r + 1) = (r + 1) p(n + 1, r + 1) = p(n, r ) with p(n, n) = 1/n!. This is a rather interesting recurrence relation, which is solved by standard methods. First iterating (2), we have Now deﬁne the probability generating function p(n, r ) = 1 r! p(n − r, 0). gn(x) = n r =0 x r p(n, r ); n ≥ 1. Multiplying (2) by x r and summing over r gives gn(x) = x n n! + n r =1 p(r, 0) x n−r (n − r )!. 108 3 Counting The sum on the right is a convolution as in Theorem 3.6.6. so multiplying by
yn and summing over n, by Theorem 3.6.6 ∞ yn gn(x) = ex y − 1 + ex y 1 Setting x = 1 and using Theorem 3.6.10 gives ∞ 1 yn p(n, 0). y 1 − y = ey − 1 + ey ∞ 1 yn p(n, 0) so that ∞ 1 yn p(n, 0) = e−y 1 − y − 1 = n ∞ yn 1 k=0 (−1)k k!. Hence, by (4) and (3), we get (1). Exercise occur. Show that for large n, it is approximately Find the probability that exactly r + s matches occur given that at least r matches 1 (r + s)! ∞ r. 1 k! If a cheque is written for each addressee and these are also placed at random in the Show that the probability that the ﬁrst letter matches its envelope, given that there are Exercise exactly r such matches, is r n. Exercise envelopes, ﬁnd: (a) The probability that exactly r envelopes contain the correct letter and cheque. (b) The probability that no envelope contains the correct letter and cheque. (c) The probability that every letter contains the wrong letter and the wrong cheque. Exercise Exercise Find the limit of each probability in 7(a), 7(b), and 7(c) as n → ∞. Use 3.4.4 to prove (1) directly. P R O B L E M S You have two pairs of red socks, three pairs of mauve socks, and four pairs with a rather attractive rainbow motif. If you pick two socks at random, what is the probability that they match? A keen student has a algebra books. b books on boundary layers, and c calculus books. If he places them on one shelf at random, what is the probability that: (a) Books on the same subject are not separated? (b) Books on the same subject are in the usual alphabetical order, but not necessarily adjacent? (c) Books on the same subject are adjacent and in alphabetical order? A pack of cards is well shufﬂed and one hand of 13 cards is dealt to each of four players. Find the probability that: (a) Each player has an ace. (b) At least one player has a complete
suit. (c) My hand is void in at least one suit. (d) Some player has all the aces. What is the most likely distribution among suits in the dealer’s hand? (4) (5) (6) (7) (8) (9 10 11 12 13 Problems 109 (b) Two pairs? (d) A ﬂush? You are dealt ﬁve cards in your hand at poker. What is the probability that you hold: Poker (a) One pair? (c) A straight? (e) A full house? Birthdays Assume people are independently equally likely to be born on any day of the year. Given a randomly selected group of r people, of whom it is known that none were born on February 29th, show that the probability that at least two of them have their birthdays either on consecutive days or on the same day is pr where pr = 1 − (365 − r − 1)! (365 − 2r )! 365−r +1, (2r < 365). (b) Four of a kind? Each die bears the symbols A, K, Q, J, 10, 9. If you roll ﬁve such dice, what is the Deduce that if r = 13, then the probability of at least two such contiguous birthdays is approximately 2, while if r = 23 then the probability of at least two such contiguous birthdays is approximately 1 9 10. You pick an integer at random between zero and 105 inclusive. What is the probability that its digits are all different? One hundred light bulbs are numbered consecutively from 1 to 100, and are off. They are wired to 100 switches in such a way that the nth switch changes the state (off to on, or on to off) of all the bulbs numbered kn; k ≥ 1. If the switches are all thrown successively, how many light bulbs are on? What is the answer if you start with M light bulbs and M switches? (a) Show that the product of any r consecutive integers is divisible by r!. (b) Show that (k!)! is divisible by (k!)(k−1)!. Poker Dice probability that your set of ﬁve symbols includes: (a) Four aces? Eight rooks are placed randomly on a chess board (with at most one on each square). What is the probability that: (a) They are all in a straight line?

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