2.4: Conditional Probability
The purpose of this section is to study how probabilities are updated in light of new information, clearly an absolutely essential topic. If you are a new student of probability, you may want to skip the technical details.
Definitions and Interpretations
The Basic Definition
As usual, we start with a random experiment modeled by a probability space \((S, \mathscr S, \P)\). Thus, \( S \) is the set of outcomes, \( \mathscr S \) the collection of events, and \( \P \) the probability measure on the sample space \( (S, \mathscr S) \). Suppose now that we know that an event \(B\) has occurred. In general, this information should clearly alter the probabilities that we assign to other events. In particular, if \(A\) is another event then \(A\) occurs if and only if \(A\) and \(B\) occur; effectively, the sample space has been reduced to \(B\). Thus, the probability of \(A\), given that we know \(B\) has occurred, should be proportional to \(\P(A \cap B)\).
However, conditional probability, given that \(B\) has occurred, should still be a probability measure, that is, it must satisfy the axioms of probability. This forces the proportionality constant to be \(1 \big/ \P(B)\). Thus, we are led inexorably to the following definition:
Let \(A\) and \(B\) be events with \(\P(B) \gt 0\). The conditional probability of \(A\) given \(B\) is defined to be \[\P(A \mid B) = \frac{\P(A \cap B)}{\P(B)}\]
The Law of Large Numbers
The definition above was based on the axiomatic definition of probability. Let's explore the idea of conditional probability from the less formal and more intuitive notion of relative frequency (the law of large numbers). Thus, suppose that we run the experiment repeatedly. For \( n \in \N_+ \) and an event \(E\), let \(N_n(E)\) denote the number of times \(E\) occurs (the frequency of \( E \)) in the first \(n\) runs. Note that \(N_n(E)\) is a random variable in the compound experiment that consists of replicating the original experiment. In particular, its value is unknown until we actually run the experiment \( n \) times.
If \(N_n(B)\) is large, the conditional probability that \(A\) has occurred, given that \(B\) has occurred, should be close to the conditional relative frequency of \(A\) given \(B\), namely the relative frequency of \(A\) for the runs on which \(B\) occurred: \(N_n(A \cap B) / N_n(B)\). But note that \[\frac{N_n(A \cap B)}{N_n(B)} = \frac{N_n(A \cap B) / n}{N_n(B) / n}\] The numerator and denominator of the main fraction on the right are the relative frequencies of \( A \cap B \) and \( B \), respectively. So by the law of large numbers again, \( N_n(A \cap B) / n \to \P(A \cap B) \) as \( n \to \infty \) and \( N_n(B) \to \P(B) \) as \( n \to \infty \). Hence \[\frac{N_n(A \cap B)}{N_n(B)} \to \frac{\P(A \cap B)}{\P(B)} \text{ as } n \to \infty\] and we are led again to the definition above.
In some cases, conditional probabilities can be computed directly , by effectively reducing the sample space to the given event. In other cases, the formula in the mathematical definition is better. In some cases, conditional probabilities are known from modeling assumptions, and then are used to compute other probabilities. We will see examples of all of these situations in the computational exercises below.
It's very important that you not confuse \(\P(A \mid B)\), the probability of \(A\) given \(B\), with \(\P(B \mid A)\), the probability of \(B\) given \(A\). Making that mistake is known as the fallacy of the transposed conditional . (How embarrassing!)
Conditional Distributions
Suppose that \(X\) is a random variable for the experiment with values in \(T\). Mathematically, \( X \) is a function from \( S \) into \( T \), and \( \{X \in A\} \) denotes the event \( \{s \in S: X(s) \in A\} \) for \(A \subseteq T \). Intuitively, \( X \) is a variable of interest in the experiment, and every meaningful statement about \( X \) defines an event. Recall that the probability distribution of \(X\) is the probability measure on \(T\) given by \[A \mapsto \P(X \in A), \quad A \subseteq T\] This has a natural extension to a conditional distribution, given an event.
If \(B\) is an event with \( \P(B) \gt 0 \), then the conditional distribution of \(X\) given \(B\) is the probability measure on \(T\) given by \[A \mapsto \P(X \in A \mid B), \quad A \subseteq T\]
Details
Recall that \( T \) will come with a \( \sigma \)-algebra of admissible subsets so that \( (T, \mathscr T) \) is a measurable space, just like the sample space \( (S, \mathscr S) \). Random variable \( X \) is required to be measurable as a function from \( S \) into \( T \). This ensures that \( \{X \in A\} \) is a valid event for each \( A \in \mathscr T \), so that the definition makes sense.
Basic Theory
Preliminary Results
Our first result is of fundamental importance, and indeed was a crucial part of the argument for the definition of conditional probability.
Suppose again that \( B \) is an event with \( \P(B) \gt 0 \). Then \(A \mapsto \P(A \mid B)\) is a probability measure on \( S \).
Proof
Clearly \( \P(A \mid B) \ge 0 \) for every event \( A \), and \( \P(S \mid B) = 1 \). Thus, suppose that \( \{A_i: i \in I\} \) is a countable collection of pairwise disjoint events. Then \[ \P\left(\bigcup_{i \in I} A_i \biggm| B\right) = \frac{1}{\P(B)} \P\left[\left(\bigcup_{i \in I} A_i\right) \cap B\right] = \frac{1}{\P(B)} \P\left(\bigcup_{i \in I} (A_i \cap B)\right) \] But the collection of events \( \{A_i \cap B: i \in I\} \) is also pairwise disjoint, so \[ \P\left(\bigcup_{i \in I} A_i \biggm| B\right) = \frac{1}{\P(B)} \sum_{i \in I} \P(A_i \cap B) = \sum_{i \in I} \frac{\P(A_i \cap B)}{\P(B)} = \sum_{i \in I} \P(A_i \mid B) \]
It's hard to overstate the importance of the last result because this theorem means that any result that holds for probability measures in general holds for conditional probability, as long as the conditioning event remains fixed. In particular the basic probability rules in the section on Probability Measure have analogs for conditional probability. To give two examples, \begin{align} \P\left(A^c \mid B\right) & = 1 - \P(A \mid B) \\ \P\left(A_1 \cup A_2 \mid B\right) & = \P\left(A_1 \mid B\right) + \P\left(A_2 \mid B\right) - \P\left(A_1 \cap A_2 \mid B\right) \end{align} By the same token, it follows that the conditional distribution of a random variable with values in \( T \), given in above , really does define a probability distribution on \( T \). No further proof is necessary. Our next results are very simple.
Suppose that \(A\) and \(B\) are events with \( \P(B) \gt 0 \).
- If \(B \subseteq A\) then \(\P(A \mid B) = 1\).
- If \(A \subseteq B\) then \(\P(A \mid B) = \P(A) / \P(B)\).
- If \(A\) and \(B\) are disjoint then \(\P(A \mid B) = 0\).
Proof
These results follow directly from the definition of conditional probability. In part (a), note that \( A \cap B = B \). In part (b) note that \( A \cap B = A \). In part (c) note that \( A \cap B = \emptyset \).
Parts (a) and (c) certainly make sense. Suppose that we know that event \( B \) has occurred. If \( B \subseteq A \) then \( A \) becomes a certain event. If \( A \cap B = \emptyset \) then \( A \) becomes an impossible event. A conditional probability can be computed relative to a probability measure that is itself a conditional probability measure. The following result is a consistency condition.
Suppose that \(A\), \(B\), and \(C\) are events with \( \P(B \cap C) \gt 0 \). The probability of \(A\) given \(B\), relative to \(\P(\cdot \mid C)\), is the same as the probability of \(A\) given \(B\) and \(C\) (relative to \(\P\)). That is, \[\frac{\P(A \cap B \mid C)}{\P(B \mid C)} = \P(A \mid B \cap C)\]
Proof
From the definition, \[\frac{\P(A \cap B \mid C)}{\P(B \mid C)} = \frac{\P(A \cap B \cap C) \big/ \P(C)}{\P(B \cap C) \big/ \P(C)} = \frac{\P(A \cap B \cap C)}{\P(B \cap C)} = \P(A \mid B \cap C)\]
Correlation
Our next discussion concerns an important concept that deals with how two events are related, in a probabilistic sense.
Suppose that \(A\) and \(B\) are events with \( \P(A) \gt 0 \) and \( \P(B) \gt 0 \).
- \(\P(A \mid B) \gt \P(A)\) if and only if \(\P(B \mid A) \gt \P(B)\) if and only if \(\P(A \cap B) \gt \P(A) \P(B)\). In this case, \( A \) and \( B \) are positively correlated .
- \(\P(A \mid B) \lt \P(A)\) if and only if \(\P(B \mid A) \lt \P(B)\) if and only if \(\P(A \cap B) \lt \P(A) \P(B)\). In this case, \( A \) and \( B \) are negatively correlated .
- \(\P(A \mid B) = \P(A)\) if and only if \(\P(B \mid A) = \P(B)\) if and only if \(\P(A \cap B) = \P(A) \P(B)\). In this case, \( A \) and \( B \) are uncorrelated or independent .
Proof
These properties following directly from the definition of conditional probability and simple algebra. Recall that multiplying or dividing an inequality by a positive number preserves the inequality.
Intuitively, if \( A \) and \( B \) are positively correlated, then the occurrence of either event means that the other event is more likely. If \(A\) and \(B\) are negatively correlated, then the occurrence of either event means that the other event is less likely. If \(A\) and \(B\) are uncorrelated, then the occurrence of either event does not change the probability of the other event. Independence is a fundamental concept that can be extended to more than two events and to random variables; these generalizations are studied in the next section on Independence. A much more general version of correlation, for random variables, is explored in the section on Covariance and Correlation in the chapter on Expected Value.
Suppose that \( A \) and \( B \) are events. Note from (4) that if \( A \subseteq B \) or \( B \subseteq A \) then \( A \) and \( B \) are positively correlated. If \( A \) and \( B \) are disjoint then \( A \) and \( B \) are negatively correlated.
Suppose that \(A\) and \(B\) are events in a random experiment.
- \(A\) and \(B\) have the same correlation (positive, negative, or zero) as \(A^c\) and \(B^c\).
- \(A\) and \(B\) have the opposite correlation as \(A\) and \(B^c\) (that is, positive-negative, negative-positive, or 0-0).
Proof
- Using DeMorgan's law and the complement law. \[ \P(A^c \cap B^c) - \P(A^c) \P(B^c) = \P\left[(A \cup B)^c\right] - \P(A^c) \P(B^c) = \left[1 - \P(A \cup B)\right] - \left[1 - \P(A)\right]\left[1 - \P(B)\right] \] Using the inclusion-exclusion law and algebra, \[ \P(A^c \cap B^c) - \P(A^c) \P(B^c) = \P(A \cap B) - \P(A) \P(B) \]
- Using the difference rule and the complement law: \[ \P(A \cap B^c) - \P(A) \P(B^c) = \P(A) - \P(A \cap B) - \P(A) \left[1 - \P(B)\right] = -\left[\P(A \cap B) - \P(A) \P(B)\right]\]
The Multiplication Rule
Sometimes conditional probabilities are known and can be used to find the probabilities of other events. Note first that if \( A \) and \( B \) are events with positive probability, then by the very definition of conditional probability, \[ \P(A \cap B) = \P(A) \P(B \mid A) = \P(B) \P(A \mid B) \] The following generalization is known as the multiplication rule of probability. As usual, we assume that any event conditioned on has positive probability.
Suppose that \((A_1, A_2, \ldots, A_n)\) is a sequence of events. Then \[\P\left(A_1 \cap A_2 \cap \cdots \cap A_n\right) = \P\left(A_1\right) \P\left(A_2 \mid A_1\right) P\left(A_3 \mid A_1 \cap A_2\right) \cdots \P\left(A_n \mid A_1 \cap A_2 \cap \cdots \cap A_{n-1}\right)\]
Proof
The product on the right a collapsing product in which only the probability of the intersection of all \(n\) events survives. The product of the first two factors is \( \P\left(A_1 \cap A_2\right) \), and hence the product of the first three factors is \( \P\left(A_1 \cap A_2 \cap A_3\right) \), and so forth. The proof can be made more rigorous by induction on \( n \).
The multiplication rule is particularly useful for experiments that consist of dependent stages, where \(A_i\) is an event in stage \(i\). Compare the multiplication rule of probability with the multiplication rule of combinatorics.
As with any other result, the multiplication rule can be applied to a conditional probability measure. In the context above, if \(E\) is another event, then \[\P\left(A_1 \cap A_2 \cap \cdots \cap A_n \mid E\right) = \P\left(A_1 \mid E\right) \P\left(A_2 \mid A_1 \cap E\right) P\left(A_3 \mid A_1 \cap A_2 \cap E\right) \cdots \P\left(A_n \mid A_1 \cap A_2 \cap \cdots \cap A_{n-1} \cap E\right)\]
Conditioning and Bayes' Theorem
Suppose that \(\mathscr{A} = \{A_i: i \in I\}\) is a countable collection of events that partition the sample space \(S\), and that \( \P(A_i) \gt 0 \) for each \( i \in I \).
The following theorem is known as the law of total probability .
If \( B \) is an event then \[\P(B) = \sum_{i \in I} \P(A_i) \P(B \mid A_i)\]
Proof
Recall that \(\{A_i \cap B: i \in I\}\) is a partition of \(B\). Hence \[ \P(B) = \sum_{i \in I} \P(A_i \cap B) = \sum_{i \in I} \P(A_i) \P(B \mid A_i) \]
The following theorem is known as Bayes' Theorem , named after Thomas Bayes:
If \( B \) is an event then \[\P(A_j \mid B) = \frac{\P(A_j) \P(B \mid A_j)}{\sum_{i \in I}\P(A_i) \P(B \mid A_i)}, \quad j \in I\]
Proof
Again the numerator is \(\P(A_j \cap B)\) while the denominator is \(\P(B)\) by the law of total probability .
These two theorems are most useful, of course, when we know \(\P(A_i)\) and \(\P(B \mid A_i)\) for each \(i \in I\). When we compute the probability of \(\P(B)\) by the law of total probability , we say that we are conditioning on the partition \(\mathscr{A}\). Note that we can think of the sum as a weighted average of the conditional probabilities \(\P(B \mid A_i)\) over \(i \in I\), where \(\P(A_i)\), \(i \in I\) are the weight factors. In the context of Bayes theorem , \(\P(A_j)\) is the prior probability of \(A_j\) and \(\P(A_j \mid B)\) is the posterior probability of \(A_j\) for \( j \in I \). We will study more general versions of conditioning and Bayes theorem in the section on Discrete Distributions in the chapter on Distributions, and again in the section on Conditional Expected Value in the chapter on Expected Value.
Once again, the law of total probability and Bayes' theorem can be applied to a conditional probability measure. So, if \(E\) is another event with \( \P(A_i \cap E) \gt 0 \) for \( i \in I \) then \begin{align} \P(B \mid E) & = \sum_{i \in I} \P(A_i \mid E) \P(B \mid A_i \cap E) \\ \P(A_j \mid B \cap E) & = \frac{\P(A_j \mid E) \P(B \mid A_j \cap E)}{\sum_{i \in I}\P(A_i \cap E) \P(B \mid A_i \cap E)}, \quad j \in I \end{align}
Examples and Applications
Basic Rules
Suppose that \(A\) and \(B\) are events in an experiment with \(\P(A) = \frac{1}{3}\), \(\P(B) = \frac{1}{4}\), \(\P(A \cap B) = \frac{1}{10}\). Find each of the following:
- \(\P(A \mid B)\)
- \(\P(B \mid A)\)
- \(\P(A^c \mid B)\)
- \(\P(B^c \mid A)\)
- \(\P(A^c \mid B^c)\)
Answer
- \(\frac{2}{5}\)
- \(\frac{3}{10}\)
- \(\frac{3}{5}\)
- \(\frac{7}{10}\)
- \(\frac{31}{45}\)
Suppose that \(A\), \(B\), and \(C\) are events in a random experiment with \(\P(A \mid C) = \frac{1}{2}\), \(\P(B \mid C) = \frac{1}{3}\), and \(\P(A \cap B \mid C) = \frac{1}{4}\). Find each of the following:
- \(\P(B \setminus A \mid C)\)
- \(\P(A \cup B \mid C)\)
- \(\P(A^c \cap B^c \mid C)\)
- \(\P(A^c \cup B^c \mid C)\)
- \(\P(A^c \cup B \ \mid C)\)
- \(\P(A \mid B \cap C)\)
Answer
- \(\frac{1}{12}\)
- \(\frac{7}{12}\)
- \(\frac{5}{12}\)
- \(\frac{3}{4}\)
- \(\frac{3}{4}\)
- \(\frac{3}{4}\)
Suppose that \(A\) and \(B\) are events in a random experiment with \(\P(A) = \frac{1}{2}\), \(\P(B) = \frac{1}{3}\), and \(\P(A \mid B) =\frac{3}{4}\).
- Find \(\P(A \cap B)\)
- Find \(\P(A \cup B)\)
- Find \(\P(B \cup A^c)\)
- Find \(\P(B \mid A)\)
- Are \(A\) and \(B\) positively correlated, negatively correlated, or independent?
Answer
- \(\frac{1}{4}\)
- \(\frac{7}{12}\)
- \(\frac{3}{4}\)
- \(\frac{1}{2}\)
- positively correlated.
Open the conditional probability experiment.
- Given \( \P(A) \), \( \P(B) \), and \( \P(A \cap B) \), in the table, verify all of the other probabilities in the table.
- Run the experiment 1000 times and compare the probabilities with the relative frequencies.
Simple Populations
In a certain population, 30% of the persons smoke cigarettes and 8% have COPD ( Chronic Obstructive Pulmonary Disease ). Moreover, 12% of the persons who smoke have COPD.
- What percentage of the population smoke and have COPD?
- What percentage of the population with COPD also smoke?
- Are smoking and COPD positively correlated, negatively correlated, or independent?
Answer
- 3.6%
- 45%
- positively correlated.
A company has 200 employees: 120 are women and 80 are men. Of the 120 female employees, 30 are classified as managers, while 20 of the 80 male employees are managers. Suppose that an employee is chosen at random.
- Find the probability that the employee is female.
- Find the probability that the employee is a manager.
- Find the conditional probability that the employee is a manager given that the employee is female.
- Find the conditional probability that the employee is female given that the employee is a manager.
- Are the events female and manager positively correlated, negatively correlated, or indpendent?
Answer
- \(\frac{120}{200}\)
- \(\frac{50}{200}\)
- \(\frac{30}{120}\)
- \(\frac{30}{50}\)
- independent
Dice and Coins
Consider the experiment that consists of rolling 2 standard, fair dice and recording the sequence of scores \(\bs{X} = (X_1, X_2)\). Let \(Y\) denote the sum of the scores. For each of the following pairs of events, find the probability of each event and the conditional probability of each event given the other. Determine whether the events are positively correlated, negatively correlated, or independent.
- \(\{X_1 = 3\}\), \(\{Y = 5\}\)
- \(\{X_1 = 3\}\), \(\{Y = 7\}\)
- \(\{X_1 = 2\}\), \(\{Y = 5\}\)
- \(\{X_1 = 3\}\), \(\{X_1 = 2\}\)
Answer
In each case below, the answers are for \( \P(A) \), \( \P(B) \), \( \P(A \mid B) \), and \( \P(B \mid A) \)
- \(\frac{1}{6}\), \(\frac{1}{9}\), \(\frac{1}{4}\), \(\frac{1}{6}\). Positively correlated.
- \(\frac{1}{6}\), \(\frac{1}{6}\), \(\frac{1}{6}\), \(\frac{1}{6}\). Independent.
- \(\frac{1}{6}\), \(\frac{1}{9}\), \(\frac{1}{4}\), \(\frac{1}{6}\). Positively correlated.
- \(\frac{1}{6}\), \(\frac{1}{6}\), \(0\), \(0\). Negatively correlated.
Note that positive correlation is not a transitive relation. From the previous exercise, for example, note that \(\{X_1 = 3\}\) and \(\{Y = 5\}\) are positively correlated, \(\{Y = 5\}\) and \(\{X_1 = 2\}\) are positively correlated, but \(\{X_1 = 3\}\) and \(\{X_1 = 2\}\) are negatively correlated (in fact, disjoint).
In dice experiment, set \(n = 2\). Run the experiment 1000 times. Compute the empirical conditional probabilities corresponding to the conditional probabilities in the last exercise.
Consider again the experiment that consists of rolling 2 standard, fair dice and recording the sequence of scores \(\bs{X} = (X_1, X_2)\). Let \(Y\) denote the sum of the scores, \(U\) the minimum score, and \(V\) the maximum score.
- Find \(\P(U = u \mid V = 4)\) for the appropriate values of \(u\).
- Find \(\P(Y = y \mid V = 4)\) for the appropriate values of \(y\).
- Find \(\P(V = v \mid Y = 8)\) for appropriate values of \(v\).
- Find \(\P(U = u \mid Y = 8)\) for the appropriate values of \(u\).
- Find \(\P[(X_ 1, X_2) = (x_1, x_2) \mid Y = 8]\) for the appropriate values of \((x_1, x_2)\).
Answer
- \(\frac{2}{7}\) for \(u \in \{1, 2, 3\}\), \(\frac{1}{7}\) for \( u = 4 \)
- \(\frac{2}{7}\) for \(y \in \{5, 6, 7\}\), \(\frac{1}{7}\) for \( y = 8 \)
- \(\frac{1}{5}\) for \( v = 4 \), \(\frac{2}{5}\) for \(v \in \{5, 6\}\)
- \(\frac{2}{5}\) for \(u \in \{2, 3\}\), \(\frac{1}{5}\) for \( u = 4 \)
- \(\frac{1}{5}\) for \((x_1, x_2) \in \{(2,6), (6,2), (3,5), (5,3), (4,4)\}\)
In the die-coin experiment, a standard, fair die is rolled and then a fair coin is tossed the number of times showing on the die. Let \(N\) denote the die score and \(H\) the event that all coin tosses result in heads.
- Find \(\P(H)\).
- Find \(\P(N = n \mid H)\) for \(n \in \{1, 2, 3, 4, 5, 6\}\).
- Compare the results in (b) with \(\P(N = n)\) for \(n \in \{1, 2, 3, 4, 5, 6\}\). In each case, note whether the events \(H\) and \(\{N = n\}\) are positively correlated, negatively correlated, or independent.
Answer
- \(\frac{21}{128}\)
- \(\frac{64}{63} \frac{1}{2^n}\) for \(n \in \{1, 2, 3, 4, 5, 6\}\)
- positively correlated for \(n \in \{1, 2\}\) and negatively correlated for \(n \in \{3, 4, 5, 6\}\)
Run the die-coin experiment 1000 times. Let \(H\) and \(N\) be as defined in the previous exercise.
- Compute the empirical probability of \(H\). Compare with the true probability in the previous exercise.
- Compute the empirical probability of \(\{N = n\}\) given \(H\), for \(n \in \{1, 2, 3, 4, 5, 6\}\). Compare with the true probabilities in the previous exercise.
Suppose that a bag contains 12 coins: 5 are fair, 4 are biased with probability of heads \(\frac{1}{3}\); and 3 are two-headed. A coin is chosen at random from the bag and tossed.
- Find the probability that the coin is heads.
- Given that the coin is heads, find the conditional probability of each coin type.
Answer
- \(\frac{41}{72}\)
- \( \frac{15}{41} \) that the coin is fair, \( \frac{8}{41} \) that the coin is biased, \( \frac{18}{41} \) that the coin is two-headed
Compare die-coin experiment and bag of coins experiment . In the die-coin experiment, we toss a coin with a fixed probability of heads a random number of times. In the bag of coins experiment, we effectively toss a coin with a random probability of heads a fixed number of times. The random experiment of tossing a coin with a fixed probability of heads \(p\) a fixed number of times \(n\) is known as the binomial experiment with parameters \(n\) and \(p\). This is a very basic and important experiment that is studied in more detail in the section on the binomial distribution in the chapter on Bernoulli Trials. Thus, the die-coin and bag of coins experiments can be thought of as modifications of the binomial experiment in which a parameter has been randomized . In general, interesting new random experiments can often be constructed by randomizing one or more parameters in another random experiment.
In the coin-die experiment, a fair coin is tossed. If the coin lands tails, a fair die is rolled. If the coin lands heads, an ace-six flat die is tossed (faces 1 and 6 have probability \(\frac{1}{4}\) each, while faces 2, 3, 4, and 5 have probability \(\frac{1}{8}\) each). Let \(H\) denote the event that the coin lands heads, and let \(Y\) denote the score when the chosen die is tossed.
- Find \(\P(Y = y)\) for \(y \in \{1, 2, 3, 4, 5, 6\}\).
- Find \(\P(H \mid Y = y)\) for \(y \in \{1, 2, 3, 4, 5, 6,\}\).
- Compare each probability in part (b) with \(\P(H)\). In each case, note whether the events \(H\) and \(\{Y = y\}\) are positively correlated, negatively correlated, or independent.
Answer
- \(\frac{5}{24}\) for \(y \in \{1, 6\}\), \(\frac{7}{48}\) for \(y \in \{2, 3, 4, 5\}\)
- \(\frac{3}{5}\) for \(y \in \{1, 6\}\), \(\frac{3}{7}\) for \(y \in \{2, 3, 4, 5\}\)
- Positively correlated for \( y \in \{1, 6\} \), negatively correlated for \( y \in \{2, 3, 4, 5\} \)
Run the coin-die experiment 1000 times. Let \(H\) and \(Y\) be as defined in the previous exercise.
- Compute the empirical probability of \(\{Y = y\}\), for each \(y\), and compare with the true probability in the previous exercise
- Compute the empirical probability of \(H\) given \(\{Y = y\}\) for each \(y\), and compare with the true probability in the previous exercise.
Cards
Consider the card experiment that consists of dealing 2 cards from a standard deck and recording the sequence of cards dealt. For \(i \in \{1, 2\}\), let \(Q_i\) be the event that card \(i\) is a queen and \(H_i\) the event that card \(i\) is a heart. For each of the following pairs of events, compute the probability of each event, and the conditional probability of each event given the other. Determine whether the events are positively correlated, negatively correlated, or independent.
- \(Q_1\), \(H_1\)
- \(Q_1\), \(Q_2\)
- \(Q_2\), \(H_2\)
- \(Q_1\), \(H_2\)
Answer
The answers below are for \( \P(A) \), \( \P(B) \), \( \P(A \mid B) \), and \( \P(B \mid A) \) where \( A \) and \( B \) are the given events
- \(\frac{1}{13}\), \(\frac{1}{4}\), \(\frac{1}{13}\), \(\frac{1}{4}\), independent.
- \(\frac{1}{13}\), \(\frac{1}{13}\), \(\frac{3}{51}\), \(\frac{3}{51}\), negatively correlated.
- \(\frac{1}{13}\), \(\frac{1}{4}\), \(\frac{1}{13}\), \(\frac{1}{4}\), independent.
- \(\frac{1}{13}\), \(\frac{1}{4}\), \(\frac{1}{13}\), \(\frac{1}{4}\), independent.
In the card experiment, set \(n = 2\). Run the experiment 500 times. Compute the conditional relative frequencies corresponding to the conditional probabilities in the last exercise.
Consider the card experiment that consists of dealing 3 cards from a standard deck and recording the sequence of cards dealt. Find the probability of the following events:
- All three cards are all hearts.
- The first two cards are hearts and the third is a spade.
- The first and third cards are hearts and the second is a spade.
Proof
- \(\frac{11}{850}\)
- \(\frac{13}{850}\)
- \(\frac{13}{850}\)
In the card experiment, set \(n = 3\) and run the simulation 1000 times. Compute the empirical probability of each event in the previous exercise and compare with the true probability.
Bivariate Uniform Distributions
Recall that Buffon's coin experiment consists of tossing a coin with radius \(r \le \frac{1}{2}\) randomly on a floor covered with square tiles of side length 1. The coordinates \((X, Y)\) of the center of the coin are recorded relative to axes through the center of the square, parallel to the sides. Since the needle is dropped randomly, the basic modeling assumption is that \( (X, Y) \) is uniformly distributed on the square \( [-1/2, 1/2]^2 \).
In Buffon's coin experiment,
- Find \(\P(Y \gt 0 \mid X \lt Y)\)
- Find the conditional distribution of \((X, Y)\) given that the coin does not touch the sides of the square.
Answer
- \(\frac{3}{4}\)
- Given \((X, Y) \in [r - \frac{1}{2}, \frac{1}{2} - r]^2\), \((X, Y)\) is uniformly distributed on this set.
Run Buffon's coin experiment 500 times. Compute the empirical probability that \(Y \gt 0\) given that \(X \lt Y\) and compare with the probability in the last exercise.
In the conditional probability experiment, the random points are uniformly distributed on the rectangle \( S \). Move and resize events \( A \) and \( B \) and note how the probabilities change. For each of the following configurations, run the experiment 1000 times and compare the relative frequencies with the true probabilities.
- \( A \) and \( B \) in general position
- \( A \) and \( B \) disjoint
- \( A \subseteq B \)
- \( B \subseteq A \)
Reliability
A plant has 3 assembly lines that produces memory chips. Line 1 produces 50% of the chips and has a defective rate of 4%; line 2 has produces 30% of the chips and has a defective rate of 5%; line 3 produces 20% of the chips and has a defective rate of 1%. A chip is chosen at random from the plant.
- Find the probability that the chip is defective.
- Given that the chip is defective, find the conditional probability for each line.
Answer
- 0.037
- 0.541 for line 1, 0.405 for line 2, 0.054 for line 3
Suppose that a bit (0 or 1) is sent through a noisy communications channel. Because of the noise, the bit sent may be received incorrectly as the complementary bit. Specifically, suppose that if 0 is sent, then the probability that 0 is received is 0.9 and the probability that 1 is received is 0.1. If 1 is sent, then the probability that 1 is received is 0.8 and the probability that 0 is received is 0.2. Finally, suppose that 1 is sent with probability 0.6 and 0 is sent with probability 0.4. Find the probability that
- 1 was sent given that 1 was received
- 0 was sent given that 0 was received
Answer
- \(12/13\)
- \(3/4\)
Suppose that \(T\) denotes the lifetime of a light bulb (in 1000 hour units), and that \(T\) has the following exponential distribution, defined for measurable \(A \subseteq [0, \infty)\):
\[\P(T \in A) = \int_A e^{-t} dt\]- Find \(\P(T \gt 3)\)
- Find \(\P(T \gt 5 \mid T \gt 2)\)
Answer
- \(e^{-3}\)
- \(e^{-3}\)
Suppose again that \(T\) denotes the lifetime of a light bulb (in 1000 hour units), but that \(T\) is uniformly distributed on the interal \([0, 10]\).
- Find \(\P(T \gt 3)\)
- Find \(\P(T \gt 5 \mid T \gt 2)\)
Answer
- \(\frac{7}{10}\)
- \(\frac{5}{8}\)
Genetics
Please refer to the discussion of genetics in the section on random experiments if you need to review some of the definitions in this section.
Recall first that the ABO blood type in humans is determined by three alleles: \(a\), \(b\), and \(o\). Furthermore, \(a\) and \(b\) are co-dominant and \(o\) is recessive. Suppose that the probability distribution for the set of blood genotypes in a certain population is given in the following table:
| Genotype | \(aa\) | \(ab\) | \(ao\) | \(bb\) | \(bo\) | \(oo\) |
|---|---|---|---|---|---|---|
| Probability | 0.050 | 0.038 | 0.310 | 0.007 | 0.116 | 0.479 |
Suppose that a person is chosen at random from the population. Let \(A\), \(B\), \(AB\), and \(O\) be the events that the person is type \(A\), type \(B\), type \(AB\), and type \(O\) respectively. Let \(H\) be the event that the person is homozygous, and let \(D\) denote the event that the person has an \(o\) allele. Find each of the following:
- \(\P(A)\), \(\P(B)\), \(\P(AB)\), \(\P(O)\), \(\P(H)\), \(\P(D)\)
- \(P(A \cap H)\), \(P(A \mid H)\), \(P(H \mid A)\). Are the events \(A\) and \(H\) positively correlated, negatively correlated, or independent?
- \(P(B \cap H)\), \(P(B \mid H)\), \(P(H \mid B)\). Are the events \(B\) and \(H\) positively correlated, negatively correlated, or independent?
- \(P(A \cap D)\), \(P(A \mid D)\), \(P(D \mid A)\). Are the events \(A\) and \(D\) positively correlated, negatively correlated, or independent?
- \(P(B \cap D)\), \(P(B \mid D)\), \(P(D \mid B)\). Are the events \(B\) and \(D\) positively correlated, negatively correlated, or independent?
- \(P(H \cap D)\), \(P(H \mid D)\), \(P(D \mid H)\). Are the events \(H\) and \(D\) positively correlated, negatively correlated, or independent?
Answer
- 0.360, 0.123, 0.038, 0.479, 0.536, 0.905
- 0.050, 0.093, 0.139. \(A\) and \(H\) are negatively correlated.
- 0.007, 0.013, 0.057. \(B\) and \(H\) are negatively correlated.
- 0.310, 0.343, 0.861. \(A\) and \(D\) are negatively correlated.
- 0.116, 0.128, 0.943. \(B\) and \(D\) are positivley correlated.
- 0.479, 0.529, 0.894. \(H\) and \(D\) are negatively correlated.
Suppose next that pod color in certain type of pea plant is determined by a gene with two alleles: \(g\) for green and \(y\) for yellow, and that \(g\) is dominant and \(y\) recessive.
Suppose that a green-pod plant and a yellow-pod plant are bred together. Suppose further that the green-pod plant has a \(\frac{1}{4}\) chance of carrying the recessive yellow-pod allele.
- Find the probability that a child plant will have green pods.
- Given that a child plant has green pods, find the updated probability that the green-pod parent has the recessive allele.
Answer
- \(\frac{7}{8}\)
- \(\frac{1}{7}\)
Suppose that two green-pod plants are bred together. Suppose further that with probability \(\frac{1}{3}\) neither plant has the recessive allele, with probability \(\frac{1}{2}\) one plant has the recessive allele, and with probability \(\frac{1}{6}\) both plants have the recessive allele.
- Find the probability that a child plant has green pods.
- Given that a child plant has green pods, find the updated probability that both parents have the recessive gene.
Answer
- \(\frac{23}{24}\)
- \(\frac{3}{23}\)
Next consider a sex-linked hereditary disorder in humans (such as colorblindness or hemophilia). Let \(h\) denote the healthy allele and \(d\) the defective allele for the gene linked to the disorder. Recall that \(h\) is dominant and \(d\) recessive for women.
Suppose that in a certain population, 50% are male and 50% are female. Moreover, suppose that 10% of males are color blind but only 1% of females are color blind.
- Find the percentage of color blind persons in the population.
- Find the percentage of color blind persons that are male.
Answer
- 5.5%
- 90.9%
Since color blindness is a sex-linked hereditary disorder, note that it's reasonable in the previous exercise that the probability that a female is color blind is the square of the probability that a male is color blind. If \(p\) is the probability of the defective allele on the \(X\) chromosome, then \(p\) is also the probability that a male will be color blind. But since the defective allele is recessive, a woman would need two copies of the defective allele to be color blind, and assuming independence, the probability of this event is \(p^2\).
A man and a woman do not have a certain sex-linked hereditary disorder, but the woman has a \(\frac{1}{3}\) chance of being a carrier.
- Find the probability that a son born to the couple will be normal.
- Find the probability that a daughter born to the couple will be a carrier.
- Given that a son born to the couple is normal, find the updated probability that the mother is a carrier.
Answer
- \(\frac{5}{6}\)
- \(\frac{1}{6}\)
- \(\frac{1}{5}\)
Urn Models
Urn 1 contains 4 red and 6 green balls while urn 2 contains 7 red and 3 green balls. An urn is chosen at random and then a ball is chosen at random from the selected urn.
- Find the probability that the ball is green.
- Given that the ball is green, find the conditional probability that urn 1 was selected.
Answer
- \(\frac{9}{20}\)
- \(\frac{2}{3}\)
Urn 1 contains 4 red and 6 green balls while urn 2 contains 6 red and 3 green balls. A ball is selected at random from urn 1 and transferred to urn 2. Then a ball is selected at random from urn 2.
- Find the probability that the ball from urn 2 is green.
- Given that the ball from urn 2 is green, find the conditional probability that the ball from urn 1 was green.
Answer
- \(\frac{9}{25}\)
- \(\frac{2}{3}\)
An urn initially contains 6 red and 4 green balls. A ball is chosen at random from the urn and its color is recorded. It is then replaced in the urn and 2 new balls of the same color are added to the urn. The process is repeated. Find the probability of each of the following events:
- Balls 1 and 2 are red and ball 3 is green.
- Balls 1 and 3 are red and ball 2 is green.

- Ball 2 is red.
- Ball 1 is red given that ball 2 is red.
Answer
- \(\frac{4}{35}\)
- \(\frac{4}{35}\)
- \(\frac{4}{35}\)
- \(\frac{3}{5}\)
- \(\frac{2}{3}\)
Think about the results in the previous exercise. Note in particular that the answers to parts (a), (b), and (c) are the same, and that the probability that the second ball is red in part (d) is the same as the probability that the first ball is red. More generally, the probabilities of events do not depend on the order of the draws. For example, the probability of an event involving the first, second, and third draws is the same as the probability of the corresponding event involving the seventh, tenth and fifth draws. Technically, the sequence of events \((R_1, R_2, \ldots)\) is exchangeable. The random process described in this exercise is a special case of Pólya's urn scheme , named after George Pólya. We sill study Pólya's urn in more detail in the chapter on Finite Sampling Models
An urn initially contains 6 red and 4 green balls. A ball is chosen at random from the urn and its color is recorded. It is then replaced in the urn and two new balls of the other color are added to the urn. The process is repeated. Find the probability of each of the following events:
- Balls 1 and 2 are red and ball 3 is green.
- Balls 1 and 3 are red and ball 2 is green.

- Ball 2 is red.
- Ball 1 is red given that ball 2 is red.
Answer
- \(\frac{6}{35}\)
- \(\frac{6}{35}\)
- \(\frac{16}{105}\)
- \(\frac{17}{30}\)
- \(\frac{9}{17}\)
Think about the results in the previous exercise, and compare with Pólya's urn . Note that the answers to parts (a), (b), and (c) are not all the same, and that the probability that the second ball is red in part (d) is not the same as the probability that the first ball is red. In short, the sequence of events \((R_1, R_2, \ldots)\) is not exchangeable.
Diagnostic Testing
Suppose that we have a random experiment with an event \(A\) of interest. When we run the experiment, of course, event \(A\) will either occur or not occur. However, suppose that we are not able to observe the occurrence or non-occurrence of \(A\) directly. Instead we have a diagnostic test designed to indicate the occurrence of event \(A\); thus the test that can be either positive for \(A\) or negative for \(A\). The test also has an element of randomness, and in particular can be in error. Here are some typical examples of the type of situation we have in mind:
- The event is that a person has a certain disease and the test is a blood test for the disease.
- The event is that a woman is pregnant and the test is a home pregnancy test.
- The event is that a person is lying and the test is a lie-detector test.
- The event is that a device is defective and the test consists of a sensor reading.
- The event is that a missile is in a certain region of airspace and the test consists of radar signals.
- The event is that a person has committed a crime, and the test is a jury trial with evidence presented for and against the event.
Let \(T\) be the event that the test is positive for the occurrence of \(A\). The conditional probability \(\P(T \mid A)\) is called the sensitivity of the test. The complementary probability \[\P(T^c \mid A) = 1 - \P(T \mid A)\] is the false negative probability. The conditional probability \(\P(T^c \mid A^c)\) is called the specificity of the test. The complementary probability \[\P(T \mid A^c) = 1 - \P(T^c \mid A^c)\] is the false positive probability. In many cases, the sensitivity and specificity of the test are known, as a result of the development of the test. However, the user of the test is interested in the opposite conditional probabilities, namely \(\P(A \mid T)\), the probability of the event of interest, given a positive test, and \(\P(A^c \mid T^c)\), the probability of the complementary event, given a negative test. Of course, if we know \( \P(A \mid T) \) then we also have \( \P(A^c \mid T) = 1 - \P(A \mid T) \), the probability of the complementary event given a positive test. Similarly, if we know \( \P(A^c \mid T^c) \) then we also have \( \P(A \mid T^c) \), the probability of the event given a negative test. Computing the probabilities of interest is simply a special case of Bayes' theorem.
The probability that the event occurs, given a positive test is \[\P(A \mid T) = \frac{\P(A) \P(T \mid A)}{\P(A) \P(T \mid A) + \P(A^c) \P(T \mid A^c)}\] The probability that the event does not occur, given a negative test is \[\P(A^c \mid T^c) = \frac{\P(A^c) \P(T^c \mid A^c)}{\P(A) \P(T^c \mid A) + \P(A^c) \P(T^c \mid A^c)}\]
There is often a trade-off between sensitivity and specificity. An attempt to make a test more sensitive may result in the test being less specific, and an attempt to make a test more specific may result in the test being less sensitive. As an extreme example, consider the worthless test that always returns positive, no matter what the evidence. Then \( T = S \) so the test has sensitivity 1, but specificity 0. At the opposite extreme is the worthless test that always returns negative, no matter what the evidence. Then \( T = \emptyset \) so the test has specificity 1 but sensitivity 0. In between these extremes are helpful tests that are actually based on evidence of some sort.
Suppose that the sensitivity \( a = \P(T \mid A) \in (0, 1)\) and the specificity \( b = \P(T^c \mid A^c) \in (0, 1) \) are fixed. Let \( p = \P(A) \) denote the prior probability of the event \( A \) and \( P = \P(A \mid T) \) the posterior probability of \( A \) given a positive test.
\( P \) as a function of \( p \) is given by \[ P = \frac{a p}{(a + b - 1) p + (1 - b)}, \quad p \in [0, 1] \]
- \( P \) increases continuously from 0 to 1 as \( p \) increases from 0 to 1.
- \( P \) is concave downward if \( a + b \gt 1 \). In this case \( A \) and \( T \) are positively correlated.
- \( P \) is concave upward if \( a + b \lt 1 \). In this case \( A \) and \( T \) are negatively correlated.
- \( P = p \) if \( a + b = 1 \). In this case, \( A \) and \( T \) are uncorrelated (independent).
Proof
The formula for \( P \) in terms of \( p \) follows from (42) and algebra. For part (a), note that \[ \frac{dP}{dp} = \frac{a (1 - b)}{[(a + b - 1) p + (1 - b)]^2} \gt 0\] For parts (b)-(d), note that \[ \frac{d^2 P}{dp^2} = \frac{-2 a (1 - b)(a + b - 1)}{[(1 + b - 1)p + (1 - b)]^3} \] If \( a + b \gt 1 \), \( d^2P/dp^2 \lt 0 \) so \( P \) is concave downward on \( [0, 1] \) and hence \( P \gt p \) for \( 0 \lt p \lt 1 \). If \( a + b \lt 1 \), \( d^2P/dp^2 \gt 0 \) so \( P \) is concave upward on \( [0, 1] \) and hence \( P \lt p \) for \( 0 \lt p \lt 1 \). Trivially if \( a + b = 1 \), \( P = p \) for \( 0 \le p \le 1 \).
Of course, part (b) is the typical case, where the test is useful. In fact, we would hope that the sensitivity and specificity are close to 1. In case (c), the test is worse than useless since it gives the wrong information about \( A \). But this case could be turned into a useful test by simply reversing the roles of positive and negative . In case (d), the test is worthless and gives no information about \( A \). It's interesting that the broad classification above depends only on the sum of the sensitivity and specificity.
Suppose that a diagnostic test has sensitivity 0.99 and specificity 0.95. Find \( \P(A \mid T) \) for each of the following values of \( \P(A) \):
- 0.001
- 0.01
- 0.2
- 0.5
- 0.7
- 0.9
Answer
- 0.0194
- 0.1667
- 0.8319
- 0.9519
- 0.9788
- 0.9944
With sensitivity 0.99 and specificity 0.95, the test in the last exercise superficially looks good. However the small value of \(\P(A \mid T)\) for small values of \(\P(A)\) is striking (but inevitable given the properties above ). The moral, of course, is that \(\P(A \mid T)\) depends critically on \(\P(A)\) not just on the sensitivity and specificity of the test. Moreover, the correct comparison is \(\P(A \mid T)\) with \(\P(A)\), as in the exercise, not \(\P(A \mid T)\) with \(\P(T \mid A)\)—Beware of the fallacy of the transposed conditional! In terms of the correct comparison, the test does indeed work well; \(\P(A \mid T)\) is significantly larger than \(\P(A)\) in all cases.
A woman initially believes that there is an even chance that she is or is not pregnant. She takes a home pregnancy test with sensitivity 0.95 and specificity 0.90 (which are reasonable values for a home pregnancy test). Find the updated probability that the woman is pregnant in each of the following cases.
- The test is positive.
- The test is negative.
Answer
- 0.905
- 0.053
Suppose that 70% of defendants brought to trial for a certain type of crime are guilty. Moreover, historical data show that juries convict guilty persons 80% of the time and convict innocent persons 10% of the time. Suppose that a person is tried for a crime of this type. Find the updated probability that the person is guilty in each of the following cases:
- The person is convicted.
- The person is acquitted.
Answer
- 0.949
- 0.341
The
Check Engine
light on your car has turned on. Without the information from the light, you believe that there is a 10% chance that your car has a serious engine problem. You learn that if the car has such a problem, the light will come on with probability 0.99, but if the car does not have a serious problem, the light will still come on, under circumstances similar to yours, with probability 0.3. Find the updated probability that you have an engine problem.
Answer
0.268
The standard test for HIV is the ELISA (Enzyme-Linked Immunosorbent Assay) test. It has sensitivity and specificity of 0.999. Suppose that a person is selected at random from a population in which 1% are infected with HIV, and given the ELISA test. Find the probability that the person has HIV in each of the following cases:
- The test is positive.
- The test is negative.
Answer
- 0.9098
- 0.00001
The ELISA test for HIV is a very good one. Let's look another test, this one for prostate cancer, that's rather bad.
The PSA test for prostate cancer is based on a blood marker known as the Prostate Specific Antigen . An elevated level of PSA is evidence for prostate cancer. To have a diagnostic test, in the sense that we are discussing here, we must decide on a definite level of PSA, above which we declare the test to be positive. A positive test would typically lead to other more invasive tests (such as biopsy) which, of course, carry risks and cost. The PSA test with cutoff 2.6 ng/ml has sensitivity 0.40 and specificity 0.81. The overall incidence of prostate cancer among males is 156 per 100000. Suppose that a man, with no particular risk factors, has the PSA test. Find the probability that the man has prostate cancer in each of the following cases:
- The test is positive.
- The test is negative.
Answer
- 0.00328
- 0.00116
Diagnostic testing is closely related to a general statistical procedure known as hypothesis testing . A separate chapter on hypothesis testing explores this procedure in detail.
Data Analysis Exercises
For the M&M data set, find the empirical probability that a bag has at least 10 reds, given that the weight of the bag is at least 48 grams.
Answer
\(\frac{10}{23}\).
Consider the Cicada data.
- Find the empirical probability that a cicada weighs at least 0.25 grams given that the cicada is male.
- Find the empirical probability that a cicada weighs at least 0.25 grams given that the cicada is the tredecula species.
Answer
- \(\frac{2}{45}\)
- \(\frac{7}{44}\)


2.2: Propositional Calculus
Logical equivalence gives us something like an “equals sign” that we can use to perform logical “calculations” and manipulations, similar to algebraic calculations and manipulations. To enable us to do such calculations, we first need a “tool chest” of basic logical equivalences to use therein.
Proposition \(\PageIndex{1}\): Rules of Propositional Calculus
Suppose \(A,B,C,E,U\) are logical statements, where \(E\) is a contradiction and \(U\) is a tautology. Then the following equivalences always hold.
-
Rules involving tautologies.
- \(\displaystyle A \lor U \Leftrightarrow U\)
- \(\displaystyle A \land U \Leftrightarrow A\)
-
Rules involving contradictions.
- \(\displaystyle A \lor E \Leftrightarrow A\)
- \(\displaystyle A \land E \Leftrightarrow E\)
-
Duality of tautologies and contradictions.
- \(\displaystyle \neg U \Leftrightarrow E\)
- \(\displaystyle \neg E \Leftrightarrow U\)
-
Double negation.
\(\displaystyle \neg \neg A \Leftrightarrow A\)
-
Idempotence.
- \(\displaystyle A \lor A \Leftrightarrow A\)
- \(\displaystyle A \land A \Leftrightarrow A\)
-
Commutativity.
- \(\displaystyle A \lor B \Leftrightarrow B \lor A\)
- \(\displaystyle A \land B \Leftrightarrow B \land A\)
-
Associativity.
- \(\displaystyle (A \lor B) \lor C \Leftrightarrow A \lor (B \lor C)\)
- \(\displaystyle (A \land B) \land C \Leftrightarrow A \land (B \land C)\)
-
Distributivity.
- \(\displaystyle A \land (B \lor C) \Leftrightarrow (A \land B) \lor (A \land C)\)
- \(\displaystyle A \lor (B \land C) \Leftrightarrow (A \lor B) \land (A \lor C)\)
- \(\displaystyle (A \lor B) \land C \Leftrightarrow (A \land C) \lor (B \land C)\)
- \(\displaystyle (A \land B) \lor C \Leftrightarrow (A \lor C) \land (B \lor C)\)
-
DeMorgan's Laws.
- \(\displaystyle \neg (A \lor B) \Leftrightarrow \neg A \land \neg B\)
- \(\displaystyle \neg (A \land B) \Leftrightarrow \neg A \lor \neg B\)
-
Constructing the conditional and biconditional.
- \(\displaystyle A \rightarrow B \Leftrightarrow \neg A \lor B\)
- \(\displaystyle A \leftrightarrow B \Leftrightarrow (A \rightarrow B) \land (B \rightarrow A)\)
Remark \(\PageIndex{1}\)
Each of these basic equivalences can be established with a truth table. See Exercise 2.5.4 .
Example \(\PageIndex{1}\): DeMorgan
Using Rule 9.b of Proposition \(\PageIndex{1}\), the following are equivalent statements.
- The triangle can't be both right and equilateral.
- The triangle is not right or it is not equilateral.
To see how the rule applies, let \(p\) represent the statement “the triangle is right” and let \(q\) represent the statement “the triangle is equilateral.” Then the first statement above is \(\neg (p \land q)\text{,}\) while the second statement above is \(\neg p \lor \neg q\text{.}\)
Now we need some new substitution rules to enable us to use the rules of Proposition \(\PageIndex{1}\) in logical calculations.
Theorem \(\PageIndex{1}\): Substitution Rules
- Replacing a substatement by an equivalent one.
Suppose \(A\) is a logical statement and \(X\) is a substatement of \(A\text{.}\) If statement \(Y\) is equivalent to \(X\text{,}\) then the new statement \(A'\) obtained from \(A\) by replacing substatement \(X\) by \(Y\) is equivalent to \(A\text{.}\) That is, if \(Y \Leftrightarrow X\) then \(A' \Leftrightarrow A\text{.}\)
- Substituting into a known equivalence.
Suppose \(A\) and \(B\) are logical statements, each of which involves substatement variables \(p_1, p_2, \ldots, p_m\text{.}\) If \(A\) and \(B\) are equivalent, then so are new statements \(A'\) and \(B'\) obtained by applying substitution \(p_i = C_i\) to both \(A\) and \(B\text{,}\) for every collection of statements \(C_1, C_2, \ldots, C_m\text{.}\)
- Proof Idea.
-
Think of \(X\) as an intermediate column in the calculation of the truth table of \(A\text{.}\) Replacing \(X\) by \(Y\) does not change this column, as the truth tables of \(X\) and \(Y\) are the same. We leave this statement for you, the reader, to consider. (Again, think of the \(C_i\) as intermediate columns in the calculations of the truth tables of \(A'\) and \(B'\text{.}\))
Example \(\PageIndex{2}\)
One of DeMorgan's Laws (Rule 9.a of Proposition \(\PageIndex{1}\)) says that \(\neg (p \lor q) \Leftrightarrow \neg p \land \neg q\text{.}\) Therefore,
\begin{equation*} \neg ((r \rightarrow t) \lor (t \rightarrow r) ) \Leftrightarrow \neg (r \rightarrow t) \land \neg (t \rightarrow r), \end{equation*}
using Rule 2 of our new Substitution Rules, with substitutions \(p = r \rightarrow t\text{,}\) \(q = t \rightarrow r\text{.}\)
Here is an example of a string of logical manipulations. It also demonstrates the use of Rule 10.a of Proposition \(\PageIndex{1}\) to manipulate an expression involving a conditional.
Example \(\PageIndex{3}\): DeMorgan with a Conditional
Consider the statement \((p_1 \lor p_2) \rightarrow q\text{.}\) We may read it as “if either \(p_1\) or \(p_2\) is true, then \(q\) will be true as well.” So it seems that each of \(p_1\) and \(p_2\) must imply \(q\) on its own. Let's see what propositional calculus says about this:
\begin{align*} (p_1 \lor p_2) \rightarrow q & \Leftrightarrow \neg (p_1 \lor p_2) \lor q & & \text{(i)}\\ & \Leftrightarrow (\neg p_1 \land \neg p_2) \lor q & & \text{(ii)}\\ & \Leftrightarrow (\neg p_1 \lor q ) \land (\neg p_2 \lor q) & & \text{(iii)}\\ & \Leftrightarrow (p_1 \rightarrow q) \land (p_2 \rightarrow q) & & \text{(iv)}, \end{align*}
with justifications
- Rule 2 of our new Substitution Rules, where we substitute \(A = p_1 \lor p_2\) into both sides of the construction of the conditional (Rule 10.a of Proposition \(\PageIndex{1}\));
- Rule 1 of our new Substitution Rules, using DeMorgan (Rule 9.a of Proposition \(\PageIndex{1}\)) on the substatement \(\neg (p_1 \lor p_2)\text{;}\)
- distributivity (Rule 8.d of Proposition \(\PageIndex{1}\)); and
- Rule 1 of our new Substitution Rules, using the construction of the conditional (Rule 10.a of Proposition \(\PageIndex{1}\)) on each of the two “factors” of the conjunction.
So our intuition about the logic of a disjunction in a conditional in this way was correct.
- A Look Ahead
-
This observation will come in handy — see Section 6.4 and Section 6.5 .


6.11: Music of Bedřich Smetana
Bedřich Smetana (b. 1824-1884) was born in Litomsyl, Bohemia while under Austrian rule (now the Czech Republic). Smetana was the son of a brewer and violinist and his father’s third wife. Smetana was a talented pianist who gave public performances from the age of six. Bohemia under Austrian rule was politically very volatile. In 1848 Smetana aligned himself with those seeking independent state- hood from Austria. After that revolution was crushed, Prague and the surrounding areas were brutally suppressed—especially those areas and people suspected of being sympathetic to Bohemian nationalism. In 1856, Smetana left for Sweden to accept a conductorship post. He hoped to follow in the footsteps of such music predecessors as Liszt. He thus expresses his admiration, “By the grace of God and with His help, I shall one day be a Liszt in technique and a Mozart in composition.”(footnote 1)
footnote 1 Taken from his diary, 23 January, 1845 found at www.quotesquotations.com/biography.
As a composer, Smetana began incorporating nationalist themes, plots, and dances in his operas and symphonic poems. He founded the Czech national school after he left Sweden and was a pioneer at incorporating Czech folk tunes, rhythms, and dances into his major works. Smetana returned to Bohemia in 1861 and assumed his role as national composer. He worked to open and establish a theatre venue in Prague where performances would be performed in their native tongue. Of his eight original operas, seven are still performed in native tongue today. One of these operas, The Bartered Bride , was and is still acclaimed. To hear Smetana’s Bartered Bride Overture, go to www.52composers.com/smetana.html and click on the link. He composed several folk dances, including polkas for orchestra. These polkas incorporated the style and levity of his Bohemian culture. To hear his Louisa’s Polka, go to www.52composers.com/smetana.html and click on the link.
Smetana also is known for composing the cycle of six symphonic poems en- titled My Country. These poems are program music, representing the beautiful Bohemian countryside, Bohemian folk dance and song rhythms, and the pageantry of Bohemian legends. The first of these symphonic poems is called Má vlast (My Fatherland) and is symbolic program music representing his birthplace. To hear Smetana’s My Fatherland , go to www.52composers.com/smetana.html and click on the link.
The second of these, Vltava, (The Moldau) is recognized as Smetana’s greatest orchestral work. Notes in the conductor’s score state
The Moldau” represents an exceptional expression of patriotic or nationalistic music. The musical poem reflects the pride, oppression, and hope of the Bohemian people. . . .
Two springs pour forth in the shade of the Bohemian Forest, one warm and gushing, the other cold and peaceful. Their waves, gaily flowing over rocky beds, join and glisten in the rays of the morning sun. The forest brook, hastening on, becomes the river Vltava (Moldau.) Coursing through Bohemia’s valleys, it grows into a mighty stream.  It flows through grass-grown pastures and lowlands where a wedding feast is being celebrated with song and dance. At night wood and water nymphs revel in its sparkling waves. Reflected on its surface are fortresses and castles—witnesses of bygone days of knightly splendor and the vanished glory of fighting times. At the St. John Rapids the stream races ahead, winding through the cataracts, heaving on a path with its foaming waves through the rocky chasm into the broad river bed— finally. Flowing on in majestic peace toward Prague—finally. Flowing on in majestic peace toward Prague and welcomed by time-honored Vysehrad (castle.) Then it vanishes far beyond the poet’s gaze.” (footnote 2)
footnote 2:Preface to the original score, Philharmonic Symphony Society of New York, The Concert Companion p. 672
|
Listening Guide For audio, go to: https://www.youtube.com/watch?v=hY0BJOcEiOM Featuring Vilem Tausky conducting the BBC Symphony |
| Composer: Bedřich Smetana (1824-1884) |
| Composition: The Moldau ( Vlatava ) |
| Date: 1874 |
| Genre: Symphonic poem |
| Form: Symphonic Poem (Tone Poem) |
| Performing Forces: piccolo, two flutes, two oboes , two clarinets, two bassoons, four French horns, two trumpets, three trombones, tuba, timpani, bass drum, triangle, cymbals, harp, strings |
|
What we want you to remember about this composition:
|
| Timing | Performing Forces, Melody, and Texture |
|
0:00 |
Two Springs.(Source of the river)
|
| 1:06 | Rippling notes moves to lower strings that lead to the main river theme. |
|
1:16 |
The River Theme.
|
| 1:31 | River theme repeated. |
| 1:46 | Melodic answer to the river theme. |
| 2:11 | The river theme is expanded (note the triangle with orchestral crescendos). |
| 2:18 | Return of the river theme. |
| 2:35 | Answer to river theme. |
| 2:52 | Expansion/elaboration of river theme. |
| 3:01 | Return of the river theme. |
|
3:20 |
Forest Hunting Scene. French horns and trumpets, hunting calls.
|
|
4:10 |
Rippling continues (in strings); dies down to gently rocking motion. Transition to next section (strings). Repeated notes in strings lead to rustic folk tune, staccato in strings and woodwinds |
|
4:38 |
Peasant Wedding. Strings present a dance-like tune (polka). Closes with repeated single note in strings |
|
6:15 |
Moonlight: Dance of Water Nymphs . Woodwinds, sustained tones.
|
|
6:35 |
Flute passage (similar to opening of work). Rippling figures in flutes; muted string theme with harp, punctuated by French horn; brass crescendo, fanfare |
| 7:02 | Muted violins in high register with a legato melody. |
| 9:06 | Intensification. |
| 9:29 | Violins present the river theme. Played in the minor mode |
| 10:17 | River theme reappears in the major mode |
| 10:29 | St. John’s Rapids. Full orchestra, ff. Brasses, timpani roll, piccolo, cymbal crashes. |
|
12:09 |
River theme, Full orchestra, Loudest dynamic/volume.
The River at its Widest Point
.
|
|
12:45 |
The Ancient Castle. The brasses and woodwinds portray Vyšehrad, the ancient castle in a hymn-like melody. Slow then Accelerates |
| 14:05 | River Dies Away, Strings slow down, lose momentum |
| Final Cadence | Two forceful closing chords. |


18.5: Diversity as Quotidian Production Practices
-
- Last updated
- Save as PDF
- Herman Gray
- University of California Press
In her study of the shift in the nature of the work object, its impact on social relations among television writers in Los Angeles, and the precarity of their work as writers, television scholar and director Felicia Henderson examines the transformation of the traditional work product, the television season, as a key unit of analysis. 20 Specifically, she focuses on the shortening of the television season from twenty-two programs to thirteen, which is made possible by changes in delivery systems, viewing platforms, viewing practices, and contract negotiations.
The virtue of Henderson’s insight is that her analysis of the writer’s room as a site of creative production (like others in this genre) dwells on the structure of creative relationships, industrial settings, and the organizing logic that defines the production and creative processes rather than on individual personalities and attitudes of creative personnel. In terms of the applicability of her approach to a concern with race and diversity, Henderson’s research commends attention to the organizational sites, creative processes, and social relations where the practices of diversity (or impediments to diversity) operate. Such an approach appreciates the fluidity and flexibility of evidence and analysis across time and space so that foundational categories, policy mandates, political stakes, and analytic conceptions can shift with the historical, technological, and political conditions in which they embedded and which they help organize and narrate. In other words, analytically it is useful to look at the conditions that structure and organize some of the foundational assumptions and questions about diversity and television (aims, means of realizing, forms of monitoring and assessing their effectiveness).
Henderson’s approach suggests that the specific conceptions, conditions, and assumptions that produce diversity (or the twenty-two-episode season as a staple unit of network television) as a desirable goal in television are not static, nor is the nexus of institutions, interests, and stakes that support or oppose its actualization. Her approach encourages an analysis of diversity (or impediments to diversity) as a dynamic and flexible set of industrial, legal, cultural, and economic practices that the study of the precarity of creativity and diversity can bring to bear on the question. In other words, the specific conceptions, conditions, and assumptions that produce diversity as a goal and practice within creative media industries like television are not fixed, and neither is the nexus of institutions and logics that organize and express them.
The challenge is moving the research focus from the founding scene of the problem of racial access and image exclusion within television to the shifting conditions that shaped television and discourses of race, including the rise of diversity, since the Kerner Commission Report. What, in other words, are the implications for quotidian practices of inequality and making race that the shorter seasons, new delivery systems, new interactive platforms, new divisions of labor, and new relations of production crystalize? What might the impact of these developments be on the very terms within which we pose the question of inequality in television that diversity glosses? Such reframing moves the issue of diversity some way from the analytic social and political scene in which it initially appeared. It shifts the industry and analytic assumption of equating diversity and social equality with access and representational parity to one where the calculus of cultural, economic, and political difference as a basis of the production of inequality is central to media industry practices.
This approach to research on television and race scrambles foundational binaries that continue to inform industrial practice, academic approaches, and media activism: inside/outside, accuracy/stereotype, author/imitation. With respect to race and difference, the terrain is considerably more complex and urges different questions that creative industry studies might help clarify: 1) How is diversity and difference framed as a labor issue and as a matter of work process and contractual management? 2) In what respect does the international division of creative labor pressure local and national formulations of diversity as matters of representation, reparation, and labor? 3) In what respects do the new international division of cultural labor, the rise of new platforms and delivery systems, and the creative arrangements that drive new projects, genres, talent bear on the question of diversity in new and unforeseen ways, especially within different national formations defined by distinct racial projects and ethnic formations? 4) These conditions could just as well open the way for the media production of diversity as cultural normativity or a technology of power/knowledge deployed to reach lifestyle niches. A critical media industry approach to inequality (rather than merely diversity) would urge that media and ethnic/racial arrangements be located and analyzed within the context of racism, racial projects, and race making nationally and globally.
On this count, John Caldwell’s insights about reflexivity, industrial knowledge, and practices and rituals among cultural producers are exemplary; 21 so too are Vicki Mayer’s considerations of the inscription of knowledge in practices at all levels of the production process, as is Sarah Banet Weiser’s work on the role of brands and branding as a mode of crafting and caring for the self in the construction of diversity. 22 Finally, Timothy Havens’s explorations of the circulation of television content about blackness in the United States and the role of industry lore about race and diversity in the creative process are especially rich. So too are Darnell Hunt’s studies of African Americans who use new media technologies to write, produce, and perform alternate and nonhegemonic conceptions of complex and intersecting minoritarian identities. 23