4.3: Philosophical Perspective of Education
-
- Last updated
- Save as PDF
- Education Department
- SUNY Oneonta via LumenLearning
There are four philosophical perspectives currently used in educational settings: essentialism, perennialism, progressivism, and social reconstructionism/critical pedagogy. Unlike the more abstract ontology and axiology, these four perspectives focus primarily on what should be taught and how it should be taught, i.e. the curriculum.
Essentialism
Essentialism adheres to a belief that a core set of essential skills must be taught to all students. Essentialists tend to privilege traditional academic disciplines that will develop prescribed skills and objectives in different content areas as well as develop a common culture. Typically, essentialism argues for a back-to-basics approach on teaching intellectual and moral standards. Schools should prepare all students to be productive members of society. Essentialist curricula focuses on reading, writing, computing clearly and logically about objective facts about the outside real world. Schools should be sites of rigor where students learn to work hard and respect authority. Because of this stance, essentialism tends to subscribe to tenets of Realism. Essentialist classrooms tend to be teacher-centered in instructional delivery with an emphasis on lecture and teacher demonstrations.
Key theorists:
William Bagley, E.D. Hirsh Jr.
Perennialism
Perennialism advocates for seeking, teaching, and learning universal truths that span across historical time periods. These truths, Perennialists argue, have everlasting importance in helping humans solve problems regardless of time and place. While Perennialism resembles essentialism at first glance, perennialism focuses on the individual development of the student rather than emphasizing skills. Perennialism supports liberal arts curricula that helps produces well-rounded individuals with some knowledge across the arts and sciences. All students should take classes in English Language Arts, foreign languages, mathematics, natural sciences, fine arts, and philosophy. Like Essentialism, Perennialism may tend to favor teacher-centered instruction; however, Perennialists do utilize student-centered instructional activities like Socratic Seminar, which values and encourages students to think, rationalize, and develop their own ideas on topics.
Key theorists:
Robert Hutchins, Mortimer Adler
Progressivism
Progressivism focuses its educational stance toward experiential learning with a focus on developing the whole child. Students learn by doing rather than being lectured to by teachers. Curriculum is usually integrated across contents instead of siloed into different disciplines. Progressivism’s stance is in stark contrast to both Essentialism and Perennialism in this manner. Progressivism follows a clear pragmatic ontology where the learner focuses on solving real-world problems through real experiences. Progressivist classrooms are student-centered where students will work in cooperative/collaborative groups to do project-based, expeditionary, problem-based, and/or service-learning activities. In progressivist classrooms, students have opportunities to follow their interests and have shared authority in planning and decision making with teachers.
Key theorists:
John Dewey, Maria Montessori
Social Reconstructionism & Critical Pedagogy
Social reconstructionism was founded as a response to the atrocities of World War II and the Holocaust to assuage human cruelty. Social reform in response to helping prepare students to make a better world through instilling democratic values. Critical pedagogy emerged from the foundation of the early social reconstructionist movement. Critical pedagogy is the application of critical theory to education. For critical pedagogues, teaching and learning is inherently a political act and they declare that knowledge and language are not neutral, nor can they be objective. Therefore, issues involving social, environmental, or economic justice cannot be separated from the curriculum. Critical pedagogy’s goal is to emancipate marginalized or oppressed groups by developing, according to Paulo Freire, conscientização, or critical consciousness in students. Critical pedagogy de-centers the traditional classroom, which positions teacher at the center. The curriculum and classroom with a critical pedagogy stance is student-centered and focuses its content on social critique and political action.
Key theorists:
Paulo Freire, bell hooks (note: bell hooks intentionally does not capitalize her name, which follows her critical stance that language, even how we write one’s own name, is political and ideological.)
Contributors and Attributions
- Foundations of Education. Authored by : SUNY Oneonta Education Department. License : CC BY: Attribution


6.8: Introduction to Another Look at Probability
What you’ll learn to do: Interpret (in context) a probability as a long-run relative frequency of an event.
Probability is a theoretical measurement of the likelihood of an event occurring. Building a solid foundation in probability helps us to better understand the data we collect in the real world and whether that data yields statistically surprising results. In statistical experiments and studies, we will estimate probabilities from collected data, and our interpretations and findings are based on probability concepts. In this next section, we begin building a solid foundation in probability and understand how to interpret a long-run relative frequency of an event.
Contributors and Attributions
CC licensed content, Shared previously
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution


6: Range and Endurance
Introduction
A Little Background
In the earliest days of powered flight the primary concern was getting the aircraft into the air and back down safely (with safely meaning the ability to limp away after the “landing”). The Wright’s famous first flight was shorter than a football field and even for a couple of years after December of 1903 they were content to circle around the family farm The Wright’s home built engines couldn’t run for long periods of time and they simply didn’t envision the need or desire for flights over distances of over a few miles. In 1908, however, Scientific American magazine challenged aviation experimenters to produce an aircraft or “aero-plane” [1] which could fly, in public view, over a distance of one mile! While the Wright’s claimed to be able to make such a flight, their obsession with secrecy as they sought military sales and their egotistical belief that no one else could approach their expertise in aviation led them to ignore the prize offered by Scientific American for the one mile public flight.
It was Glenn Curtiss, a builder of motorcycle engines and holder of numerous world speed records in motorcycle racing, who, in July of 1908, made the first public one-mile flight. Curtiss, who had worked with Alexander Bell and others to develop their own airplanes, made the flight with newspaper reporters watching and with movie cameras recording the flight. Curtiss became the top aviator in America and the Wrights were furious, leading to numerous legal suits as Wilbur and Orville sought to prove in the courts that Curtiss and Bell had infringed on their patents. Curtiss went on to outperform the Wrights and others in aviation meets in America and Europe. The Wright’s subsequent patent suits aimed at reserving for themselves the sole rights to design and build airplanes in the United States stagnated aircraft development in America and shifted the scene of aeronautical progress to Europe where it remained until after World War I.
As aircraft and aviation continued to develop, range and endurance became the primary objective in aircraft design. In war, bombers needed long ranges to reach enemy targets beyond the front lines and by the end of World War I huge bombers had been developed in several countries. After the war, European governments subsidized the conversion of these giants into passenger aircraft. Some larger planes had even been built as passenger carrying vehicles before the conflict. Sikorski’s early designs are good examples. In the United States, however, with little government interest in promoting air travel for the public until the late 1920’s, long range aircraft development was more fantasy than fact. In 1927 Lindberg’s trans-Atlantic flight captured the public’s imagination and interest in long range flight increased. Lindberg’s flight, like that of Curtiss, was prompted by a prize from the printed media, illustrating the role of newspapers and magazines in spurring technological progress.
By the late 1930’s the public began to see flight as a way to travel long distances in short times, national and international airline routes had developed, and planes like the “China Clipper” set standards for range and endurance. World War II forced people and governments to think in global terms leading to wartime development of bombers capable of non‑stop flight over thousands of miles and to post‑war trans‑continental and trans‑ocean aircraft. Since 1903 we have seen aircraft ranges go from feet to non‑stop circling the globe and endurances go from minutes to days!
6.1 Fuel Usage and Weight
In studying range and endurance we must, for the first time in this course, consider fuel usage. In the aircraft of Curtiss and the Wrights, it was not uncommon for the engine to quit from mechanical problems or overheating before the fuel ran out. In today’s aircraft, range and endurance depend on the amount of fuel on board. When the last drop of fuel is gone the plane has reached its limit for range and endurance. One could, of course, include the glide range and endurance after the aircraft runs out of fuel, but an airline that operated that way would attract few passengers!
Fuel usage depends on engine design, throttle settings, altitude and a number of other factors. It is, however, not the purpose of this text to study engine fuel efficiency or the pilot’s use of the throttle. We will assume that we are given an engine with certain specifications for efficiency and fuel use and that the throttle setting is that specified in the aircraft handbook or manual for optimum range or endurance at the chosen altitude. It is assumed that the student will take a separate course on propulsion to study the origins of the figures used here for these parameters.
Our primary concern in fuel usage will be the change in the weight of the aircraft with time. Many of our performance equations used in previous chapters include the aircraft weight. In those chapters we treated weight as a constant. Weight is, in reality, constant only for the glider or sailplane. For other aircraft the weight is always changing, always decreasing as the fuel is burned. This means that the aerodynamic performance of the airplane changes during the flight. This does not, however, negate the value of the methods used earlier to study cruise and climb. Those calculations will normally be done using the maximum gross weight of the airplane which will lead to a conservative or “worst case” analysis of those performance parameters. We can also use the methods developed earlier to look at the “instantaneous” capabilities of the aircraft at a given weight, realizing that at a later time in flight and at a lower weight, the performance may be different.
In considering range and endurance it is imperative that we consider weight as a variable, changing from maximum gross weight at take‑off to an empty fuel tank weight at the end of the flight. To do this we will deal with fuel usage in terms of the weight of the fuel (as opposed to fuel volume, in gallons, which we normally use for automobiles). When our concern is endurance we are interested in the change in weight of the fuel per unit time
\[\dfrac{dW_f}{dt}\]
and, when range is the concern we want to know how the weight of fuel decreases with distance traveled.
\[\dfrac{dW_f}{dS}\]
Aircraft engine manufacturers like to specify engine fuel usage in terms of specific fuel consumption . For jet engines this becomes a thrust specific fuel consumption and for prop aircraft, a power specific fuel consumption . Since thrust and power bring different units into the equations we must consider the two cases separately.
6.2 Range and Endurance: Jet
We speak of the engine output of a jet engine in terms of thrust; therefore, we speak of the fuel usage of the jet engine in terms of a thrust specific fuel consumption , C t . C t is the mass of fuel consumed per unit time per unit thrust. The unit of time should be seconds and the unit for thrust should be in pounds or Newtons of thrust.
[C t ] = (sl/sec)/lb thrust , or = (kg/sec)/N thrust
The above is the proper definition of thrust specific fuel consumption, however, it is not really exactly what we need for our calculations. We would prefer a definition based on weight of fuel consumed instead of the mass . We will thus define a weight specific fuel consumption , t , as the weight of fuel used per unit time per unit thrust. This gives units of (time) -1 .
\[\left[\gamma_{\mathrm{t}}=\mathrm{gC}_{\mathrm{t}}\right]=\left(\mathrm{lb}_{\text {fuel }} / \mathrm{sec}\right) / \mathrm{lb}_{\text {thrust }}, \text { or }=\left(\mathrm{N}_{\text {fuel }} / \mathrm{sec}\right) / \mathrm{N}_{\text {thrust }}=(\mathrm{sec})^{-1}\]
The reader should be aware that many aircraft performance texts and propulsion texts are very vague regarding the units of specific fuel consumption. Some even define it in terms of mass and give it units of 1 / sec ., making it dimensionally incorrect. Part of the confusion, particularly in older American propulsion texts, lies in the use of the pound‑mass as the unit of mass. This gives a combination of pounds‑mass divided by pounds‑force, which, in reality gives sec 2 / ft . The situation is then further complicated by the author seemingly throwing in a term called g c which is supposed to resolve the lb m /lb f issue. At any rate it is very important that the engineer using specific fuel consumption carefully consider the units involved before beginning the solution of a range or endurance problem. A correctly specified weight specific fuel consumption will have units of sec -1 and will do so without the use of anything called g c.
It should be added that one might also encounter specific fuel consumption which has been calculated using mass of fuel (kg) and thrust in kilograms, a non-SI unit which is used in much of the world (virtually no one in the world knows what a Newton is or how to use it). If this is done the numerical value should be the same as obtained using the weight specific fuel consumption definition above.
Some sources of specific fuel consumption data use units of ( hours) -1 since the hour is a more convenient unit of time. The use of seconds is, however, correct in any standard unit system and the student may be well advised to convert hours to seconds before beginning calculation even though this will ultimately result in the calculation of endurance in seconds, giving rather large numbers for answers.
To find endurance we want the rate of fuel weight (W f ) change per unit time which can be written in terms of the thrust specific fuel consumption
\[\mathrm{dW}_{\mathrm{f}} / \mathrm{dt}=\gamma_{\mathrm{t}} \mathrm{T}\]
And, in straight and level flight where thrust equals drag
\[\mathrm{d} \mathrm{W}_{\mathrm{f}} / \mathrm{dt}=\gamma_{\mathrm{t}} \mathrm{D}\]
For maximum endurance we want to minimize the above term. This clearly shows that for maximum endurance the jet plane must be flown at minimum drag conditions. We will look at how to find that endurance after taking a brief look at range.
To find the flight range we must look at the rate of change of fuel weight with distance of flight. We might pause a little at this point to realize that this may be more complicated than endurance because range will depend on more than simply the aerodynamic performance of the airplane. It will also require consideration of the wind speed. An airplane can fly forever at a speed of 100 mph into a 100 mph head‑wind and still have a range of zero! For now, however, we will put those worries aside and look at the simple mathmatics with which we begin consideration of the problem. We looked above at the rate of weight change with time. We can combine this with the change of distance with time (speed) to get the rate of change of weight with distance.
\[\mathrm{dW}_{\mathrm{f}} / \mathrm{dS}=\left(\mathrm{dW}_{\mathrm{f}} / \mathrm{dt}\right) /(\mathrm{dS} / \mathrm{dt})=\gamma_{\mathrm{t}}(\mathrm{T} / \mathrm{V})=\gamma_{\mathrm{t}}(\mathrm{D} / \mathrm{V})\]
Note that we still assume D = T or straight and level flight.
From the above it is obvious that maximum range will occur when the drag divided by velocity ( D/V ) is a minimum. This is not a condition which we have studied earlier but we can get some idea of where this occurs by looking at the plot of drag versus velocity for an aircraft.
On this plot a line drawn from the origin to intersect the drag curve at any point has a tangent equal to the drag at the point of intersection divided by the velocity at that point. The minimum possible value of D/V for the aircraft represented by the drag curve must then be found when the line is just tangent to the drag curve. This point will give the velocity for maximum range. Note that it is a higher speed than that for minimum drag (which, in turn was higher than the speed for minimum power).
In the above we have found the conditions needed to achieve maximum range and endurance for a jet aircraft. We have not yet found equations for the actual range or endurance. To find these we need to return to the time and distance differentials and integrate them. For time we have
We now wish to put the equations in a form which includes the weight of the aircraft instead of the weight of the fuel. Since the change in weight of the aircraft in flight is equal and opposite the weight of fuel consumed
dW = – dW fuel
we have
Finally, integrating over time to find the endurance gives
In a similar manner the range is found from the distance differential
or
In the above equations we must know how the aircraft velocity, thrust and specific fuel consumption vary with aircraft weight. At this point we need to make some assumptions about the way the flight is to be conducted. This is sometimes called the “flight schedule”.
6.3 Approximate Solutions for Range and Endurance for a Jet
The first assumption to be made in finding range and endurance equations is that the flight will be essentially straight and level. In order to give ourselves some leeway we will call this “quasi ‑level” flight, our desire being merely to use the L = W. T = D relations.
T = D = W(D/L) = W(C D /C L ) ,
and we can also use
Substituting these into the range and endurance relationships above give
At this point we need to make some further assumptions about the flight schedule in order to simplify the integration of these equations. For example, in the endurance equation, if we assume that the flight is made at constant angle of attack, we are assuming
that the lift and drag coefficients are constant for the entire flight. If we also assume that the specific fuel consumption is constant for the flight the only variable left in the integral is weight itself and the integral becomes:
The range integral contains an additional variable, the density of the atmosphere. It is still possible to make a couple of combinations of assumptions which will result in simple integration and realistic flight conditions. The first case will be to assume cruise at both constant altitude and constant angle of attack giving both density and the lift and drag coefficients as constants in the integration.
A second simple case combines the assumptions of constant angle of attack and constant speed, which can be used with the earlier form of the range equation.
giving
Note that the last equation above is simply the endurance equation multiplied by the velocity. This should not be surprising since this is the case where velocity is constant.
In the final equations above for range and endurance we should note that if standard units are used with specific fuel consumptions in sec. -1 , range will be given in feet or meters and endurance in seconds. We may find it easier to ascertain the degree to which our answers are realistic if we convert these answers to miles or kilometers and hours.
In finding the above equations for range and endurance we have looked only at special cases which would result in simple integrations. If we know more complicated flight schedules we can determine the functional relationships between the lift and drag coefficients, velocity, density, etc. and weight loss during flight and insert them into the original integrals to solve for range and endurance. The above cases are, however, very close to actual operational cruise conditions for long range aircraft and will probably suffice for an introductory study of aircraft performance. Let’s take a look at those simple cases.
Both range cases included our endurance assumption of constant angle of attack and specific fuel consumption. The first case combined these assumptions with specification of constant altitude. This appears to be the simplest case to actually fly but to see what it actually means we need to go back to the straight and level flight velocity relation
V = [2W/( ρ SC L )] 1/2
or V∝W 1/2 , when ρ an d C L are both constant.
If altitude (density) and angle of attack (lift coefficient) are both constant it is obvious that the velocity must change as the weight changes. In other words, for this flight schedule as fuel is burned and the weight of the aircraft decreases, the flight speed must decrease in proportion to the square root of the weight .
The other case, constant speed combined with constant angle of attack, is seen from the velocity relation above to require that density decrease in proportion to the weight.
W/ ρ = const. when V and C L are constant.
This means that as the aircraft burns off fuel, the aircraft will slowly move to higher altitudes where the density is lower . This is commonly known as the drift ‑up flight schedule. This is actually very similar to the way that commercial airliners fly long distance routes. Those of you who have been on such flights will recall the pilot announcing that “we are now cruising at 35,000 feet and will climb to 39,000 feet after crossing the Mississippi” or some such plan. While the FAA will not allow aircraft to simply “drift‑up” as they fly from coast‑to‑coast, they will allow schedules which incrementally approximate the drift‑up technique.
It must be noted that the two range equations above will give two different answers for the same amount of fuel. Also note that the equations are based on only the cruise portion of the flight. An actual flight will include take‑off, climb to cruise altitude, descent and landing in addition to cruise. Allowance also must be made for reserve fuel to handle emergency situations and “holds” imposed by air traffic controllers.
The biggest assumption used in all the integrations above is that of constant angle of attack. While this fits our conditions for optimum cases such as maximum endurance
occurring at maximum lift‑to drag ratio (minimum drag), it may not fit real flight very well. While the pilot can easily monitor his or her airspeed and altitude, the airplane’s angle of attack is not as easily monitored and directly controlled.
The equations above for range and endurance are valid for any flight condition which falls within the assumptions made in their derivation. If we have a Boeing 747 flying at an angle of attack of eight degrees and a speed of 250 miles per hour these equations can be used to find the range and endurance even though this is obviously not an optimum speed and angle of attack. Should we wish to determine the optimum range or endurance we must use the values of lift and drag coefficient and the velocity which we found earlier to be needed for these optimums.
Earlier we found that for maximum endurance the aircraft needs to fly at minimum drag conditions. Our actual endurance equation confirms this, showing endurance as a function of the lift‑to‑drag coefficient ratio which will be a maximum if drag is a minimum.
We also found that range would be optimum if the drag divided by velocity was a minimum. The correlation between this condition and the range equations derived is not as obvious as that of minimum drag with the endurance equation. Using the straight and level flight force relations which can be manipulated to show
D = W[D/L] = W[C D /C L ]
the quantity V/D can be written
V/D = [V/ W][ C L /C D ] .
Now using the velocity relation for straight and level flight
we find
Therefore, we find that the maximum range occurs when, for a given weight and altitude
C L 1/2 /C D is a maximum.
If we assume a parabolic drag polar with constant C D0 and K we can write
C L 1/2 /C D = C L 1/2 / [C D0 + KC L 2 ]
To find when this combination of terms is at a maximum we can take its derivative with respect to its variable ( C L ) and set it equal to zero.
Solving this gives
½( C D0 + KC L 2 )C L 1/2 – (2KC L 2 ) / C L 1/2 = 0
or
C D0 + KC L 2 -4KC L 2 = 0
then
C D0 = 3KC L 2
and, finally
C L = [C D0 / 3K] 1/ 2 .
Thus, for maximum range
Using this in the drag polar gives the value of drag coefficient for maximum range
C Dmax R = C D0 + KC LmaxR 2 = C D0 + KC D0 /(3K) = (4/ 3)C D 0 .
These are referred to as the conditions for “instantaneous” maximum range. The term instantaneous is used because the calculations are for a given weight and we know that weight is changing during the flight. In other words, at any point during the flight, at the weight and altitude at that point, the lift and drag coefficients found above will give the best range.
6.4 Range and Endurance: Prop
We will now look at range and endurance for propeller driven aircraft in which the engine performance is normally expressed in terms of power instead of thrust. An examination of range and endurance for aircraft which have performance measured in terms of power (propeller aircraft) is made by defining a power specific fuel consumption similar to the thrust specific fuel consumption used for jets. The power specific fuel consumption is defined as the mass of fuel consumed per unit time per unit shaft power. The units are slugs per unit power per second in the English system or kilogram per unit power per second in SI units.
[C p ] = sl /(power-sec) or kg/(power-sec)
The power units used are horsepower in the English system and watts in SI units.
Just as we did in the jet (thrust) case, we will often find an alternate definition of specific fuel consumption given in terms of the weight of fuel consumed instead of the mass.
While the proper time unit is seconds we will often find such data given for an engine in terms of hours. We will develop our equations in terms of the fundamental units (seconds for time) and, as in the jet case, assume “quasi‑level” flight which has
P avail = P req = DV
In dealing with prop engines we must consider the propulsive efficiency, η p , which relates the shaft power, P s , coming from the engine itself to the power effectively used by the prop to transfer momentum to the air.
As for the jet, to find endurance we must consider
and for range are interested in
From the above equations it is obvious that, for a given specific fuel consumption and efficiency, the rate of fuel use is a minimum (instantaneous endurance is a maximum) when the power required (DV) is a minimum. It is also obvious that the fuel use per amount of distance traveled is a minimum (instantaneous range is a maximum) when the drag is a minimum.
So we again run into our old friends minimum power required and minimum drag as conditions needed for optimum flight. We already know how to find these graphically from power versus velocity plots as shown below. This graphical determination of minimum power and minimum drag speeds is valid for any drag polar, even if not parabolic.
At this point we should pause and say: “Hey, wait just a minute! It was only a couple of pages back that you said that maximum endurance occurred at minimum drag conditions. Now you say it’s maximum range that I get at minimum drag conditions. Make up your mind, for Pete’s sake!”
The problem is that in one case we are talking about jets and the other, prop aircraft. This means that we must be very careful to see which type of plane we are dealing with before starting any calculations. It is very easy to get into a big rush and get the two cases mixed up (especially in the heat of battle on a test or exam!).
Now, as we did for the jet, we can develop integrals to determine the range or endurance for any flight situation. For endurance we have
and for range
6.5 Approximate Solutions for Range and Endurance for a Prop Aircraft
Once more we will assume “quasi‑level” flight and manipulate the terms in our force balance relations to give
\[D = W\left[\dfrac{D}{L}\right] = W\left[\dfrac{C_D}{C_L}\right]\]
This makes the endurance integral
\[\mathrm{E}=-\int_{W_{1}}^{\mathrm{W} 2} \frac{\eta_{\mathrm{p}}}{\gamma_{\mathrm{p}}} \frac{1}{\mathrm{~V}} \frac{\mathrm{C}_{\mathrm{L}}}{\mathrm{C}_{\mathrm{D}}} \frac{\mathrm{d} \mathrm{W}}{\mathrm{W}}\]
Using the straight and level velocity relation
\[\mathrm{V}=\left[2 \mathrm{~W} /\left(\rho \mathrm{SC}_{\mathrm{L}}\right)\right]^{1 / 2}\]
we get
\[\mathbf{E}=-\int_{\mathbf{W} 1}^{\mathrm{W} 2} \frac{\eta_{\mathrm{p}}}{\gamma_{\mathrm{p}}}[\rho \mathrm{S} / 2]^{1 / 2} \frac{\mathrm{C}_{\mathrm{L}}^{3 / 2}}{\mathrm{C}_{\mathrm{D}}} \frac{\mathrm{dW}}{\mathrm{W}^{3 / 2}}\]
The range integral can be written in a similar fashion as
\[R=-\int_{W_{1}}^{W_{2}} \frac{\eta_{P}}{\gamma_{P}} \frac{C_{L}}{C_{D}} \frac{d W}{W}\]
Now we need to consider the same flight schedules examined in the jet case. Constant angle of attack flight will give constant lift and drag coefficients and constant altitude will give constant density. We will also assume constant specific fuel consumption.
For range we need only to use the constant angle of attack assumption to give a simple integral. The resulting range is
\[R=\frac{\eta_{P}}{\gamma_{P}} \frac{C_{L}}{C_{D}} \ln \left(\frac{W_{1}}{W_{2}}\right), \quad(\text { const } \alpha)\]
For endurance we will consider two cases. The first holds both altitude and angle of attack constant, giving
\[E=-\frac{\eta_{P}}{\gamma_{P}} \sqrt{\frac{\rho S}{2}} \frac{C_{L}^{3 / 2}}{C_{D}} \int_{W_{1}}^{W_{2}} \frac{d W}{W^{3 / 2}}\]
which integrates to
\[E=\frac{\eta_{P}}{\gamma_{P}} \sqrt{2 \rho S} \frac{C_{L}^{3 / 2}}{C_{D}}\left(\frac{1}{\sqrt{W_{2}}}-\frac{1}{\sqrt{W_{1}}}\right),\left(\begin{array}{l}
\text { const } \alpha \\
\text { const } \rho
\end{array}\right)\]
The second case has angle of attack and velocity constant
\[E=-\frac{\eta_{P}}{\gamma_{P}} \frac{1}{V} \frac{C_{L}}{C_{D}} \int_{W_{1}}^{W_{2}} \frac{d W}{W}\]
or
\[E=\frac{\eta_{P}}{\gamma_{P}} \frac{1}{V} \frac{C_{L}}{C_{D}} \ln \left(\frac{W_{1}}{W_{2}}\right), \quad\left(\begin{array}{l}
\text { const } \alpha \\
\text { const } V
\end{array}\right)\]
This is the “drift up” flight schedule.
6.6 Wind Effects
The above range and endurance equations for both jet or prop aircraft were derived assuming no atmospheric winds. The speeds in the equations are the airspeeds, not speeds over the ground. If there is a wind the airspeed is, of course, not equal to the speed over the ground.
Endurance calculations are not altered by the presence of an atmospheric wind. If our concern is how long the aircraft can stay in the air at a given airspeed and altitude and we don’t particularly care if it is making progress over the ground we need not worry about winds. We are doing endurance calculations based only on the aerodynamic behavior of the airplane at a given speed and altitude in a mass of air.
Range is related to speed across the ground rather than the airspeed; thus, if there is a wind our range equation results need to be re-evaluated to account for the wind. The logic of this is simple: a headwind will slow progress over the ground and reduce range while a tailwind will increase range. What is not so obvious is how to correct the calculations to account for this wind. Since our usual concern is to find the maximum range , we will examine the correction for wind effects only for this optimum situation.
Maximum range for a jet was found to occur when D / V was a minimum while, for a prop, maximum range occurred at minimum drag conditions. The velocities for both cases can be determined graphically by finding the point of tangency for a line drawn from the zero velocity origin on either the drag versus velocity curve in the jet case or the power required versus velocity curve for the prop plane. We can use an extension of this graphical approach to find the speed for best range with either a head wind or a tail wind.
The important first step in determining optimum range in the presence of an atmospheric wind is to find a new airspeed for best range with a wind. This new speed will then be used to calculate a new value of the optimum range. The new value of best range airspeed is found as illustrated in the figures below. The first task is to draw a conventional drag versus velocity (for a jet) or power required versus velocity (for a prop) plot. To this plot is added a new origin, displaced to the left by the value of a tailwind or to the right by the magnitude of the tailwind. A line is then drawn from the displaced origin, tangent to the drag or power curve and the point of tangency locates the new velocity for optimum range with a wind. The magnitude of this new optimum range velocity is read with respect to the original origin (not the displaced origin). This speed is an airspeed, not a ground speed.
This new optimum range velocity is then used to find a new range value from the same equations developed previously. Using the new velocity, new values of lift and drag coefficients are first calculated and these new coefficients and velocity are used to find the optimum range with the wind. To this new range must be added another range which results purely from the aircraft’s time of exposure (endurance) to the wind. This endurance is also found using the newly found optimum range velocity and associated coefficients. The final corrected range for maximum range in a wind is
R with wind = R Corrected + V w E Corrected for a tailwind
or
R with wind = R Corrected – V w E Corrected for a headwind
6.7 Let the Buyer Beware
Airplane manufacturers, like those of automobiles and other products, like to do anything they can to make their product look good and sometimes they hope that the buyer doesn’t look too closely at the contradictions in their specifications and advertising. A car may be advertised as having seating for five, an EPA fuel economy rating of 38 mpg, the ability to go 542 miles on a single tank of gas and a top speed of 120 miles per hour. Most of us, however, know not to expect that car to go 542 miles on a single tank of gas while carrying 5 people at a speed of 120 mph!
What about airplanes? Is this product of an industry which is regulated at every step by the FAA just as subject to contradiction in specifications as a car?
Let’s look at a few simple examples taken from a general aviation Aircraft Fleet Directory of a few years back. A Cessna 150, the most widely used two place aircraft in the country, quotes a range of 815 nautical miles on 32 gallons (210 pounds) of fuel. The plane has an empty weight (no pilot, passenger, baggage or fuel) of 1104 pounds and a maximum gross takeoff weight of 1600 pounds. This means that with the full fuel tanks needed for maximum range there is only a 286 pound allowance for both pilot and passenger, hardly enough for two adults and luggage! This is why one of the favorite questions of flight examiners who are preparing for a private pilot check‑ride in a Cessna 150 involves weight and balance of the aircraft and why sometimes pilots may have to actually pump fuel out of an airplane before takeoff.
A Cessna 172, the most popular four place aircraft in the world, is a little better than the 150 cited above. It has an empty weight of 1387 pounds and to reach its advertised range of 742 miles it has a fuel tank which holds 288 pounds of gas. This gives a total weight for airplane and fuel of 1675 pounds. The maximum gross takeoff weight of the 172 is 2300 pounds, leaving 625 pounds allowance for four passengers and their stuff; an average of 156 pounds each!
With another Cessna product, the all around best of their 4 seat line , the Skylane, things are a little better. Its listed empty weight of 1707 pounds, range of 979 nautical miles on 474 pounds of fuel and max gross weight of 2950 pounds leave 769 pounds for pilot, passengers and accessories (192 pounds each). Finally an airplane for real people!
Lest the naive get the idea that this is only a problem for small single engine airplanes, let’s look at one more example, the eight place Learjet 25C. It claims a range of 2472 miles, just the ticket for the rich young business tycoon to fill with seven of her closest friends for a transcontinental weekend jaunt. The listed fuel capacity of 7393 pounds, adds to the quoted “zero fuel” weight of 11,400 pounds to give a 18,793 pound airplane. So how much is left for those 8 passengers? The listed max gross weight of the Learjet 25C is 15,000 pounds! With a full tank of gas the airplane is over its maximum allowable takeoff weight! With a 160 pound pilot and no other passengers or payload this airplane can carry enough fuel for a real range of about 1150 miles, less than half that advertised. Why claim a range of almost 2500 miles? Well, the fuel tanks are big enough to carry the needed fuel. If only the airplane could get off the ground!
Homework 6
1. An aircraft weighs 56,000 pounds and has 900ft 2 wing area. Its drag polar equation is given by C D = 0.016 + 0.04C L 2 . The plane has a turbojet engine with constant thrust at any given altitude as shown below:
Table 6.1: Question 1
| altitude (ft) | 0 | 5000 | 10,000 | 15,000 | 20,000 | 25,000 | 30,000 |
| thrust (lb) | 6420 | 5810 | 5200 | 4590 | 4000 | 3360 | 2700 |
a. Find the minimum thrust required for straight and level flight and the corresponding true airspeeds at sea level and at 30,000 ft.
b. Find the minimum power required and the corresponding true airspeeds at sea level and 30,000 ft.
2. For the aircraft above:
a. plot thrust and drag vs V e for straight and level flight.
b. plot altitude vs V emax and V max for straight and level flight.
Figure 6.6: Maximum (& Min) Speed for Straight and Level Flight Versus Altitude
c. find the altitude for maximum true airspeed.
d. find the maximum obtainable altitude.
e. compare V at minimum drag from the plot and the calculation.
f. calculate (L/D) max.
References
Figure 6.1: Kindred Grey (2021). “Finding Velocity for Maximum Range .” CC BY 4.0 . Adapted from James F. Marchman (2004). CC BY 4.0 . Available from https://archive.org/details/6.1-updated
Figure 6.2: Kindred Grey (2021). “Velocities for Minimum Power and Drag.” CC BY 4.0 . Adapted from James F. Marchman (2004). CC BY 4.0 . Available from https://archive.org/details/6.2-updated
Figure 6.3: Kindred Grey (2021). “Speed for Best Range with Wind (Jet).” CC BY 4.0 . Adapted from James F. Marchman (2004). CC BY 4.0 . Available from https://archive.org/details/6.3-updated
Figure 6.4: Kindred Grey (2021). “Speed for Best Range with Wind (Prop).” CC BY 4.0 . Adapted from James F. Marchman (2004). CC BY 4.0 . Available from https://archive.org/details/6.4-updated
Figure 6.5: Kindred Grey (2021). “Thrust and Drag Versus V_e For Straight and Level Flight.” CC BY 4.0 . Adapted from James F. Marchman (2004). CC BY 4.0 . Available from https://archive.org/details/hw-6-part-1
Figure 6.6: Kindred Grey (2021). “Maximum (& Min) Speed for Straight and Level Flight Versus Altitude.” CC BY 4.0 . Adapted from James F. Marchman (2004). CC BY 4.0 . Available from https://archive.org/details/hw-6-part-2
- The term “aero-plane” originally referred to a wing, a geometrically planar surface meant to support a vehicle in flight through the air. By 1903 the term had become associated with the entire flying vehicle. By the 1920s the American press and magazines had changed the word to “airplanes”; however, it is still common in Britain to see “aeroplane” used in books and papers. ↵
<!– pb_fixme –><!– pb_fixme –>


3.4: Probability and Compound Events
- Compute probability in compound events involving "and"
- Compute probability in compound events involving "or"
- Compute probability in compound events involving "and" and "or"
- Use complements for computing the probability of compound events
- Determine probability from two-way tables
- Extend the use of probability rules to other compound events
Review and Preview Probability and Compound Events
In Chapter \(3,\) we have discussed basic probability and counting concepts to prepare us for our future work with inferential statistics. To use the classical approach for determining the probability of an event \(A,\) we need to know the size of the sampe space, \(n,\) and we need to know that each of those outcomes in the sample space is equally likely to occur. Then, if event \(A \) can occur in \(x\) ways in that sample space, \(P\left( A \right)\) \(= \frac{x}{n}.\) We also discussed the empirical/experimental approach where we collect data (preferably a large data set to satisfy our Law of Large Numbers more completely) related to our situation of interest. Then, in that data set of size \(n,\) if \(x\) of the data satisfy the description of event \(A,\) we state \(P\left( A \right)\) \(= \frac{x}{n}.\) However, we must remember that the empirically based probability value only estimates the actual probability when the data is from a sample set. For example, in Section \(2.1,\) we had a relative frequency table on M&M colors built on data from a package of \(55\) M&M's, as shown below.
| Color | Relative Frequency |
|---|---|
| Brown | \(\frac{17}{55}\approx0.309\) |
| Red | \(\frac{18}{55}\approx0.327\) |
| Yellow | \(\frac{7}{55}\approx0.127\) |
| Green | \(\frac{7}{55}\approx0.127\) |
| Blue | \(\frac{2}{55}\approx0.036\) |
| Orange | \(\frac{4}{55}\approx0.073\) |
From this empirical evidence, we have probability estimates for all M&M colors. If we picked a single M&M from the population of all M&M's ever made, we would estimate, empirically, \(P(\small\text{RED }\)\(\small\text{M&M})\) \(\approx0.327\) and \(P(\small\text{BLUE }\)\(\small\text{M&M})\) \(\approx0.073.\)
We have also discussed several counting rules to help us determine the size of an entire sample space and the number of outcomes matching an event description. We established various multiplication rules for counting and an addition rule for counting. Each of these rules has conditions for their use. For example, to use the general permutation rule, we had \(n\) distinct objects in which we were interested in how many ways we could select \(r\) of those objects where the order of selection mattered. Knowing the conditions allowed us to quickly determine if a given event could be counted by a specific counting method. We now extend the multiplication and addition rules of counting to similar multiplication and addition rules for probabilities of compound events.
Probability with Compound Events Involving "and"
As discussed, some event descriptions can be compound: descriptions involving multiple events. We have discussed two phrases used to combine events: "and" as well as "or." For example, in rolling two dice, the event of \(\small{\text{ROLLING }}\)\(\small{\text{A }}\)\(\small{\text{TWO }}\)\(\small{\text{ON }}\)\(\small{\text{THE }}\)\(\small{\text{FIRST }}\)\(\small{\text{DIE}}\)\(\normalsize{\text{ and }}\)\(\small{\text{ROLLING }}\)\(\small{\text{A }}\)\(\small{\text{FIVE }}\)\(\small{\text{ON }}\)\(\small{\text{THE }}\)\(\small{\text{SECOND }}\)\(\small{\text{DIE}}\) is a compound "and" description. As another example, the event of \(\small{\text{PERSON }}\)\(\small{\text{IS }}\)\(\small{\text{OVER }}\)\(\small{\text{SIXTY }}\)\(\small{\text{YEARS }}\)\(\small{\text{OLD}}\)\(\normalsize{\text{ or }}\)\( \small{\text{IS }}\)\(\small{\text{UNDER }}\)\(\small{\text{TWELVE }}\)\(\small{\text{YEARS }}\)\(\small{\text{OLD}}\) is a compound "or" description. The event of \(\small{\text{THREE }}\)\(\small{\text{MANAGERS}}\)\( \normalsize{\text{ and }}\)\( \small{\text{SEVEN }}\)\(\small{\text{HOURLY }}\)\(\small{\text{EMPLOYEES}}\)\( \normalsize{\text{ or }}\)\( \small{\text{FOUR }}\)\(\small{\text{MANAGERS}}\)\(\normalsize{\text{ and }}\)\( \small{\text{SIX }}\)\(\small{\text{HOURLY }}\)\(\small{\text{EMPLOYEES}}\) is a compound description involving both "and" as well as "or" descriptions. Although compound event descriptions may get complicated, we aim to simplify determining the probability of such compound events by developing computation methods similar to our counting methods.
We will deal first with the "and" type of event descriptions. Our previous work with probability established that with fair dice, \(P(\small{\text{TWO }}\)\(\small{\text{ON }}\)\(\small{\text{FIRST }}\)\(\small{\text{DIE}})\) \(=\frac{1}{6}\) and \(P(\small{\text{FIVE }}\)\(\small{\text{ON }}\)\(\small{\text{SECOND }}\)\(\small{\text{DIE}})\) \(=\frac{1}{6}.\) Our work with the sample space of all outcomes of two fair dice showed \(P(\small{\text{TWO }}\)\(\small{\text{ON }}\)\(\small{\text{FIRST }}\)\(\small{\text{DIE}}\)\( \normalsize{\text{ and }}\)\( \small{\text{FIVE }}\)\(\small{\text{ON }}\)\(\small{\text{SECOND }}\)\(\small{\text{DIE}})\) \(=\frac{1}{36}.\) We notice that the product of our two event probabilities, \(\frac{1}{6} \cdot \frac{1}{6}\) \( =\frac{1}{36},\) is the same value as the compound "and" event probability. A multiplication operation happens within probability, just as in counting. Also shown in Section \(3.3,\) we must carefully check if the events separated by "and" depend on each other. Remember the example of the deck of cards \(\small{\text{SPADE }}\)\(\small{\text{FIRST}}\)\(\normalsize{\text{ and }}\)\(\small{\text{BLACK }}\)\(\small{\text{CARD }}\)\(\small{\text{SECOND}}.\) When the events are dependent, the count changes for the second event causing the probability to change. \(P( \small{\text{SPADE }}\)\(\small{\text{FIRST}}\)\(\normalsize{\text{ and }}\)\(\small{\text{BLACK }}\)\(\small{\text{CARD }}\)\(\small{\text{SECOND}})\) \( = \frac{13}{52} \cdot \frac{25}{51}.\) We must notice if the probability of the second event changes when the first event has occurred, as in the card example. With this added condition, we still see multiplication between the two event probabilities. We have a useful general Probability Multiplication Rule . Given events \(A\) and \(B,\) \[\nonumber \begin{align*} P\left( A \text{ and }B\right) &= P \left( A \right) \cdot P \left( B | A \right) \\ &=P \left(B\right) \cdot P\left( A|B\right)\end{align*} \]Recall from Section \(3.3\) that \(B|A\) is read as "event \(B\) given event \(A\) has occurred." Such a probability involving the "given" condition, \( P\left(B|A\right)\) is called a Conditional Probability .
There are events in which \( P\left( B|A\right)\) \(=P\left(B\right)\) or equivalently in which \( P\left( A|B\right)\) \(=P\left(A\right)\); the occurrence of one event does not affect the occurrence of the other event.  In events where \( P\left( B|A\right)\) \(=P\left(B\right),\) we say that the two events \(A\) and \(B\) are independent of each other. But, in our example of selecting a spade card first and then a black card second, the occurrence of the first event did impact the occurrence of the second event. In events where \( P\left( B|A\right) \ne P\left(B\right),\) we say that the two events \(A\) and \(B\) are dependent of each other. The conditional probability analysis is not needed if the independence of the events is already known, we need only multiply the two simple event probabilities together, a Probability Multiplication Rule for Known Independent Events : \[\nonumber \begin{align*} P\left( A \text{ and }B\right) &= P \left( A \right) \cdot P \left( B \right) \\ &=P \left(B\right) \cdot P\left( A \right).\end{align*}\]We reiterate, use of this rule requires us to know the independence of the events involved before computing our compound probability value. We see in later content that the independence of events can be required to apply specific statistical analyses. Generally, it is best to assume events are dependent and be pleasantly surprised when we can show the events are independent.
Answer the following compound probability questions. Recognize that the given event description is a compound event that can be broken down into multiple events.
 What is the probability of getting a \(6\) on the die and an ace on the card draw?
- Answer
-
This is a compound "and" event description. We also notice that the two simple events are independent of each other \[ \nonumber \begin{align*} P\left( A \text{ and } B \right) &= P\left( \small{\text{SIX ON DIE}}\normalsize\text{ and }\small{\text{ACE ON CARD DRAW | SIX ON DIE}}\right) \\ &= P\left( \small{\text{SIX ON DIE}}\normalsize\text{ and }\small{\text{ACE ON CARD DRAW}}\right) \\ &= P\left( \small{\text{SIX ON DIE}}\right) \cdot P\left(\small{\text{ACE ON CARD DRAW}}\right) \\ &=\frac{1}{6} \cdot \frac{4}{52} \\ &= \frac{1}{78} \approx 1.2821\% \end{align*} \]
-
We shuffle a standard deck of playing cards so the cards are randomly placed through the deck.
- We draw two cards. What is the probability of getting two face cards if the first card is replaced randomly in the deck before drawing the second?
- We draw two cards. What is the probability of getting two face cards if the first card is not replaced in the deck before drawing the second?
- Answer
-
- We do not want to build the sample space in this situation: there are \(52 \cdot 52\) \(= 2,704\) different outcomes. We notice that if the first card is replaced randomly in the deck before drawing the second, the two events are independent. We also notice that the deck has \(12\) face cards. By our multiplication rule: \[ \nonumber \begin{align*}P\left( A \text{ and } B \right) &= P \left( \small{\text{FACE CARD ON FIRST DRAW}}\normalsize\text{ and }\small{\text{FACE CARD ON SECOND DRAW}}\right) \\ &= P \left( \small{\text{FACE ON FIRST FIRST}} \right) \cdot P \left( \small{\text{FACE ON SECOND | FACE CARD ON FIRST}} \right) \\ &=P \left( \small{\text{FACE ON FIRST FIRST}} \right) \cdot P \left( \small{\text{FACE ON SECOND}} \right) \\ &\frac{12}{52} \cdot \frac{12}{52} \\ &= \frac{144}{2,704} = \frac{9}{169} \\ &\approx 5.3254\% \end{align*} \]We notice this is not an unusual event.
- If the first card is not replaced in the deck before drawing the second, the second event's probability depends on the first event occurring because the number of face cards still in the deck will be down to \(11.\) The number of cards in the entire deck will be \(51\) By our multiplication rule:\[\nonumber\begin{align*}P\left( A \text{ and } B \right) &= P\left(\small{\text{FACE CARD ON FIRST DRAW}}\normalsize\text{ and }\small{\text{FACE CARD ON SECOND DRAW}}\right) \\ &=P\left(\small{\text{FACE ON FIRST FIRST}}\right)\cdot P\left(\small{\text{FACE ON SECOND | FACE ON FIRST}}\right)\\&=\frac{12}{52}\cdot \frac{11}{51}\\&=\frac{132}{2,652}=\frac{11}{221}\\ &\approx 4.9774\%\end{align*}\]Although the probability measure does not differ from the first situation drastically, it is different. The event is a little less likely to occur if the first card is not replaced; this is something we might have expected.
-
Suppose we have our bag of \(55\) M&M candies with the color distribution as given previously:
Table \(\PageIndex{2}\): Frequencies and Relative Frequencies of Sampled M&M's Color Relative Frequency Brown \(\frac{17}{55}\) Red \(\frac{18}{55}\) Yellow \(\frac{7}{55}\) Green \(\frac{7}{55}\) Blue \(\frac{2}{55}\) Orange \(\frac{4}{55}\) We randomly grab four candies, one at a time, without replacement. What is the probability that we have a red candy first, a green candy second, a brown candy third, and a brown candy fourth?
- Answer
-
These events are not independent. For example, the probability of getting a green candy second will vary depending on whether we had a red candy first. By our multiplication rule:\[\nonumber\begin{align*} P\left(\small{\text{RED }}\normalsize1^{st}\text{ and }\small{\text{GREEN }} \normalsize2^{nd}\text{ and }\small{\text{BROWN }}\normalsize3^{rd} \text{ and } \small{\text{BROWN }}\normalsize 4^{th}\right) &=P\left(\small{\text{RED}}\right)\cdot P\left(\small{\text{GREEN | RED }}\normalsize1^{st} \right)\cdot P \left(\small{\text{BROWN | RED}}\normalsize \text{ and } \small{\text{ GREEN}}\normalsize{\text{ previously}}\right)\\ &~~~\cdot P\left(\small{\text{BROWN | RED}}\normalsize \text{ and } \small{\text{ GREEN}}\normalsize \text{ and } \small{\text{BROWN}}\normalsize{\text{ previously}}\right) \\[5pt]&=\frac{18}{55}\cdot \frac{7}{54} \cdot \frac{17}{53} \cdot \frac{16}{52} \\&=\frac{34,272}{8,185,320}=\frac{476}{113,685} \\ &\approx 0.4187\%\end{align*}\]
-
Senior citizens make up about \(18\%\) of the U.S. adult population, according to the Pew Research Center website.
- What is the probability of randomly selecting two U.S. adults who are both senior citizens?
- What is the probability of randomly selecting two U.S. adults, the first not a senior citizen and the second is a senior citizen?
- What is the probability of randomly selecting three U.S. adults such that the first two are not senior citizens and the third is a senior citizen
- Answer
-
- As the U.S. population of senior citizens is very large, we will assume independence in the events, realizing our probability measures are approximations. By our multiplication rule: \[ \nonumber \begin{align*}P\left( A \text{ and } B \right) &= P \left( \small{\text{SENIOR CITIZEN}}\normalsize\text{ and }\small{\text{SENIOR CITIZEN}}\right) \\ &= P \left( \small{\text{SENIOR CITIZEN}} \right) \cdot P \left( \small{\text{SENIOR CITIZEN}} \right) \\ &= 18\% \cdot 18\% \\ &\approx 3.24\% \end{align*} \]We do emphasize that assumption of independence is not always reasonable, but is used at times when dealing with large population/sample sizes as probability measures do not get impacted significantly in value. We can see this in an example of comparison on, say, a dependence measure of \(\frac{2456783}{34595200} \cdot \frac{2456782}{34595199}\) \(\approx 0.504314831996\% \) versus assumed independence measure of \(\frac{2456783}{34595200} \cdot \frac{2456783}{34595200}\) \(\approx 0.50431502269\%.\) Although the probability values are technically different, the practical interpretation for real-life application would not usually have practical meaning in the difference. We do note that we reflect carefully before assuming independence as it can lead to drastic consequences.
- Again, as the U.S. population is very large, we will assume independence in the events. By our complement rule, we note that about \(82\%\) of the U.S. adult population is less than \(65\) years old. By our multiplication rule: \[ \nonumber \begin{align*} P \left( \small{\text{NOT A SENIOR CITIZEN}}\normalsize\text{ and }\small{\text{SENIOR CITIZEN}}\right) &= P \left( \small{\text{NOT A SENIOR CITIZEN}} \right) \cdot P \left( \small{\text{SENIOR CITIZEN}} \right) \\ &= 82\% \cdot 18\% \\ &\approx 14.76\% \end{align*} \]
- Again, as the U.S. population is very large, we will assume independence in the events. By our complement rule, we note that about \(82\%\) of the U.S. adult population is less than \(65\) years old. By our multiplication rule: \[ \nonumber \begin{align*} P \left( \small{\text{NOT A SENIOR CITIZEN}}\normalsize\text{ and }\small{\text{NOT A SENIOR CITIZEN}}\normalsize\text{ and }\small{\text{SENIOR CITIZEN}}\right) &= 82\% \cdot 82\% \cdot 18\% \\ &\approx 12.10\% \end{align*} \]
Probability with Compound Events Involving "or"
Now that we have handled probability measures on events separated by "and," we turn to probability and events separated by "or." For example, we might ask, "In rolling a single fair die, what is the probability of rolling a two or a three?" In the case of our M&M's, we might ask, "What is the probability of the next M&M we randomly remove from a bag being a blue or an orange M&M?" As discussed in Section \(3.3,\) the "or" compound description is tied to addition. As a reminder, our developed Addition Rule for Counting was given as\[\nonumber \text{# of outcomes in }(A \text{ or }B)=\text{# of outcomes in }A+\text{# of outcomes in }B-\text{# of outcomes in }(A \text{ and }B)\] Our developed rule reminds us to check for the outcomes that matched the descriptions of both events and to subtract the set of the twice-counted outcomes.
Similar to how the Multiplication Rule for Counting extends to the Multiplication Rule for Probability, the Addition Rule for Counting will also extend to an Addition Rule for Probability .\[\nonumber P \left( A \text{ or }B \right)=P \left( A \right) + P \left( B \right) - P \left( A \text{ and }B \right) \]In answering the question, "In rolling a single fair die, what is the probability of rolling a two or a three?", we can apply this addition rule to produce\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{ROLLING A TWO}} \normalsize\text{ or } \small{\text{ROLLING A THREE}} \right) \\ &= P \left( \small{\text{ROLLING A TWO}} \right) + P \left( \small{\text{ROLLING A THREE}} \right) - P \left( \small{\text{ROLLING A TWO}} \normalsize\text{ and } \small{\text{ROLLING A THREE}} \right) \\ &= \frac{1}{6} + \frac{1}{6} - \frac{0}{6} \\ &= \frac{2}{6} = \frac{1}{3} \approx 33.3333\% \end{align*} \]Since the sample space of rolling a single fair die is small in size, one could likely answer this question much faster by knowing that only two of the six outcomes in the sample space meet the description of being "\(\small\text{EITHER }\)\(\small\text{A }\)\(\small\text{TWO }\)\(\normalsize\text{or }\)\(\small\text{A }\)\(\small\text{THREE}.\) But as we see in the text exercises, this addition rule can be very useful when the sample spaces are large.
Answer the following compound probability questions. Recognize that the given event description is a compound event that can be broken down into multiple events.
-
A standard deck of playing cards is well-shuffled for randomness. Determine the following probabilities about a single card draw.
- Find the probability that a randomly drawn card is a king or a queen.
- Find the probability that a randomly drawn card is black or an ace.
- Find the probability that a randomly drawn card is a spade or a face card.
- Answer
-
- This is a compound "or" event description, as the event of interest is a king or queen card.\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{CARD IS A KING}} \normalsize\text{ or } \small{\text{CARD IS A QUEEN}} \right) \\ &= P \left( \small{\text{KING}} \right) + P \left( \small{\text{QUEEN}} \right) - P \left( \small{\text{KING}} \normalsize\text{ and } \small{\text{QUEEN}} \right) \\ &= \frac{4}{52} + \frac{4}{52} - \frac{0}{52} \\ &= \frac{8}{52} = \frac{2}{13} \approx 15.3846\% \end{align*} \]We notice that, because \( P \left( \small{\text{KING}} \normalsize \text{ and } \small{\text{QUEEN}} \right)\) \(= 0\%,\) these two events are mutually exclusive. There is a little over \(15\%\) probability that a random card draw will produce a king or a queen. Although not highly probable, we would not consider this outcome unusual.
- This is a compound "or" event description, as the event of interest is a black card or an ace.\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{CARD IS BLACK}} \normalsize \text{ or } \small{\text{CARD IS AN ACE}} \right) \\ &= P \left( \small{\text{BLACK}} \right) + P \left( \small{\text{ACE}} \right) - P \left( \small{\text{BLACK}} \normalsize\text{ and } \small{\text{ACE}} \right) \\ &= \frac{26}{52} + \frac{4}{52} - \frac{2}{52} \\ &= \frac{28}{52} = \frac{7}{13} \approx 53.8462\% \end{align*} \]We notice that, because \( P ( \small{\text{BLACK}}\)\( \normalsize \text{ and }\)\( \small{\text{ACE}})\) \( \ne 0\%\), these two events are not mutually exclusive. There is almost a \(54\%\) chance that a random card draw will produce a black or an ace card; such an outcome is reasonably probable.
- This is a compound "or" event description, as the event of interest is a card that is a spade or a face card. We note "\(\small{\text{FACE CARD}}\)" is also an "or" event since "\(\small{\text{FACE CARD}}\)" means "\(\small{\text{KING}} \normalsize \text{ or } \small{\text{QUEEN}} \normalsize \text{ or } \small{\text{JACK}}\)". Sometimes, event descriptions can be rephrased to make a compound event more evident as an "and" or an "or" type event\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{CARD IS SPADE}} \normalsize\text{ or } \small{\text{FACE CARD}} \right) \\ &= P \left( \small{\text{SPADE}} \right) + \left( P \left( \small{\text{KING}} \right) + P \left( \small{\text{QUEEN}} \right) + P \left( \small{\text{JACK}} \right) \right)- P \left( \small{\text{SPADE}} \normalsize \text{ and } \small{\text{FACE CARD}} \right) \\ &= \frac{13}{52} + \left( \frac{4}{52} + \frac{4}{52} + \frac{4}{52} \right)- \frac{3}{52} \\ &= \frac{22}{52}\\ &= \frac{11}{26} \approx 42.3077\% \end{align*} \]We again notice, because \( P( \small{\text{SPADE}}\)\( \normalsize \text{ and }\)\( \small{\text{FACE }}\)\(\small{\text{CARD}} )\) \( \ne 0\%\), these two events are not mutually exclusive. There is a little over \(42\%\) probability that a random card draw will produce a spade or a face card.
- An online clothing store has a liberal return policy since customers cannot try on the items before purchasing. From survey data gathered in the return process from their customers, \(21\%\) of all purchased items are returned due to the item being too small/tight, while \(4\%\) are returned due to the item being too big/loose. If a customer-purchased item is randomly selected from all purchases, what is the probability that the item will be returned due to being too big/loose or too small/tight?
- Answer
-
This is a compound "or" event description as the purchased item must be returned due to being too big/loose or too small/tight. We notice that these two events are mutually exclusive; we assume that a clothing item cannot simultaneously be too big and too small\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{TOO BIG/LOOSE}} \normalsize\text{ or } \small{\text{TOO SMALL/TIGHT}} \right) \\ &= P \left( \small{\text{TOO BIG/LOOSE}} \right) + P \left( \small{\text{TOO SMALL/TIGHT}} \right) - P \left( \small{\text{TOO BIG/LOOSE}} \normalsize\text{ and } \small{\text{TOO SMALL/TIGHT}} \right) \\ &= 21\% + 4\% - 0\% = 25\% \end{align*} \]Twenty-five percent of all customer-purchased items are returned due to being either too big/loose or too small/tight. Such a significant return rate for these two issues will likely require the store to find ways to help customers find a better fit for their orders.
-
We return to our bag of \(55\) M&M candies.
Table \(\PageIndex{3}\): Frequencies and Relative Frequencies of Sampled M&M's Color Relative Frequency Brown \(\frac{17}{55}\) Red \(\frac{18}{55}\) Yellow \(\frac{7}{55}\) Green \(\frac{7}{55}\) Blue \(\frac{2}{55}\) Orange \(\frac{4}{55}\)
- What is the probability of randomly drawing one candy of a primary color, red, yellow, or blue?
- What is the probability of randomly drawing one candy that is not of one of those primary colors?
- Answer
-
- By our addition rule and noticing the events are mutually exclusive, we compute\[ \nonumber \begin{align*} P \left( \small{\text{PRIMARY COLOR M&M}} \right) &=P \left( \small{\text{RED M&M}}\normalsize\text{ or }\small{\text{YELLOW M&M}}\normalsize\text{ or }\small{\text{BLUE M&M}}\right) \\ &= P \left( \small{\text{RED}} \right) + P \left( \small{\text{YELLOW}} \right) + P \left( \small{\text{YELLOW}} \right) \\ &= \frac{18}{55} + \frac{7}{55} + \frac{2}{55} \\ &= \frac{27}{55} \approx 49.0909\%. \end{align*} \]
- We could apply our addition rule again relative to the colors of brown, green, and orange. But we notice this is also a complement to the previous event.\[ \nonumber \begin{align*}P \left( \small{\text{NOT PRIMARY COLOR M&M}} \right) &= 1 - P \left( \small{\text{PRIMARY COLOR M&M}} \right) \\ &\approx 1 - 0.490909 \\ &= .509091 = 50.9091\% \end{align*} \]We notice that the probability of randomly selecting a primary color M&M is practically the same as getting a non-primary color. Stated equivalently, the proportion of M&M's is approximately \(50\%\) for each color grouping.
Extended Concepts on Compound Events
As mentioned, we sometimes have event descriptions that include multiple compounding actions. Thankfully, we do not need any extra computation rules for these. However, we do have to read very carefully to properly understand the compound event descriptions and ask for clarification if needed.
We must clarify a small detail about our "AND" rule. In our previous examples using compound "AND" event descriptions, we worked with descriptions involving "sequential trials" of two or more trials: tossing two dice, drawing four M&M candies, and randomly selecting two people. What if we are describing an event with the word "and" but in a single trial: Selecting a single M&M that is both red and brown, drawing a single card that is red and a face card? We use a compound "and," but does our multiplication rule apply in a single trial event? We can often reflect on our sample space and naturally handle the "and" within a single trial. Our general multiplication rule still works, but the two events separated by "and" will often be dependent in a single trial situation. We must carefully consider the conditional probability on the second described event as we apply the multiplication rule.
Let us first examine the probability of selecting a single M&M that is both red and brown from our bag of \(55\) M&M's. Based on the given information (and our experience eating M&M's), we can reason that there are no M&M's that meet that description, so the probability is \(\frac{0}{55}\) \(=0\%.\) By reflecting on the sample space, we determine the probability value; no special probability rule is necessary, but let us see that the multiplication rule still works.\[ \nonumber \begin{align*} P\left(\small{\text{RED }}\normalsize\text{ and }\small{\text{BROWN}} \right) &=P\left(\small{\text{RED}}\right)\cdot P\left(\small{\text{BROWN | RED }} \right) \\&=\frac{18}{55}\cdot \frac{0}{18} \\&=\frac{\cancel{18}}{55}\cdot \frac{0}{\cancel{18}} \\&=\frac{0}{55}=0\%\end{align*} \]We can see that in handling the conditional probability of selecting a single M&M that is \(\small{\text{BROWN}}\) given that we are restricted to the \(\small{\text{RED}}\) M&M's produces the probability value \(\frac{0}{18};\) none of the red M&M's are brown. Our multiplication rule for "AND" still works for even single trial cases. It is acceptable and encouraged to reflect on the sample space and not use our multiplication rule to determine such probability values. We should use sound reasoning, not blind use of formulas, to determine measures.
To see another example, find the probability of drawing a single card that is both red and a face card. By thinking about the sample space of a deck of cards, one might quickly reason that there are six cards in the deck of fifty-two that are red face cards. The probability is \(\frac{6}{52}\) \(=\frac{3}{26}\) \(\approx 11.5385\%.\) Let us check that our multiplication rule produces this value as well.\[ \nonumber \begin{align*} P\left(\small{\text{RED }}\normalsize\text{ and }\small{\text{FACE CARD}} \right) &=P\left(\small{\text{RED}}\right)\cdot P\left(\small{\text{FACE CARD | RED }} \right) \\&=\frac{26}{52}\cdot \frac{6}{26} \\&=\frac{\cancel{26}}{52}\cdot \frac{6}{\cancel{26}} \\&=\frac{6}{52}\approx 11.5385\%\end{align*} \]We can see that in handling the conditional probability of selecting a single \(\small{\text{FACE }}\) \(\small{\text{CARD}}\) given that we are restricted to the \(\small{\text{RED}}\) cards produces the probability value \(\frac{6}{52}\) since six of the red cards are face cards. Using conditional probability with the multiplication rule can be handy in more challenging event descriptions.
Let's try a few single trial "and" compound event exercises.
Answer the following single-trial compound probability questions. Recognize that the given event description is a compound event that can be broken down into multiple events.
- We draw a single card from a well-shuffled standard deck of playing cards. What is the probability of getting a black ace?
- Answer
-
This is a compound "and" event description as the card must be both black and an ace.\[ \nonumber \begin{align*} P\left( \small{\text{BLACK CARD}}\normalsize\text{ and }\small{\text{ACE CARD}}\right) &= P\left( \small{\text{BLACK CARD}}\right) \cdot P\left(\small{\text{ACE CARD | BLACK CARD}}\right) \\ &=\frac{26}{52} \cdot \frac{2}{26} \\ &= \frac{2}{52} =\frac{1}{26} \\ &\approx 3.8462\% \end{align*} \]We could have answered this by reflecting on the sample space of the \(52\) outcomes, realizing there are \(2\) black aces, but our multiplication rule does work.
-
Based on \(2023-2024\) information from the U.S. Center for Disease Control
website
, about \(70\%\) of senior citizens (65 years and older) in the U.S. get the flu vaccine, whereas about \(40%\) of those adults under \(65\) years old get vaccinated. Senior citizens make up about \(18\%\) of the U.S. adult population, according to the Pew Research Center website.
- What is the probability of randomly selecting one U.S. adult who is a senior citizen and has had the flu shot?
- What is the probability of randomly selecting one U.S. adult who is not a senior citizen and has not had the flu shot?
- Answer
-
- We can use our multiplication rule. \[ \nonumber \begin{align*} P \left( \small{\text{SENIOR CITIZEN}}\normalsize\text{ and }\small{\text{FLU SHOT}}\right) &= P \left( \small{\text{SENIOR CITIZEN}} \right) \cdot P \left( \small{\text{FLU SHOT | SENIOR CITIZEN}} \right) \\ &= 18\% \cdot 70\% \\ &= 0.18 \cdot 0.70 \\ &= 0.126 \\ &= 12.6\% \end{align*} \]In \(2023-2024,\) about \(12.6\%\) of the U.S. adult population consisted of senior citizens who had taken the flu shot.
- We can use our multiplication rule. \[ \nonumber \begin{align*}P \left( \small{\text{NOT A SENIOR CITIZEN}}\normalsize\text{ and }\small{\text{NO FLU SHOT}}\right) &= P \left( \small{\text{NOT A SENIOR CITIZEN}} \right) \cdot P \left( \small{\text{NO FLU SHOT | NOT A SENIOR CITIZEN}} \right) \\ &= 82\% \cdot 60\%\\ &= 49.2\% \end{align*} \]In \(2023-2024,\) about \(49.2\%\) of the U.S. adult population consisted of non-senior citizens who had not taken the flu shot. Notice that this tells health officials about the demographics of those who did not have the flu shot.
Now, we focus on investigating our event descriptions with multiple compound events. No new probability calculation rules are necessary; we must apply our current understanding to slightly new contexts. For example, suppose we wish to find the probability of first throwing a fair die with a result of an even number and then drawing a card from a standard deck of playing cards that is either a red or a face card. This compound event description involves both "and" and "or." Reading carefully, we break down the description: a first event involving throwing a fair die, then a second event involving drawing a card. We notice these two events are independent. In the card draw event, we note that drawing a red card and a face card are not mutually exclusive. We have\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{DIE TOSS OF EVEN}} \normalsize\text{ and } \left[ \small{\text{CARD DRAW OF RED}}\normalsize\text{ or }\small{\text{FACE}} \right] \right)\\ &= P \left( \small{\text{DIE TOSS OF EVEN}} \right) \cdot P \left( \small{\text{CARD DRAW RED}} \normalsize\text{ or } \small{\text{FACE}} \right) \\ &= P \left( \small{\text{DIE TOSS OF EVEN}} \right) \cdot \left( P \left( \small{\text{CARD DRAW RED}} \right) + P \left( \small{\text{CARD DRAW FACE}} \right) - P \left( \small{\text{CARD DRAW RED}}\normalsize\text{ and } \small{\text{FACE}} \right) \right) \\ &= \frac{3}{6} \cdot \left( \frac{26}{52} + \frac{12}{52} - \frac{6}{52} \right) \\ &= \frac{1}{2} \cdot \left( \frac{32}{52} \right) \\ &= \frac{1}{2} \cdot \frac{8}{13} = \frac{4}{13} \approx 30.7692\%. \end{align*} \]
Determine the probabilities of each of the following.
- We draw five cards from a well-shuffled standard deck of playing cards. What is the probability of getting a flush (all cards share the same suit)?
- Answer
-
This is a compound "and" event description, as the five cards must all be of one suit. We note that the suit does not matter, so the first card drawn can be of any suit; the remaining cards must match the first. The events are not independent.\[ \nonumber \begin{align*} &\quad \, P\left( \small{\text{FLUSH}} \right) \\ &= P \left(\small{\text{ANY SUIT CARD }}\normalsize1^{st}\text{ and }\small{\text{SAME SUIT CARD }}\normalsize2^{nd} \text{ and }\small{\text{SAME SUIT CARD }}\normalsize3^{rd} \text{ and }\small{\text{SAME SUIT CARD }}\normalsize 4^{th} \text{ and }\small{\text{SAME SUIT CARD }}\normalsize 5^{th}\right) \\ &= P \left(\small{\text{ANY SUIT CARD }}\normalsize1^{st}\right) \cdot P\left(\small{\text{SAME SUIT CARD }}\normalsize2^{nd} | 1^{st}\small{\text{ CARD SUIT}}\right) \cdot P\left(\small{\text{SAME SUIT CARD }}\normalsize3^{rd} | 1^{st},2^{nd}\small{\text{ CARD SUIT}}\right) \\ &\quad \quad \quad \cdot P\left(\small{\text{SAME SUIT CARD }}\normalsize 4^{th} | 1^{st},2^{nd},3^{rd}\small{\text{ CARD SUIT}}\right) \cdot P\left(\small{\text{SAME SUIT CARD }}\normalsize 5^{th} | 1^{st},2^{nd},3^{rd},4^{th}\small{\text{ CARD SUIT}}\right) \\ &=\frac{52}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{9}{48} \\ &= \frac{33}{16660} \approx 0.1981\% \end{align*} \]It should not be surprising that flush of five cards from a well-shuffled deck is a very unlikely event.

- Answer
-
We approach this in two different ways. The first may initially sound easier or more natural, but the second brings us different, valuable insights. Accurate rephrasing of a given situation can make the probability calculation different yet produce the same results.
 Our event description is of the "or" type. Since we can toss a \(6\) and get a head on the coin in one execution of the situation, we note that our events are not mutually exclusive.  Now, we can use our complement and multiplication rule.\[ \nonumber \begin{align*} & \quad \, P \left( 6\small{ \text{ ON DIE}}\normalsize\text{ or }\small{\text{HEAD ON COIN}}\right) \\ &= \normalsize1 - \left( P \left( \small{\text{NOT A }\normalsize6\small\text{ ON DIE}}\normalsize \text{ and } \small{\text{NOT A HEAD ON COIN}}\right) \right) \\ &= \normalsize 1 - \left( P \left( \small{\text{NOT A }\normalsize6\small\text{ ON DIE}} \right) \cdot P \left( \small{\text{NOT A HEAD ON COIN}}\right) \right) \\ &=1 - \left( \frac{5}{6} \cdot \frac{1}{2} \right) \\ &= 1 - \frac{5}{12} = \frac{7}{12} \approx 58.3333\% \end{align*} \]We do notice we get the same answer although computed through a very different approach.
- If we roll a fair die three times, what is the probability that one or more throws will come up with a \(1?\)
- Answer
-
On the surface, this sounds easy, but that changes once we think about the possible ways to have one or more of the three approaches produce a \(1.\) We could have the first toss produce a \(1\) while the other two do not, or we can have the second produce a \(1\) and the other two not, or we can have the first two tosses both produce a \(1\) and the third not, or many other possibilities. We must determine many probabilities to use our rule with "or." However, there is an easier way. Notice the complement of "at least one of the throws will be a \(1\)" is given by "none of the three throws produce a \(1.\)"
We obtain the following by our complement and multiplication rule, utilizing the independence of the die throws.\[ \nonumber \begin{align*} &\quad \, P\left( \small{ \text{AT LEAST ONE - }}\normalsize 1 \small{\text{ AMONG THREE TOSSES}} \right) \\ &=\normalsize 1 - P \left( \small{ \text{NOT }\normalsize 1 \small \text{ ON FIRST DIE}}\normalsize\text{ and }\small{ \text{NOT }\normalsize 1\small \text{ ON SECOND DIE}}\normalsize\text{ and }\small{\text{NOT } \normalsize 1 \small \text{ ON THIRD DIE}} \right) \\ &= 1 - \left[ P \left( \small{ \text{NOT }\normalsize 1 \small \text{ ON FIRST DIE}} \right) \cdot P \left( \small{ \text{NOT }\normalsize 1 \small \text{ ON SECOND DIE}} \right) \cdot P \left( \small{ \text{NOT }\normalsize 1 \small \text{ ON THIRD DIE}} \right) \right] \\ &=1 - \left( \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \right) \\ &= 1 - \frac{125}{216} = \frac{91}{216} \approx 42.1296\% \end{align*} \]
- Based on \(2023-2024\) information from the U.S. Center for Disease Control website , about \(70\%\) of senior citizens (65 years and older) in the U.S. get the flu vaccine, whereas about \(40%\) of those adults under \(65\) years old get vaccinated. Senior citizens make up about \(18\%\) of the U.S. adult population, according to the Pew Research Center website. What is the probability of randomly selecting two U.S. adults who are both senior citizens and recipients of the flu vaccine?
- Answer
-
Due to the large population size, we assume the selection of individuals is independent. We believe that if the first selected adult is a senior citizen, the second selected adult still has a \(18\%\) chance of being a senior citizen; likewise, we assume that the probability that the second person chosen is vaccinated is the same as the first person.
Two-Way Tables and Probability
Sometimes, data frequencies for a collected single variable are disaggregated , separated into mutually exclusive subgroups. For example, quiz data frequencies for a class may be separated by the frequency of those who passed the quiz versus those who did not. Alternatively, perhaps the data is separated into those students who are in extra-curricular school activities and those who are not. If done well, this allows a comparison between subgroups.
Let us go one step further in this discussion: disaggregating data in two ways. These results are displayed in a Two-Way Table or a Contingency Table . We take the quiz data of a class of \(20\) students first given in Section \(2.4\) and separate by passing, \(6\) or above, and non-passing scores, below \(6,\) then by student involvement in extra-curricular school activities resulting in the following two-way table.
| Disaggregation | Passed Quiz | Failed Quiz |
|
Not Involved in Extra-Curricular
School Activities |
\(7\) | \(1\) |
|
Involved in Extra-Curricular
School Activities |
\(9\) | \(3\) |
When reading the table, we must pay attention to the row and column headings. The value \(3\) in the table implies that \(3\) of the \(20\) students were involved in extra-curricular activities and failed the quiz. We often include row and column totals for each disaggregation when working with two-way tables.
| Disaggregation | Passed Quiz | Failed Quiz | Row Totals |
|
Not Involved in Extra-Curricular
School Activities |
\(7\) | \(1\) | \(8\) |
|
Involved in Extra-Curricular
School Activities |
\(9\) | \(3\) | \(12\) |
| Column Totals | \(16\) | \(4\) | \(20\) |
We can see a total of \(20\) student data points. Of those, there were \(16\) that passed the quiz, while there were \(8\) not involved in extra-curricular activities.
We can now answer several probability/proportion questions concerning these results. What is the probability that a randomly selected student from the class was involved in extra-curricular school activities? By our classical probability approach, \(P( \small{\text{IN }}\)\(\small{\text{EXTRA }}\)\(\small{\text{CURRICULAR}})\) \( = \frac{12}{20}\) \( = \frac{3}{5}\) \(=60\%.\)
What is the probability that a randomly selected student from the class failed the quiz and was not involved in extra-curricular activities? Since randomly selecting a student from the twenty, \(P( \small{\text{FAILED}}\)\(\small{\text{ QUIZ}} \normalsize \text{ and } \small{\text{NOT }}\)\(\small{\text{IN }}\)\(\small{\text{EXTRA }}\)\(\small{\text{CURRICULAR}} ) \) \(= \frac{1}{20}\) \(= 5\%.\) This is because the table shows only \(1\) student in both the \(\small{\text{FAILED }}\)\(\small{\text{ QUIZ}}\) column as well as the \( \small{\text{NOT }}\)\(\small{\text{ IN }}\)\(\small{\text{ EXTRA }}\)\(\small{\text{ CURRICULAR}} \) row. We could also apply our multiplication rule, noting the events are not independent.\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{FAILED QUIZ}} \normalsize\text{ and } \small{\text{NOT IN EXTRA CURRICULAR}} \right) \\ &= P \left( \small{\text{FAILED QUIZ}} \right) \cdot P \left( \small{\text{NOT IN EXTRA CURRICULAR | FAILED QUIZ}} \right) \\ &=\frac{4}{20} \cdot \frac{1}{4} = \frac{1}{20} = 5\% \end{align*} \]In this situation, the first approach is much easier due to the table information.
Answer the following probability questions about a given two-way table.
-
A polygraph device is being studied for its accuracy in lie detection. A group of \(200\) randomly drawn participants were divided into two groups: those instructed not to lie and those who were supposed to lie. The technician running the device did not know each participant's group. Each participant was tested with the device, and a positive or negative detection of a lie was returned. The results are given in the following contingency table.
As an example, \(71\) of the \(200\) participants did not lie, and the detector did not sense a lie occurring.Disaggregation Positive Lie
Detection ReadingNegative Lie
Detection ReadingParticipant Did Not Lie \(9\) \(71\) Participant Did Lie \(85\) \(35\)
- What is the probability that a randomly selected participant had a positive lie detection reading?
- What proportion of the participants had a false positive reading (the device detected a lie, but the participant did not lie)?
- What is the probability of randomly selecting two participants in sequence so that both had a positive lie detection reading?
- What is the probability of randomly selecting a participant who lied but had a negative detection reading or who did not lie but had a positive detection reading? In other words, for what proportion of the trials was the device inaccurate?
- What is the probability of randomly selecting two participants in sequence so that the first lied with a positive detection reading and the second did not lie with a negative detection reading?
- Answer
-
We first add column and row totals to our table.
Disaggregation Positive Lie
Detection ReadingNegative Lie
Detection ReadingRow Totals Participant Did Not Lie \(9\) \(71\) \(80\) Participant Did Lie \(85\) \(35\) \(120\) Column Totals \(94\) \(106\) \(200\) Now, we answer the probability/proportion questions using appropriate probability rules.
- From the two-way table, we notice there were \(94\) participants that had positive lie detection readings. So, \( P( \small{\text{POSITIVE }}\)\(\small{\text{ DETECTION}})\) \( = \frac{94}{200}\) \( = \frac{47}{100}\) \( = 47\%.\)
-
From the table, we can notice there were \(9\) participants had a false positive reading. So, we can quickly produce results of \( P( \small{\text{FALSE }}\)\(\small{\text{POSITIVE }}\)\(\small{\text{DETECTION}})\) \( = \frac{9}{200} \) \(= 4.5\%.\)
We instead might have noticed this was an "and" compound event: \( \small{\text{DID }}\)\(\small{\text{NOT }}\)\(\small{\text{LIE}}\normalsize\text{ and } \small{\text{POSITIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION}}\). Therefore, it would also have been appropriate (though more cumbersome) to measure the proportion using our general multiplication rule.\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{DID NOT LIE}} \normalsize\text{ and } \small{\text{POSITIVE LIE DETECTION}} \right) \\ &= P \left( \small{\text{DID NOT LIE}} \right) \cdot P \left( \small{\text{POSITIVE LIE DETECTION | DID NOT LIE}} \right) \\ &=\frac{80}{200} \cdot \frac{9}{80} \\ &= \frac{9}{200} = 4.5\% \end{align*} \]
We mention two items in reflection of this exercise. First, we notice how the table is used for a conditional probability such as \( P( \small{\text{POSITIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION | DID }}\)\(\small{\text{NOT }}\)\(\small{\text{LIE}})\) \(=\frac{9}{80}.\) The given condition of \(\small{\text{DID }}\)\(\small{\text{NOT }}\)\(\small{\text{LIE}}\) requires us to use only the information in the table row of "Participants Did Not Lie."
This exercise demonstrates that multiple approaches might be taken to a probability/proportion question, but there is only one correct answer. What is important is to have sound reasoning in our approach. After that, experience is the best teacher in finding the easiest approaches without making mistakes in our reasoning.
- We notice the event description can be re-worded as \( \small{\text{FIRST }}\)\(\small{\text{PARTICIPANT }}\)\(\small{\text{HAD }}\)\(\small{\text{POSITIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION}} \normalsize \text{ and } \small{\text{SECOND }}\)\(\small{\text{PARTICIPANT }}\)\(\small{\text{HAD }}\)\(\small{\text{POSITIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION}}\), so we use the general multiplication rule.\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{FIRST PARTICIPANT HAD POSITIVE LIE DETECTION}} \normalsize \text{ and } \small{\text{SECOND PARTICIPANT HAD POSITIVE LIE DETECTION}} \right) \\ &= P \left( \small{\text{FIRST HAD POSITIVE DETECTION}} \right) \cdot P \left( \small{\text{SECOND HAD POSITIVE DETECTION | FIRST HAD POSTIVE DETECTION}} \right) \\ &=\frac{94}{200} \cdot \frac{93}{199} \\ &= \frac{8,742}{39,800} = \frac{4371}{19,900} \approx 21.9648\% \end{align*} \]We notice how these sequential events are not independent since we cannot pick the same participant twice.
- We notice the event description appears to primarily be an "or" compound event. That is, our event of interest requires selection of one participant that \( \small{\text{LIED }}\)\(\small{\text{WITH }}\)\(\small{\text{NEGATIVE }}\)\(\small{\text{DETECTION}} \normalsize \text{ or } \small{\text{DID }}\)\(\small{\text{NOT }}\)\(\small{\text{LIE }}\)\(\small{\text{WITH }}\)\(\small{\text{POSITIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION}},\) so we use the general addition rule noting these two events are mutually exclusive.\[ \nonumber \begin{align*} &\quad \, P \left( \small{\text{LIED WITH NEGATIVE DETECTION}} \normalsize \text{ or } \small{\text{DID NOT LIE WITH POSITIVE LIE DETECTION}} \right) \\ &= P \left( \small{\text{LIED WITH NEGATIVE DETECTION}} \right) + P \left( \small{\text{DID NOT LIE WITH POSITIVE LIE DETECTION}} \right) \\ &=\frac{35}{200} + \frac{9}{200} \\ &= \frac{44}{200} = \frac{11}{50} = 22\% \end{align*} \]We observe that a test with a \(22\%\) inaccuracy rate is not very dependable. Also, we excluded the subtraction in the general addition rule for possible double counting of \( \small{\text{LIED }}\)\(\small{\text{WITH }}\)\(\small{\text{NEGATIVE }}\)\(\small{\text{DETECTION}} \normalsize \text{ and } \small{\text{DID }}\)\(\small{\text{NOT }}\)\(\small{\text{LIE }}\)\(\small{\text{WITH }}\)\(\small{\text{POSITIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION}} \) since we recognized the events were mutually exclusive.
- The event description is an "and" compound type as we want sequential selection of two participants with \( \small{\text{FIRST }}\)\(\small{\text{PARTICIPANT }}\)\(\small{\text{LIED }}\)\(\small{\text{WITH }}\)\(\small{\text{POSITIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION}} \normalsize \text{ and } \small{\text{SECOND }}\)\(\small{\text{PARTICIPANT }}\)\(\small{\text{DID }}\)\(\small{\text{NOT }}\)\(\small{\text{LIE }}\)\(\small{\text{WITH }}\)\(\small{\text{NEGATIVE }}\)\(\small{\text{LIE }}\)\(\small{\text{DETECTION}}\).


11.2: Solutions by Graphing
Systems of Equations
Systems of Equations
A collection of two linear equations in two variables is called a system of linear equations in two variables , or more briefly, a system of equations . The pair of equations
\(\left\{\begin{array}{r}
5 x-2 y=5 \\
x+y=8
\end{array}\right.\)
is a system of equations. The brace { is used to denote that the two equations occur together (simultaneously).
Solution to A System of Equations
Solution to a System
We know that one of the infinitely many solutions to one linear equation in two variables is an ordered pair. An ordered pair that is a solution to both of the equations in a system is called a solution to the system of equations . For example, the ordered pair \((3, 5)\) is a solution to the system
\(\left\{\begin{array}{r}
5 x-2 y=5 \\
x+y=8
\end{array}\right.\)
since \((3, 5)\) is a solution to both equations.
\(\begin{array}{flushleft}
5x - 2y &= 5 & & x + y &= 8\\
5(3) - 2(5) &= 5 & \text{ Is this correct? } & 3 + 5 &= 8 & \text{ Is this correct? }\\
15 - 10 &= 5 & \text{ Is this correct? } & 8 &= 8 & \text{ Yes, this is correct. }\\
5 &= 5 & \text{ Yes, this is correct. }
\end{array}\)
Graphs of Systems of Equations
One method of solving a system of equations is by graphing. We know that the graph of a linear equation in two variables is a straight line. The graph of a system will consist of two straight lines. When two straight lines are graphed, one of three possibilities may result.
The lines intersect at the point \(a, b\). This point \((a, b)\) is the solution to the corresponding system.
The lines are parallel. They do not intersect. The system has no solution.
The lines are coincident (one on the other). They intersect at infinitely many points. The system has infinitely many solutions.
Independent, Inconsistent, and Dependent Systems
Independent Systems
Systems in which the lines intersect at precisely one point are called
independent systems
. In applications, independent systems can arise when the collected data are accurate and complete. For example,
The sum of two numbers is 10 and the product of the two numbers is 21. Find the numbers.
In this application, the data are accurate and complete. The solution is 7 and 3.
Inconsistent Systems
Systems in which the lines are parallel are called
inconsistent systems
. In applications, inconsistent systems can arise when the collected data are contradictory. For example,
The sum of two even numbers is 30 and the difference of the same two numbers is 0. Find the numbers.
The data are contradictory. There is no solution to this application.
Dependent Systems
Systems in which the lines are coincident are called
dependent systems
. In applications, dependent systems can arise when the collected data are incomplete. For example.
The difference of two numbers is 9 and twice one number is 18 more than twice the other.
The data are incomplete. There are infinitely many solutions.
The Method of Solving A System Graphically
To solve a system of equations graphically: Graph both equations.
- If the lines intersect, the solution is the ordered pair that corresponds to the point of intersection. The system is independent.
- If the lines are parallel, there is no solution. The system is inconsistent.
- If the lines are coincident, there are infinitely many solutions. The system is dependent.
Sample Set A
Solve each of the following systems by graphing.
\(\left\{\begin{array}{r}
2x + y = 5 \\
x+y=2
\end{array}\right.\)
Write each equation in slope-intercept form.
1) \(y = 2x + 5\)
2) \(y = -x + 2\)
Graph each of these equations:
The lines appear to intersect at the point \((-1, 3)\). The solution to this system is \((-1, 3)\), or
\(x = -1, y = 3\).
Check: Substitute \(x - 01, y = 3\) into each equation.
1)
\(\begin{array}{flushleft}
-2x + y &= 5\\
-2(-1) + 3 &= 5 & \text{ Is this correct? }\\
2 + 3 &= 5 & \text{ Is this correct? } \\
5 &= 5 & \text{ Yes, this is correct }
\end{array}\)
2)
\(\begin{array}{flushleft}
x + y &= 2\\
-1 + 3 &= 2 & \text{ Is this correct? }\\
2 &= 2 & \text{ Yes, this is correct. }
\end{array}\)
\(\left\{\begin{array}{r}
-x + y = -1 \\
-x + y = 2
\end{array}\right.\)
Write each equation in slope-intercept form.
1) \(y = 2x + 5\)
2) \(y = -x + 2\)
Graph each of these equations.
These lines are parallel. This system has no solution. We denote this fact by writing
inconsistent
.
We are sure that these lines are parallel because we notice that they have the same slope, \(m = 1\) for both lines. The lines are not coincident because the y-intercepts are different.
\(\left\{\begin{array}{r}
-2x + 3y -2 \\
-6x + 9y = -6
\end{array}\right.\)
Write each equation in slope-intercept form.
1) \(y = \dfrac{2}{3}x - \dfrac{2}{3}\)
2) \(y = \dfrac{2}{3}x - \dfrac{2}{3}\)
Both equations are the same. This system has infinitely many solutions. We write dependent .
Practice Set A
Solve each of the following systems by graphing. Write the ordered pair solution or state that the system is inconsistent, or dependent.
\(\left\{\begin{array}{r}
2x + y = 1 \\
-x + y = -5
\end{array}\right.\)
- Answer
-
\(x=2,y=−3\)
\(\left\{\begin{array}{r}
-2x + 3y = 6 \\
6x - 9y = -18
\end{array}\right.\)
- Answer
-
dependent
\(\left\{\begin{array}{r}
3x + 5y = 15 \\
9x + 15y = 15
\end{array}\right.\)
- Answer
-
inconsistent
\(\left\{\begin{array}{r}
y = -3 \\
x + 2y = -4
\end{array}\right.\)
- Answer
-
\(x=2,y=−3\)
Exercises
For the following problems, solve the systems by graphing. Write the ordered pair solution, or state that the system is inconsistent or dependent.
\(\left\{\begin{array}{r}
x + y = -5 \\
-x + y = 1
\end{array}\right.\)
- Answer
-
\((−3,−2)\)
\(\left\{\begin{array}{r}
x + y = 4 \\
x + y = 0
\end{array}\right.\)
\(\left\{\begin{array}{r}
-3x + y = 5 \\
-x + y = 3
\end{array}\right.\)
- Answer
-
\((−1,2)\)
\(\left\{\begin{array}{r}
x - y = -6 \\
x + 2y = 0
\end{array}\right.\)
\(\left\{\begin{array}{r}
3x + y = 0 \\
4x - 3y = 12
\end{array}\right.\)
- Answer
-
\((\dfrac{12}{13}, -\dfrac{36}{13})\)
\(\left\{\begin{array}{r}
-4x + y = 7 \\
-3x + y = 2
\end{array}\right.\)
\(\left\{\begin{array}{r}
2x + 3y = 6 \\
3x + 4y = 6
\end{array}\right.\)
- Answer
-
These coordinates are hard to estimate. This problem illustrates that the graphical method is not always the most accurate.
\((−6,6)\)
\(\left\{\begin{array}{r}
x + y = -3 \\
4x + 4y = -12
\end{array}\right.\)
\(\left\{\begin{array}{r}
2x - 3y = 1 \\
4x - 6y = 4
\end{array}\right.\)
- Answer
-
inconsistent
\(\left\{\begin{array}{r}
x + 2y = 3 \\
-3x - 6y = -9
\end{array}\right.\)
\(\left\{\begin{array}{r}
x - 2y = 6 \\
3x - 6y = 18
\end{array}\right.\)
- Answer
-
dependent
\(\left\{\begin{array}{r}
2x + 3y = 6 \\
-10x - 15y = 30
\end{array}\right.\)
Exercises For Review
Express \(0.000426\) in scientific notation.
- Answer
-
\(4.26 + 10^{-4}\)
Find the product \((7x - 3)^2\).
Supply the missing word. The _____ of a line is a measure of the steepness of the line.
- Answer
-
slope
Supply the missing word. An equation of the form \(ax^2 + bx + c = 0\), a \not = 0\), is called a equation.
Construct the graph of the quadratic equation \(y = x^2 - 3\)
- Answer
-


12.8.7: Sample of Re-engineering
selected template will load here
This action is not available.
The process of purchasing the correct textbooks in a timely manner for college classes has always been problematic. And now, with online bookstores such as Amazon competing directly with the college bookstore for students’ purchases, the college bookstore is under pressure to justify its existence.
But college bookstores have one big advantage over their competitors: they have access to students’ data. In other words, once a student has registered for classes, the bookstore knows exactly what books that student will need for the upcoming term. To leverage this advantage and take advantage of new technologies, the bookstore wants to implement a new process that will make purchasing books through the bookstore advantageous to students. Though they may not be able to compete on price, they can provide other advantages, such as reducing the time it takes to find the books and the ability to guarantee that the book is the correct one for the class. In order to do this, the bookstore will need to undertake a process redesign.
The goal of the process redesign is simple: capture a higher percentage of students as customers of the bookstore. After diagramming the existing process and meeting with student focus groups, the bookstore comes up with a new process. In the new process, the bookstore utilizes information technology to reduce the amount of work the students need to do in order to get their books. In this new process, the bookstore sends the students an e-mail with a list of all the books required for their upcoming classes. By clicking a link in this e-mail, the students can log into the bookstore, confirm their books, and purchase the books. The bookstore will then deliver the books to the students.