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3.6 The Multiplicative Rule and Independent Events

The probability of an intersection of two events can be calculated with the multiplicative rule, which employs the conditional probabilities we defined in the previous section. Actually, we’ve already developed the formula in another context. Recall that the conditional probability of B given A is

[&P|pbo|B|pipe|A|pbc||=|*frac*{P|pbo|A|inter|B|pbc|}{P|pbo|A|pbc|} &]

P (B | A) = \frac{P (A \cap B)}{P (A)}

$P (B | A) = \frac{P (A \cap B)}{P (A)}$

Multiplying both sides of this equation by P(A), we obtain a formula for the probability of the intersection of events A and B. This formula is often called the multiplicative rule of probability.

Multiplicative Rule of Probability

$P (A \cap B) = P (A) P (B | A)$ $P (A \cap B) = P (A) P (B | A)$ or, equivalently, $P (A \cap B) = P (B) P (A | B)$ $P (A \cap B) = P (B) P (A | B)$

Example 3.16 The Multiplicative Rule—a Famous Psychological Experiment

Problem

In a classic psychology study conducted in the early 1960s, Stanley Milgram performed a series of experiments in which a teacher is asked to shock a learner who is attempting to memorize word pairs whenever the learner gives the wrong answer. The shock levels increase with each successive wrong answer. (Unknown to the teacher, the shocks are not real.) Two events of interest are

[&A:|thn||cbo|~rom~The teacher |odquote|applies|cdquote| a severe shock|thn||pbo|~normal~450|thn|~rom~volts|pbc|~normal~.|cbc| &]

[&B:|thn||cbo|~rom~The learner protests verbally prior to receiving the shock~normal~.|cbc| &]
$\begin{array}{l} A : {The teacher “ applies ” a severe shock (450 volts) .} \\ B : {The learner protests verbally prior to receiving the shock .} \end{array}$ $\begin{array}{l} A : {The teacher “ applies ” a severe shock (450 volts) .} \\ B : {The learner protests verbally prior to receiving the shock .} \end{array}$

A recent application of Milgram’s shock study revealed that $P (B) = .5$ $P (B) = .5$ and $P (A | B) = .7 .$ $P (A | B) = .7 .$ On the basis of this information, what is the probability that a learner will protest verbally and a teacher will apply a severe shock? That is, find $P (A \cap B) .$ $P (A \cap B) .$

Solution

We want to calculate $P (A \cap B) .$ $P (A \cap B) .$ Using the formula for the multiplicative rule, we obtain

[&P|pbo|A|inter|B|pbc||=|P|pbo|B|pbc| P|pbo|A|pipe|B|pbc||=||pbo|.5|pbc||pbo|.7|pbc||=|.35 &]
$P (A \cap B) = P (B) P (A | B) = (.5) (.7) = .35$ $P (A \cap B) = P (B) P (A | B) = (.5) (.7) = .35$

Thus, about 35% of the time, the learner will give verbal protest and the teacher will apply a severe shock.

Look Back

The multiplicative rule can be expressed in two ways: $P (A \cap B) =$ $P (A \cap B) =$ $P (A) P (B | A),$ $P (A) P (B | A),$ or $P (A \cap B) = P (B) P (A | B) .$ $P (A \cap B) = P (B) P (A | B) .$ Select the formula that involves a given event for which you know the probability (e.g., event B in the example).

Now Work Exercise 3.81

Intersections often contain only a few sample points. In this case, the probability of an intersection is easy to calculate by summing the appropriate sample point probabilities. However, the formula for calculating intersection probabilities is invaluable when the intersection contains numerous sample points, as the next example illustrates.

Example 3.17 Applying The Multiplicative Rule—Study of Welfare Workers

Problem

A county welfare agency employs 10 welfare workers who interview prospective food stamp recipients. Periodically, the supervisor selects, at random, the forms completed by two workers and subsequently audits them for illegal deductions. Unknown to the supervisor, three of the workers have regularly been giving illegal deductions to applicants. What is the probability that both of the workers chosen have been giving illegal deductions?

Solution

Define the following two events:

[&*AS*A: *AP*|cbo|~rom~First worker selected gives illegal deductions~normal~.|cbc| &]

[&*AS*B: *AP*|cbo|~rom~Second worker selected gives illegal deductions~normal~.|cbc| &]
$\begin{array}{l} A : {First worker selected gives illegal deductions .} \\ B : {Second worker selected gives illegal deductions .} \end{array}$ $\begin{array}{l} A : {First worker selected gives illegal deductions .} \\ B : {Second worker selected gives illegal deductions .} \end{array}$

We want to find the probability that both workers selected have been giving illegal deductions. This event can be restated as ${First worker gives illegal deductions a n d second worker gives illegal deductions .} .$ ${First worker gives illegal deductions a n d second worker gives illegal deductions .} .$ Thus, we want to find the probability of the intersection $A \cap B .$ $A \cap B .$ Applying the multiplicative rule, we have

[&P|pbo|A|inter|B|pbc||=|P|pbo|A|pbc|P|pbo|B|pipe|A|pbc| &]
$P (A \cap B) = P (A) P (B | A)$ $P (A \cap B) = P (A) P (B | A)$

To find P(A), it is helpful to consider the experiment as selecting 1 worker from the 10. Then the sample space for the experiment contains 10 sample points (representing the 10 welfare workers), in which the 3 workers giving illegal deductions are denoted by the symbol $I (I_{1}, I_{2}, I_{3})$ $I (I_{1}, I_{2}, I_{3})$ and the 7 workers not giving illegal deductions are denoted by the symbol $N (N_{1}, \dots, N_{7}) .$ $N (N_{1}, \dots, N_{7}) .$ The resulting Venn diagram is illustrated in Figure 3.16.

Figure 3.16

Venn diagram for finding P(A)

Since the first worker is selected at random from the 10, it is reasonable to assign equal probabilities to the 10 sample points. Thus, each sample point has a probability of $\frac{1}{10}$ $\frac{1}{10}$ . The sample points in event A are ${I_{1}, I_{2}, I_{3}}$ ${I_{1}, I_{2}, I_{3}}$ —the three workers who are giving illegal deductions. Thus,

[&P|pbo|A|pbc||=|P|pbo|I_{1}|pbc||+|P|pbo|I_{2}|pbc||+|P|pbo|I_{3}|pbc||=|*frac*{1}{10}|+|*frac*{1}{10}|+|*frac*{1}{10}|=|*frac*{3}{10} &]
$P (A) = P (I_{1}) + P (I_{2}) + P (I_{3}) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}$ $P (A) = P (I_{1}) + P (I_{2}) + P (I_{3}) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}$

To find the conditional probability $P (B | A),$ $P (B | A),$ we need to alter the sample space S. Since we know that A has occurred (i.e., the first worker selected is giving illegal deductions), only 2 of the 9 remaining workers in the sample space are giving illegal deductions. The Venn diagram for this new sample space is shown in Figure 3.17.

Figure 3.17
Venn diagram for finding $P (B | A)$ $P (B | A)$

Each of these nine sample points is equally likely, so each is assigned a probability of $\frac{1}{9} .$ $\frac{1}{9} .$ Since the event $B | A$ $B | A$ contains the sample points ${I_{1}, I_{2}},$ ${I_{1}, I_{2}},$ we have

[&P|pbo|B|pipe|A|pbc||=|P|pbo|I_{1}|pbc||+|P|pbo|I_{2}|pbc||=|*frac*{1}{9}|+|*frac*{1}{9}|=|*frac*{2}{9} &]
$P (B | A) = P (I_{1}) + P (I_{2}) = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}$ $P (B | A) = P (I_{1}) + P (I_{2}) = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}$

Substituting $P (A) = \frac{3}{10}$ $P (A) = \frac{3}{10}$ and $P (B | A) = \frac{2}{9}$ $P (B | A) = \frac{2}{9}$ into the formula for the multiplicative rule, we find that

[&P|pbo|A|inter|B|pbc||=|P|pbo|A|pbc|P|pbo|B|pipe|A|pbc||=||3(|*frac*{3}{10}|3)||3(|*frac*{2}{9}|3)||=|*frac*{6}{90}|=|*frac*{1}{15} &]
$P (A \cap B) = P (A) P (B | A) = (\frac{3}{10}) (\frac{2}{9}) = \frac{6}{90} = \frac{1}{15}$ $P (A \cap B) = P (A) P (B | A) = (\frac{3}{10}) (\frac{2}{9}) = \frac{6}{90} = \frac{1}{15}$

Thus, there is a 1-in-15 chance that both workers chosen by the supervisor have been giving illegal deductions to food stamp recipients.

Look Back

The key words both and and in the statement “both A and B occur” imply an intersection of two events. This, in turn, implies that we should multiply probabilities to obtain the answer.

Now Work Exercise 3.77

The sample-space approach is only one way to solve the problem posed in Example 3.17. An alternative method employs the tree diagram (introduced in Example 3.1). Tree diagrams are helpful for calculating the probability of an intersection.

Figure 3.18
Tree diagram for Example 3.17

To illustrate, a tree diagram for Example 3.17 is displayed in Figure 3.18. The tree begins at the far left with two branches. These branches represent the two possible outcomes N (no illegal deductions) and I (illegal deductions) for the first worker selected. The unconditional probability of each outcome is given (in parentheses) on the appropriate branch. That is, for the first worker selected, $P (N) = \frac{7}{10}$ $P (N) = \frac{7}{10}$ and $P (I) = \frac{3}{10} .$ $P (I) = \frac{3}{10} .$ (These unconditional probabilities can be obtained by summing sample point probabilities as in Example 3.17.)

The next level of the tree diagram (moving to the right) represents the outcomes for the second worker selected. The probabilities shown here are conditional probabilities, since the outcome for the first worker is assumed to be known. For example, if the first worker is giving illegal deductions (I), the probability that the second worker is also giving illegal deductions (I) is $\frac{2}{9}$ $\frac{2}{9}$ , because, of the 9 workers left to be selected, only 2 remain who are giving illegal deductions. This conditional probability, $\frac{2}{9},$ $\frac{2}{9},$ is shown in parentheses on the bottom branch of Figure 3.18.

Finally, the four possible outcomes of the experiment are shown at the end of each of the four tree branches. These events are intersections of two events (outcome of first worker and outcome of second worker). Consequently, the multiplicative rule is applied to calculate each probability, as shown in Figure 3.18. You can see that the intersection ${I \cap I}$ ${I \cap I}$ —the event that both workers selected are giving illegal deductions—has probability $\frac{6}{90} = \frac{1}{15},$ $\frac{6}{90} = \frac{1}{15},$ which is the same as the value obtained in Example 3.17.

In Section 3.5, we showed that the probability of event A may be substantially altered by the knowledge that an event B has occurred. However, this will not always be the case; in some instances, the assumption that event B has occurred will not alter the probability of event A at all. When this occurs, we say that the two events A and B are independent events.

Events A and B are independent events if the occurrence of B does not alter the probability that A has occurred; that is, events A and B are independent if

P (A | B) = P (A)

$P (A | B) = P (A)$

When events A and B are independent, it is also true that

P (B | A) = P (B)

$P (B | A) = P (B)$

Events that are not independent are said to be dependent.

Example 3.18 Checking for Independence— Die-Tossing Experiment

Problem

Consider the experiment of tossing a fair die, and let

[&*AS*A*AP*|=||cbo|~rom~Observe an even number~normal~.|cbc| &]

[&*AS*B*AP*|=||cbo|~rom~Observe a number less than or equal to~normal~ 4.|cbc| &]
$\begin{array}{l} A & = & {Observe an even number .} \\ B & = & {Observe a number less than or equal to 4.} \end{array}$ $\begin{array}{l} A & = & {Observe an even number .} \\ B & = & {Observe a number less than or equal to 4.} \end{array}$

Are A and B independent events?

Solution

The Venn diagram for this experiment is shown in Figure 3.19. We first calculate

Figure 3.19

Venn diagram for die-toss experiment

[&*AS*P|pbo|A|pbc|*AP*|=|P|pbo|2|pbc||+|P|pbo|4|pbc||+|P|pbo|6|pbc||=|*frac*{1}{2} &]

[&*AS*P|pbo|B|pbc|*AP*|=|P|pbo|1|pbc||+|P|pbo|2|pbc||+|P|pbo|3|pbc||+|P|pbo|4|pbc||=|*frac*{4}{6}|=|*frac*{2}{3} &]

[&*AS*P|pbo|A|inter|B|pbc|*AP*|=|P|pbo|2|pbc||+|P|pbo|4|pbc||=|*frac*{2}{6}|=|*frac*{1}{3} &]
$\begin{array}{l} P (A) & = & \begin{matrix} P (2) + P (4) + P (6) = \frac{1}{2} \end{matrix} \\ P (B) & = & \begin{matrix} P (1) + P (2) + P (3) + P (4) = \frac{4}{6} = \frac{2}{3} \end{matrix} \\ P (A \cap B) & = & \begin{matrix} P (2) + P (4) = \frac{2}{6} = \frac{1}{3} \end{matrix} \end{array}$ $\begin{array}{l} P (A) & = & \begin{matrix} P (2) + P (4) + P (6) = \frac{1}{2} \end{matrix} \\ P (B) & = & \begin{matrix} P (1) + P (2) + P (3) + P (4) = \frac{4}{6} = \frac{2}{3} \end{matrix} \\ P (A \cap B) & = & \begin{matrix} P (2) + P (4) = \frac{2}{6} = \frac{1}{3} \end{matrix} \end{array}$

Now, assuming that B has occurred, we see that the conditional probability of A given B is

[&P|pbo|A|pipe|B|pbc||=|*frac*{P|pbo|A|inter|B|pbc|}{P|pbo|B|pbc|}|=|*frac*{^{1}|sfrac|_{3}}{^{2}|sfrac|_{3}}|=|*frac*{1}{2}|=|P|pbo|A|pbc| &]
$P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} = P (A)$ $P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} = P (A)$

Thus, assuming that the occurrence of event B does not alter the probability of observing an even number, that probability remains $\frac{1}{2}$ $\frac{1}{2}$ . Therefore, the events A and B are independent.

Look Back

Note that if we calculate the conditional probability of B given A, our conclusion is the same:

[&P|pbo|B|pipe|A|pbc||=|*frac*{P|pbo|A|inter|B|pbc|}{P|pbo|A|pbc|}|=|*frac*{^{1}|sfrac|_{3}}{^{1}|sfrac|_{2}}|=|*frac*{2}{3}|=|P|pbo|B|pbc| &]

P (B | A) = \frac{P (A \cap B)}{P (A)} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} = P (B)

$P (B | A) = \frac{P (A \cap B)}{P (A)} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} = P (B)$

Example 3.19 Checking for Independence—Consumer Complaint Study

Problem

Refer to the consumer product complaint study in Example 3.15. The percentages of complaints of various types during and after the guarantee period are shown in Table 3.6 (p. 147). Define the following events:

[&*AS*A: *AP*|cbo|~rom~The cause of the complaint is the appearance of the product~normal~.|cbc| &]

[&*AS*B: *AP*|cbo|~rom~The complaint occurred during the guarantee term~normal~.|cbc| &]
$\begin{array}{l} A : {The cause of the complaint is the appearance of the product .} \\ B : {The complaint occurred during the guarantee term .} \end{array}$ $\begin{array}{l} A : {The cause of the complaint is the appearance of the product .} \\ B : {The complaint occurred during the guarantee term .} \end{array}$

Are A and B independent events?

Solution

Events A and B are independent if $P (A | B) = P (A) .$ $P (A | B) = P (A) .$ In Example 3.15, we calculated $P (A | B)$ $P (A | B)$ to be .51, and from Table 3.6, we see that

[&P|pbo|A|pbc||=|.32|+|.03|=|.35 &]
$P (A) = .32 + .03 = .35$ $P (A) = .32 + .03 = .35$

Therefore, $P (A | B)$ $P (A | B)$ is not equal to P(A), and A and B are dependent events.

Now Work Exercise 3.73c

Biography Blaise Pascal (1623–1662)

Solver of the Chevalier’s Dilemma

As a precocious child growing up in France, Blaise Pascal showed an early inclination toward mathematics. Although his father would not permit Pascal to study mathematics before the age of 15 (removing all math texts from his house), at age 12 Blaise discovered on his own that the sum of the angles of a triangle are two right angles.

Pascal went on to become a distinguished mathematician, as well as a physicist, a theologian, and the inventor of the first digital calculator. Most historians attribute the beginning of the study of probability to the correspondence between Pascal and Pierre de Fermat in 1654. The two solved the Chevalier’s Dilemma, a gambling problem related to Pascal by his friend and Paris gambler the Chevalier de Mere. The problem involved determining the expected number of times one could roll two dice without throwing a double 6. (Pascal proved that the “break-even” point was 25 rolls.)

To gain an intuitive understanding of independence, think of situations in which the occurrence of one event does not alter the probability that a second event will occur. For example, new medical procedures are often tested on laboratory animals. The scientists conducting the tests generally try to perform the procedures on the animals so that the results for one animal do not affect the results for the others. That is, the event that the procedure is successful on one animal is independent of the result for another. In this way, the scientists can get a more accurate idea of the efficacy of the procedure than if the results were dependent, with the success or failure for one animal affecting the results for other animals.

As a second example, consider an election poll in which 1,000 registered voters are asked their preference between two candidates. Pollsters try to use procedures for selecting a sample of voters so that the responses are independent. That is, the objective of the pollster is to select the sample so that one polled voter’s preference for candidate A does not alter the probability that a second polled voter prefers candidate A.

Now consider the world of sports. Do you think the results of a batter’s successive trips to the plate in baseball, or of a basketball player’s successive shots at the basket, are independent? For example, if a basketball player makes two successive shots, is the probability of making the next shot altered from its value if the result of the first shot is not known? If a player makes two shots in a row, the probability of a third successful shot is likely to be different from what we would assign if we knew nothing about the first two shots. Why should this be so? Research has shown that many such results in sports tend to be dependent because players (and even teams) tend to get on “hot” and “cold” streaks, during which their probabilities of success may increase or decrease significantly.

We will make three final points about independence.

Point 1: The property of independence, unlike the property of mutual exclusivity, generally cannot be shown on, or gleaned from, a Venn diagram. This means that you can’t trust your intuition. In general, the only way to check for independence is by performing the calculations of the probabilities in the definition.
Point 2: Suppose that events A and B are mutually exclusive, as shown in Figure 3.20, and that both events have nonzero probabilities. Are these events independent or dependent? That is, does the assumption that B occurs alter the probability of the occurrence of A? It certainly does, because if we assume that B has occurred, it is impossible for A to have occurred simultaneously. That is, $P (A | B) = 0.$ $P (A | B) = 0.$ Thus, mutually exclusive events are dependent events, since $P (A) \neq P (A | B) .$ $P (A) \neq P (A | B) .$
Point 3: The probability of the intersection of independent events is very easy to calculate. Referring to the formula for calculating the probability of an intersection, we find that

[&P|pbo|A|inter|B|pbc||=|P|pbo|A|pbc|P|pbo|B|pipe|A|pbc| &]
$P (A \cap B) = P (A) P (B | A)$ $P (A \cap B) = P (A) P (B | A)$

Thus, since $P (B | A) = P (B)$ $P (B | A) = P (B)$ when A and B are independent, we have the following useful rule:

Figure 3.20
Mutually exclusive events are dependent events

Probability of Intersection of Two Independent Events

If events A and B are independent, then the probability of the intersection of A and B equals the product of the probabilities of A and B; that is,

[&P|pbo|A|inter|B|pbc||=|P|pbo|A|pbc|P|pbo|B|pbc| &]

P (A \cap B) = P (A) P (B)

$P (A \cap B) = P (A) P (B)$

The converse is also true: If $P (A \cap B) = P (A) P (B),$ $P (A \cap B) = P (A) P (B),$ then events A and B are independent.

In the die-toss experiment, we showed in Example 3.18 that the two events $A : {Observe an even number .}$ $A : {Observe an even number .}$ and $B : {Observe a number less than or equal to 4.}$ $B : {Observe a number less than or equal to 4.}$ are independent if the die is fair. Thus,

[&P|pbo|A|inter|B|pbc||=|P|pbo|A|pbc|P|pbo|B|pbc||=||3(|*frac*{1}{2}|3)||3(|*frac*{2}{3}|3)||=|*frac*{1}{3} &]

P (A \cap B) = P (A) P (B) = (\frac{1}{2}) (\frac{2}{3}) = \frac{1}{3}

$P (A \cap B) = P (A) P (B) = (\frac{1}{2}) (\frac{2}{3}) = \frac{1}{3}$

This result agrees with the one we obtained in the example:

[&P|pbo|A|inter|B|pbc||=|P|pbo|2|pbc||+|P|pbo|4|pbc||=|*frac*{2}{6}|=|*frac*{1}{3} &]

P (A \cap B) = P (2) + P (4) = \frac{2}{6} = \frac{1}{3}

$P (A \cap B) = P (2) + P (4) = \frac{2}{6} = \frac{1}{3}$

Example 3.20 Probability of Independent Events Occurring Simultaneously—Divorced Couples Study

Problem

Recall from Example 3.5 (p. 125) that the American Association for Marriage and Family Therapy (AAMFT) found that 25% of divorced couples are classified as “fiery foes” (i.e., they communicate through their children and are hostile toward each other).
1. What is the probability that in a sample of 2 divorced couples, both are classified as “fiery foes”?
2. What is the probability that in a sample of 10 divorced couples, all 10 are classified as “fiery foes”?

Solution

Let $F_{1}$ $F_{1}$ represent the event that divorced couple 1 is classified as a pair of “fiery foes” and $F_{2}$ $F_{2}$ represent the event that divorced couple 2 is also classified as a pair of “fiery foes.” Then the event that both couples are “fiery foes” is the intersection of the two events, $F_{1} {\cap F}_{2} .$ $F_{1} {\cap F}_{2} .$ On the basis of the AAMFT survey, which found that 25% of divorced couples are “fiery foes,” we could reasonably conclude that $P (F_{1}) = .25$ $P (F_{1}) = .25$ and $P (F_{2}) = .25 .$ $P (F_{2}) = .25 .$ However, in order to compute the probability of $F_{1} {\cap F}_{2}$ $F_{1} {\cap F}_{2}$ from the multiplicative rule, we must make the assumption that the two events are independent. Since the classification of any divorced couple is not likely to affect the classification of another divorced couple, this assumption is reasonable. Assuming independence, we have

[&P|pbo|F_{1}|inter|F_{2}|pbc||=|P|pbo|F_{1}|pbc|P|pbo|F_{2}|pbc||=||pbo|.25|pbc||pbo|.25|pbc||=|.0625 &]
$P (F_{1} {\cap F}_{2}) = P (F_{1}) P (F_{2}) = (.25) (.25) = .0625$ $P (F_{1} {\cap F}_{2}) = P (F_{1}) P (F_{2}) = (.25) (.25) = .0625$
To see how to compute the probability that 10 of 10 divorced couples will all be classified as “fiery foes,” first consider the event that 3 of 3 couples are “fiery foes.” If $F_{3}$ $F_{3}$ represents the event that the third divorced couple are “fiery foes,” then we want to compute the probability of the intersection $F_{1} {\cap F}_{2} {\cap F}_{3} .$ $F_{1} {\cap F}_{2} {\cap F}_{3} .$ Again assuming independence of the classifications, we have

[&P|pbo|F_{1}|inter|F_{2}|inter|F_{3}|pbc||=|P|pbo|F_{1}|pbc|P|pbo|F_{2}|pbc|P|pbo|F_{3}|pbc||=||pbo|.25|pbc||pbo|.25|pbc||pbo|.25|pbc||=|.015625 &]
$P (F_{1} {\cap F}_{2} {\cap F}_{3}) = P (F_{1}) P (F_{2}) P (F_{3}) = (.25) (.25) (.25) = .015625$ $P (F_{1} {\cap F}_{2} {\cap F}_{3}) = P (F_{1}) P (F_{2}) P (F_{3}) = (.25) (.25) (.25) = .015625$

Similar reasoning leads us to the conclusion that the intersection of 10 such events can be calculated as follows:

[&P|pbo|F_{1}|inter|F_{2}|inter|F_{3}*N*[2%0]|interns| |elipns||interns| F_{10}|pbc||=|P|pbo|F_{1}|pbc|P|pbo|F_{2}|pbc||elipns|P|pbo|F_{10}|pbc||=||pbo|.25|pbc|^{10}|=|.000001 &]
$P (F_{1} {\cap F}_{2} {\cap F}_{3} \cap \dots \cap F_{10}) = P (F_{1}) P (F_{2}) \dots P (F_{10}) = (.25)^{10} = .000001$ $P (F_{1} {\cap F}_{2} {\cap F}_{3} \cap \dots \cap F_{10}) = P (F_{1}) P (F_{2}) \dots P (F_{10}) = (.25)^{10} = .000001$

Thus, the probability that 10 of 10 divorced couples sampled are all classified as “fiery foes” is 1 in 1 million, assuming that the probability of each couple being classified as “fiery foes” is .25 and that the classification decisions are independent.

Look Back

The very small probability in part b makes it extremely unlikely that 10 of 10 divorced couples are “fiery foes.” If this event should actually occur, we should question the probability of .25 provided by the AAMFT and used in the calculation—it is likely to be much higher. (This conclusion is another application of the rare-event approach to statistical inference.)

Now Work Exercise 3.89

Statistics in Action Revisited

The Probability of Winning Cash 3 or Play 4

In addition to the biweekly Lotto 6/53, the Florida Lottery runs several other games. Two popular daily games are Cash 3 and Play 4. In Cash 3, players pay $1 to select three numbers in order, where each number ranges from 0 to 9. If the three numbers selected (e.g., 2–8–4) match exactly the order of the three numbers drawn, the player wins $500. Play 4 is similar to Cash 3, but players must match four numbers (each number ranging from 0 to 9). For a $1 Play 4 ticket (e.g., 3–8–3–0), the player will win $5,000 if the numbers match the order of the four numbers drawn.

During the official drawing for Cash 3, 10 table tennis balls numbered 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are placed into each of three chambers. The balls in the first chamber are colored pink, the balls in the second chamber are blue, and the balls in the third chamber are yellow. One ball of each color is randomly drawn, with the official order as pink–blue–yellow. In Play 4, a fourth chamber with orange balls is added, and the official order is pink–blue–yellow–orange. Since the draws of the colored balls are random and independent, we can apply an extension of the probability rule for the intersection of two independent events to find the odds of winning Cash 3 and Play 4. The probability of matching a numbered ball being drawn from a chamber is 1/10; therefore,

[&*AS*P|pbo|~rom~Win Cash ~normal~3|pbc|*AP*|=|P|pbo|~rom~match pink ~normal~and ~rom~match blue~norm~ &]

[&*AS*~F-~*AP*|=|~F+~and ~rom~match yellow|pbc|~norm~ &]

[&*AS**AP*|=|P|pbo|~rom~match pink|pbc||multi|~normal~P|pbo|~rom~match blue|pbc||multi|~norm~ &]

[&*AS*~F-~*AP*|=|~F+~P|pbo|~rom~match yellow|pbc|~norm~ &]

[&*AS**AP*|=||pbo|1|sol|10|pbc||pbo|1|sol|10|pbc||pbo|1|sol|10|pbc||=|1|sol|1000|=|.001 &]

[&*AS*P|pbo|~rom~Win Play~normal~ 4|pbc|*AP*|=|P|pbo|~rom~match pink~normal~ and ~rom~match blue~norm~ &]

[&*AS*~F-~*AP*|=|~F+~~it~and~rom~ match yellow~normal~ and ~rom~match orange|pbc|~norm~ &]

[&*AS**AP*|=|P|pbo|~rom~match pink|pbc||multi|~normal~P|pbo|~rom~match blue|pbc||multi|~norm~ &]

[&*AS*~F-~*AP*|=|~F+~P|pbo|~rom~match yellow|pbc||multi|~normal~P|pbo|~rom~match orange|pbc|~norm~ &]

[&*AS**AP*|=||pbo|1|sol|10|pbc||pbo|1|sol|10|pbc||pbo|1|sol|10|pbc| &]

[&*AS**AP*|=|1|sol|10,000|=|.0001 &]

\begin{array}{l} P (Win Cash 3) & = & P (match pink a n d match blue \\ a n d match yellow) \\ = & P (match pink) \times P (match blue) \times \\ P (match yellow) \\ = & (1 / 10) (1 / 10) (1 / 10) = 1 / 1000 = .001 \\ P (Win Play 4) & = & P (match pink a n d match blue) \\ a n d match yellow a n d match orange \\ = & P (match pink) \times P (match blue) \times \\ P (match yellow) \times P (match orange) \\ = & (1 / 10) (1 / 10) (1 / 10) \\ = & 1 / 10, 000 = .0001 \end{array}

$\begin{array}{l} P (Win Cash 3) & = & P (match pink a n d match blue \\ a n d match yellow) \\ = & P (match pink) \times P (match blue) \times \\ P (match yellow) \\ = & (1 / 10) (1 / 10) (1 / 10) = 1 / 1000 = .001 \\ P (Win Play 4) & = & P (match pink a n d match blue) \\ a n d match yellow a n d match orange \\ = & P (match pink) \times P (match blue) \times \\ P (match yellow) \times P (match orange) \\ = & (1 / 10) (1 / 10) (1 / 10) \\ = & 1 / 10, 000 = .0001 \end{array}$

Although the odds of winning one of these daily games is much better than the odds of winning Lotto 6/53, there is still only a 1 in 1,000 chance (for Cash 3) or 1 in 10,000 chance (for Play 4) of winning the daily game. And the payoffs ($500 or $5,000) are much smaller. In fact, it can be shown that you will lose an average of 50¢ every time you play either Cash 3 or Play 4!

Exercises 3.68–3.103

Understanding the Principles

3.68 Explain the difference between an unconditional probability and a conditional probability.
3.69 Give the multiplicative rule of probability for
1. two independent events.
2. any two events.
3.70 Give the formula for finding $P (B | A) .$ $P (B | A) .$
3.71 Defend or refute each of the following statements:
1. Dependent events are always mutually exclusive.
2. Mutually exclusive events are always dependent.
3. Independent events are always mutually exclusive.

Learning the Mechanics

3.72 For two events A and B, $P (A) = .4, P (B) = .2,$ $P (A) = .4, P (B) = .2,$ and $P (A | B) = .6 .$ $P (A | B) = .6 .$
1. Find $P (A \cap B) .$ $P (A \cap B) .$
2. Find $P (B | A) .$ $P (B | A) .$
3.73 For two events A and B, $P (A) = .4, P (B) = .2,$ $P (A) = .4, P (B) = .2,$ and $P (A \cap B) = .1 .$ $P (A \cap B) = .1 .$
1. Find $P (A | B) .$ $P (A | B) .$
2. Find $P (B | A) .$ $P (B | A) .$
3. Are A and B independent events?
3.74 An experiment results in one of three mutually exclusive events A, B, and C. It is known that $P (A) = .30,$ $P (A) = .30,$ $P (B) = .55,$ $P (B) = .55,$ and $P (C) = .15,$ $P (C) = .15,$ Find each of the following probabilities:
1. $P (A \cup B)$ $P (A \cup B)$
2. $P (A \cap B)$ $P (A \cap B)$ )
3. $P (A | B)$ $P (A | B)$
4. $P (B \cup C)$ $P (B \cup C)$ )
5. Are B and C independent events? Explain.
3.75 For two independent events A and B, $P (A) = .4$ $P (A) = .4$ and $P (B) = .2 .$ $P (B) = .2 .$
1. Find $P (A \cap B) .$ $P (A \cap B) .$
2. Find $P (A | B) .$ $P (A | B) .$
3. Find $P (A \cup B) .$ $P (A \cup B) .$
3.76 An experiment results in one of five sample points with the following probabilities: $P (E_{1}) = .22, P (E_{2}) = .31,$ $P (E_{1}) = .22, P (E_{2}) = .31,$ $P (E_{3}) = .15, P (E_{4}) = .22,$ $P (E_{3}) = .15, P (E_{4}) = .22,$ and $P (E_{5}) = .1 .$ $P (E_{5}) = .1 .$ The following events have been defined:

$\begin{aligned} A : {E_{1}, E_{3}} \end{aligned}$ $\begin{aligned} A : {E_{1}, E_{3}} \end{aligned}$

$\begin{aligned} B : {E_{2}, E_{3}, E_{4}} \end{aligned}$ $\begin{aligned} B : {E_{2}, E_{3}, E_{4}} \end{aligned}$

$\begin{aligned} C : {E_{1}, E_{5}} \end{aligned}$ $\begin{aligned} C : {E_{1}, E_{5}} \end{aligned}$

Find each of the following probabilities:
1. P(A)
2. P(B)
3. $P (A \cap B)$ $P (A \cap B)$
4. $P (A | B)$ $P (A | B)$
5. $P (B \cap C)$ $P (B \cap C)$
6. $P (C | B)$ $P (C | B)$
7. Consider each pair of events A and B, A and C, and B and C. Are any of the pairs of events independent? Why?
3.77 Consider the experiment defined by the accompanying Venn diagram, with the sample space S containing five sample points. The sample points are assigned the following probabilities: $P (E_{1}) = .1, P (E_{2}) = .1,$ $P (E_{1}) = .1, P (E_{2}) = .1,$ $P (E_{3}) = .2, P (E_{4}) = .5, P (E_{5}) = .1 .$ $P (E_{3}) = .2, P (E_{4}) = .5, P (E_{5}) = .1 .$
1. Calculate P(A), P(B), and $P (A \cap B)$ $P (A \cap B)$ .
2. Suppose we know that event A has occurred, so the reduced sample space consists of the three sample points in $A : E_{1}, E_{2},$ $A : E_{1}, E_{2},$ and $E_{3} .$ $E_{3} .$ Use the formula for conditional probability to determine the probabilities of these three sample points given that A has occurred. Verify that the conditional probabilities are in the same ratio to one another as the original sample point probabilities and that they sum to 1.
3. Calculate the conditional probability $P (B | A)$ $P (B | A)$ in two ways: First, sum $P (E_{2} | A)$ $P (E_{2} | A)$ and $P (E_{3} | A),$ $P (E_{3} | A),$ since these sample points represent the event that B occurs given that A has occurred. Second, use the formula for conditional probability:
  
  [&P|pbo|B|pipe|A|pbc||=|*frac*{P|pbo|A|inter|B|pbc|}{P|pbo|A|pbc|} &]
  $P (B | A) = \frac{P (A \cap B)}{P (A)}$ $P (B | A) = \frac{P (A \cap B)}{P (A)}$
  
  Verify that the two methods yield the same result.
3.78 A sample space contains six sample points and events A, B, and C, as shown in the Venn diagram below. The probabilities of the sample points are $P (1) = .20, P (2) = .05,$ $P (1) = .20, P (2) = .05,$ $P (3) = .30, P (4) = .10, P (5) = .10,$ $P (3) = .30, P (4) = .10, P (5) = .10,$ and $P (6) = .25 .$ $P (6) = .25 .$
1. Which pairs of events, if any, are mutually exclusive? Why?
2. Which pairs of events, if any, are independent? Why?
3. Find $P (A \cup B)$ $P (A \cup B)$ by adding the probabilities of the sample points and then by using the additive rule. Verify that the answers agree. Repeat for $P (A \cup C) .$ $P (A \cup C) .$
3.79 Two fair dice are tossed, and the following events are defined:

$\begin{array}{l} A : {The sum of the numbers showing is odd .} \\ B : {The sum of the numbers showing is 9, 11, or 12 .} \end{array}$ $\begin{array}{l} A : {The sum of the numbers showing is odd .} \\ B : {The sum of the numbers showing is 9, 11, or 12 .} \end{array}$

Are events A and B independent? Why?
3.80 A box contains two white, two red, and two blue poker chips. Two chips are randomly chosen without replacement, and their colors are noted. Define the following events:

[&*AS*A: *AP*|cbo|~rom~Both chips are of the same color~normal~.|cbc| &]

[&*AS*B: *AP*|cbo|~rom~Both chips are red~normal~.|cbc| &]

[&*AS*C: *AP*|cbo|~rom~At least one chip is red or white~normal~.|cbc| &]
$\begin{array}{l} A : {Both chips are of the same color .} \\ B : {Both chips are red .} \\ C : {At least one chip is red or white .} \end{array}$ $\begin{array}{l} A : {Both chips are of the same color .} \\ B : {Both chips are red .} \\ C : {At least one chip is red or white .} \end{array}$

Find $P (B | A), P (B | A^{c}), P (B | C), P (A | C),$ $P (B | A), P (B | A^{c}), P (B | C), P (A | C),$ and $P (C | A^{c}) .$ $P (C | A^{c}) .$

Applet Exercise 3.5

Use the applet entitled Simulating the Probability of Rolling a 6 to simulate conditional probabilities. Begin by running the applet twice with $n = 10,$ $n = 10,$ without resetting between runs. The data on your screen represent 20 rolls of a die. The diagram above the Roll button shows the frequency of each of the six possible outcomes. Use this information to find each of the following probabilities:

1. The probability of 6 given that the outcome is 5 or 6
2. The probability of 6 given that the outcome is even
3. The probability of 4 or 6 given that the outcome is even
4. The probability of 4 or 6 given that the outcome is odd

Applying the Concepts—Basic

3.81 Blood diamonds. According to Global Research News (Mar. 4, 2014), one-fourth of all rough diamonds produced in the world are blood diamonds. (Any diamond that is mined in a war zone—often by children—in order to finance a warlord’s activity, an insurgency, or an invading army’s effort is considered a blood diamond.) Also, 90% of the world’s rough diamonds are processed in Surat, India, and of these diamonds, one-third are blood diamonds.
1. Find the probability that a rough diamond is not a blood diamond.
2. Find the probability that a rough diamond is processed in Surat and is a blood diamond.
3.82 Do social robots walk or roll? Refer to the International Conference on Social Robotics (Vol. 6414, 2010) study of the trend in the design of social robots, Exercise 3.17 (p. 129). Recall that in a random sample of 106 social robots, 63 were built with legs only, 20 with wheels only, 8 with both legs and wheels, and 15 with neither legs nor wheels. If a social robot is designed with wheels, what is the probability that the robot also has legs?
3.83 Crop damage by wild boars. Refer to the Current Zoology (Apr. 2014) study of crop damage by wild boars, Exercise 3.18 (p. 129). Recall that researchers identified 157 incidents of crop damage in the study area caused by wild boars over a five-year period. The table giving the types of crops destroyed and corresponding percentage of incidents is reproduced below. The researchers also determined that of the cereal crops damaged, 71% involved wheat, barley, or oats. For one randomly selected incident, what is the likelihood that the wild boars damaged a wheat, barley, or oats cereal crop?

Type Percentage

Cereals 45%

Orchards 5

Legumes 20

Vineyards 15

Other crops 15

Total 100%
3.84 Cardiac stress testing. In addition to standard exercise electrocardiography (SEE), two widely applied methods of stress testing of cardiac patients are stress echocardiography (SE) and single-photon emission computed tomography (SPECT). These methods were evaluated in the American Journal of Medical Quality (Mar./Apr. 2014). A sample of 866 heart catheterization patients in the Penn State Hershey Medical Center all received some form of stress testing prior to the catheterization. Of these, SE was performed on 272 patients, SPECT on 560 patients, and SEE on 38 patients. Four patients received both SE and SPECT but not SEE. Consider a patient randomly selected from those in the study.
1. What is the probability that the patient received SE prior to the catheterization?
2. What is the probability that the patient received both SE and SPECT prior to the catheterization?
3. Given that the patient received SPECT, what is the probability the patient also received SE prior to the catheterization?
3.85 National firearms survey. The Harvard School of Public Health conducted a study of privately held firearm stock in the United States. In a representative household telephone survey of 2,770 adults, 26% reported that they own at least one gun (Injury Prevention, Jan. 2007). Of those that own a gun, 5% own a handgun. Suppose one of the 2,770 adults surveyed is randomly selected.
1. What is the probability that the adult owns at least one gun?
2. What is the probability that the adult owns at least one gun and the gun is a handgun?
3.86 Guilt in decision making. Refer to the Journal of Behavioral Decision Making (Jan. 2007) study of the effect of guilt emotion on how a decision maker focuses on the problem, Exercise 3.59 (p. 142). The results (number responding in each category) for the 171 study participants are reproduced in the accompanying table. Suppose one of the 171 participants is selected at random.

Alternate View

Emotional State Choose Stated Option Do Not Choose Stated Option Totals

Guilt 45 12 57

Anger 8 50 58

Neutral 7 49 56

Totals 60 111 171

Based on Gangemi, A., and Mancini, F. “Guilt and focusing in decision-making.” Journal of Behavioral Decision Making, Vol. 20, Jan. 2007 (Table 2).
1. a. Given the respondent is assigned to the guilty state, what is the probability that the respondent chooses the stated option?
2. b.If the respondent does not choose to repair the car, what is the probability of the respondent being in the anger state?
3. c.Are the events ${repair the car}$ ${repair the car}$ and ${guilty state}$ ${guilty state}$ independent?
3.87 Speeding linked to fatal car crashes. According to the National Highway Traffic Safety Administration’s National Center for Statistics and Analysis (NCSA), “Speeding is one of the most prevalent factors contributing to fatal traffic crashes” (NHTSA Technical Report, Aug. 2005). The probability that speeding is a cause of a fatal crash is .3. Furthermore, the probability that speeding and missing a curve are causes of a fatal crash is .12. Given that speeding is a cause of a fatal crash, what is the probability that the crash occurred on a curve?
3.88 Appraisals and negative emotions. According to psychological theory, each negative emotion one has is linked to some thought (appraisal) about the environment. This theory was tested in the journal Cognition and Emotion (Vol. 24, 2010). Undergraduate students were asked to respond to events that occurred within the past 15 minutes. Each student rated the negative emotions he or she felt (e.g., anger, sadness, fear, or guilt) and gave an appraisal of the event (e.g., unpleasant or unfair). A total of 10,797 responses were evaluated in the study, of which 594 indicated a perceived unfairness appraisal. Of these unfairness appraisals, the students reported feeling angry in 127 of them. Consider a response randomly selected from the study.
1. What is the probability that the response indicates a perceived unfairness appraisal?
2. Given a perceived unfairness appraisal, what is the probability of an angry emotion?

Type	Percentage
Cereals	45%
Orchards	5
Legumes	20
Vineyards	15
Other crops	15
Total	100%

Emotional State	Choose Stated Option	Do Not Choose Stated Option	Totals
Guilt	45	12	57
Anger	8	50	58
Neutral	7	49	56
Totals	60	111	171

Applying the Concepts—Intermediate

3.89 Health risks to beachgoers. Refer to the University of Florida study of health risks to beachgoers, Exercise 3.24 (p. 130). According to the study, 6 out of 1,000 people exposed to wet sand for a 10-minute period will acquire gastroenteritis and 7 out of 1,000 people exposed to ocean water for a 10-minute period will acquire gastroenteritis. Suppose you and a friend go to the beach. Over the next 10 minutes, you play in the wet sand while your friend swims in the ocean. What is the probability that at least one of you acquires gastroenteritis? (What assumption did you need to make in order to find the probability?)
3.90 Sleep apnea and sleep stage transitioning. Refer to the sleep apnea study published in Chance (Winter 2009), Exercise 3.57 (p. 141). Recall that the various stages of sleep for a large group of sleep apnea patients were monitored in 30-second “epochs,” and the sleep stage for the previous and current epoch was determined. A summary table is reproduced in the next column. One of the epochs in the study is selected.

Alternate View

Previous Sleep Stage

Current Stage Non-REM REM Wake

Non-REM 31,880 160 1,733

REM 346 7,609 175

Wake 1,588 358 6,079

Totals 33,814 8,127 7,987

Based on Caffo, B. S., et al. “An overview of observational sleep research with application to sleep stage transitioning.” Chance, Vol. 22, No. 1, Winter 2009 (Table 2).
1. Given that the previous sleep stage for the epoch was the Wake state, what is the probability that the current sleep stage is REM?
2. Given that the current sleep stage for the epoch is REM, what is the probability that the previous sleep stage was not the Wake state?
3. Are the events {previous stage is REM} and {current stage is REM} mutually exclusive?
4. Are the events {previous stage is REM} and {current stage is REM} independent?
5. Are the events {previous stage is Wake} and {current stage is Wake} independent?
3.91 Firefighters’ use of gas detection devices. Two deadly gases that can be present in fire smoke are hydrogen cyanide and carbon monoxide. Fire Engineering (Mar. 2013) reported the results of a survey of 244 firefighters conducted by the Fire Smoke Coalition. The purpose of the survey was to assess the base level of knowledge of firefighters regarding the use of gas detection devices at the scene of a fire. The survey revealed the following: Eighty percent of firefighters had no standard operating procedures (SOP) for detecting/monitoring hydrogen cyanide in fire smoke; 49% had no SOP for detecting/monitoring carbon monoxide in fire smoke. Assume that 94% of firefighters had no SOP for detecting either hydrogen cyanide or carbon monoxide in fire smoke. What is the probability that a firefighter has no SOP for detecting hydrogen cyanide and no SOP for detecting carbon monoxide in fire smoke?

	Previous Sleep Stage
Non-REM	31,880	160	1,733
REM	346	7,609	175
Wake	1,588	358	6,079
Totals	33,814	8,127	7,987

3.92 Compensatory advantage in education. According to Sociology of Education (Apr. 2014), compensatory advantage occurs when individuals from privileged backgrounds are less dependent on prior negative outcomes than individuals from disadvantaged families. The researcher applied this theory to educational success. Let S_t represent the event that a student is successfully promoted to the next grade level in year t, and let F_t represent the event that a student fails to be promoted to the next grade level in year t. Now consider two probabilities: $P_{1} = the probability$ $P_{1} = the probability$ that a student is successfully promoted to the next grade level in year $t + 1$ $t + 1$ given that the student was successfully promoted the previous year (i.e., in year t) and $P_{2} = the probability$ $P_{2} = the probability$ that a student is successfully promoted to the next grade level in year $t + 1$ $t + 1$ given that the student failed in the previous year. Compensatory advantage theory states that the difference between the two probabilities, $P_{1} - P_{2}$ $P_{1} - P_{2}$ , is greater for disadvantaged families (say, the lower-class L) than for privileged families (say, the upper-class U).

Write P₁ as a conditional probability using symbols.
Write P₂ as a conditional probability using symbols.

The tables on p. 157 give the grade promotion results (number of students) for both upper- and lower-class students at a particular school. Does the compensatory advantage theory hold at this school?

Results for Exercise 3.92

Upper Class	Successfully promoted (S_t )	Failure (F_t )
Successfully promoted $(S_{t + 1})$ $(S_{t + 1})$	45	8
Failure $(F_{t + 1})$ $(F_{t + 1})$	5	2
Totals	50	10

Lower Class	Successfully promoted (S_t )	Failure (F_t )
Successfully promoted $(S_{t + 1})$ $(S_{t + 1})$	90	15
Failure $(F_{t + 1})$ $(F_{t + 1})$	10	5
Totals	100	20

3.93 Are you really being served red snapper? Red snapper is a rare and expensive reef fish served at upscale restaurants. Federal law prohibits restaurants from serving a cheaper, look-alike variety of fish (e.g., vermillion snapper or lane snapper) to customers who order red snapper. Researchers at the University of North Carolina used DNA analysis to examine fish specimens labeled “red snapper” that were purchased from vendors across the country (Nature, July 15, 2004). The DNA tests revealed that 77% of the specimens were not red snapper, but the cheaper, look-alike variety of fish.
1. Assuming that the results of the DNA analysis are valid, what is the probability that you are actually served red snapper the next time you order it at a restaurant?
2. If there are five customers at a restaurant, all who have ordered red snapper, what is the probability that at least one customer is actually served red snapper?
3.94 Fighting probability of fallow deer bucks. Refer to the Aggressive Behavior (Jan./Feb. 2007) study of fallow deer bucks fighting during the mating season, presented in Exercise 3.61 (p. 142). Recall that researchers recorded 167 encounters between two bucks, one of which clearly initiated the encounter with the other. A summary of the fight status of the initiated encounters is provided in the accompanying table. Suppose we select 1 of these 167 encounters and note the outcome (fight status and winner).
1. Given that a fight occurs, what is the probability that the initiator wins?
2. Given no fight, what is the probability that the initiator wins?
3. Are the events “no fight” and “initiator wins” independent?
  
  Alternate View
  
  Initiator Wins No Clear Winner Initiator Loses Totals
  
  Fight 26 23 15 64
  
  No Fight 80 12 11 103
  
  Totals 106 35 26 167
  
  Based on Bartos, L. et al. "Estimation of the probability of fighting in fallow deer (Dama dama) during the rut." Aggressive Behavior, Vol. 33, Jan./Feb. 2007, pp. 7–13.
NZBIRDS 3.95 Extinct New Zealand birds. Refer to Consider the Evolutionary Ecology Research (July 2003) study of the patterns of extinction in the New Zealand bird population, presented in Exercise 2.24 (p. 42). Consider the data on extinction status (extinct, absent from island, present) for the 132 bird species. The data are summarized in the accompanying MINITAB printout. Suppose you randomly select 10 of the 132 bird species (without replacement) and record the extinction status of each.
1. What is the probability that the first species you select is extinct? (Note: $Extinct = Yes$ $Extinct = Yes$ on the MINITAB printout.)
2. Suppose the first 9 species you select are all extinct. What is the probability that the 10th species you select is extinct?
3.96 Muscle, fat, and bone issues while aging. In Ageing Research Reviews (May 2014), nutritionists published a study of the health issues faced by aging adults. Three of the most common health conditions are obesity (high body fat mass), osteoporosis (low bone mineral density in neck and spine), and sarcopenia (low skeletal muscle mass index). The researchers provided names for aging adults who suffer from one or more of these conditions. These are shown in the accompanying Venn diagram. Statistics from the CDC are used to estimate the following probabilities:
- $P (Obesity) = .35$ $P (Obesity) = .35$
- $P (Osteoporosis) = .30$ $P (Osteoporosis) = .30$
- $P (Sarcopenia) = .15$ $P (Sarcopenia) = .15$
- $P (Osteopenic Obesity) = .01$ $P (Osteopenic Obesity) = .01$
- $P (Osteopenic Sarcopenia) = .03$ $P (Osteopenic Sarcopenia) = .03$
- $P (Sarcopenic Obesity) = .05$ $P (Sarcopenic Obesity) = .05$
- $P (Osteo-sarcopenic Obesity) = .02$ $P (Osteo-sarcopenic Obesity) = .02$
1. Find the probability that an adult suffers from both obesity and sarcopenia.
2. Find the probability that an adult suffers from either obesity or sarcopenia.
3.97 Ambulance response time. Geographical Analysis (Jan. 2010) presented a study of Emergency Medical Services (EMS) ability to meet the demand for an ambulance. In one example, the researchers presented the following scenario. An ambulance station has one vehicle and two demand locations, A and B. The probability that the ambulance can travel to a location in under eight minutes is .58 for location A and .42 for location B. The probability that the ambulance is busy at any point in time is .3.
1. Find the probability that EMS can meet demand for an ambulance at location A.
2. Find the probability that EMS can meet demand for an ambulance at location B.
3.98 Intrusion detection systems. A computer intrusion detection system (IDS) is designed to provide an alarm whenever there is unauthorized access into a computer system. A probabilistic evaluation of a system with two independently operating intrusion detection systems (a double IDS) was published in the Journal of Research of the National Institute of Standards and Technology (Nov.–Dec. 2003). Consider a double IDS with system A and system B. If there is an intruder, system A sounds an alarm with probability .9 and system B sounds an alarm with probability .95. If there is no intruder, the probability that system A sounds an alarm (i.e., a false alarm) is .2 and the probability that system B sounds an alarm is .1.
1. Use symbols to express the four probabilities just given.
2. If there is an intruder, what is the probability that both systems sound an alarm?
3. If there is no intruder, what is the probability that both systems sound an alarm?
4. Given that there is an intruder, what is the probability that at least one of the systems sounds an alarm?
3.99 Detecting traces of TNT. University of Florida researchers in the Department of Materials Science and Engineering have invented a technique that rapidly detects traces of TNT (Today, Spring 2005). The method, which involves shining a laser on a potentially contaminated object, provides instantaneous results and gives no false positives. In this application, a false positive would occur if the laser detected traces of TNT when, in fact, no TNT were actually present on the object. Let A be the event that the laser light detects traces of TNT. Let B be the event that the object contains no traces of TNT. The probability of a false positive is 0. Write this probability in terms of A and B, using symbols such as “ $\cup$ $\cup$ ”, “ $\cap$ $\cap$ ”, and “ $|$ $|$ ”.

	Initiator Wins	No Clear Winner	Initiator Loses	Totals
Fight	26	23	15	64
No Fight	80	12	11	103
Totals	106	35	26	167

Applying the Concepts—Advanced

3.100 Random mutation of cells. Chance (Spring 2010) presented an article on the random mutation hypothesis developed by microbiologists. Under this hypothesis, when a wild-type organic cell (e.g., a bacteria cell) divides, there is a chance that at least one of the two “daughter” cells is a mutant. When a mutant cell divides, both offspring will be mutant. The schematic in the next column shows a possible pedigree from a single cell that has divided. Note that one “daughter” cell is mutant () and one is a normal cell ().
1. Consider a single, normal cell that divides into two offspring. List the different possible pedigrees.
2. Assume that a “daughter” cell is equally likely to be mutant or normal. What is the probability that a single, normal cell that divides into two offspring will result in at least one mutant cell?
3. Now assume that the probability of a mutant “daughter” cell is .2. What is the probability that a single, normal cell that divides into two offspring will result in at least one mutant cell?
4. The schematic below shows a possible second-generation pedigree from a single cell that has divided. Note that the first-generation mutant cell automatically produces two mutant cells in the second generation. List the different possible second-generation pedigrees. (Hint: Use your answer to part a.)
5. Assume that a “daughter” cell is equally likely to be mutant or normal. What is the probability that a single, normal cell that divides into two offspring will result in at least one mutant cell after the second generation?
3.101 Testing a psychic’s ability. Consider an experiment in which 10 identical small boxes are placed side by side on a table. A crystal is placed at random inside one of the boxes. A self-professed “psychic” is asked to pick the box that contains the crystal.
1. If the “psychic” simply guesses, what is the probability that she picks the box with the crystal?
2. If the experiment is repeated seven times, what is the probability that the “psychic” guesses correctly at least once?
3. A group called the Tampa Bay Skeptics tested a self-proclaimed “psychic” by administering the preceding experiment seven times. The “psychic” failed to pick the correct box all seven times (Tampa Tribune, Sept. 20, 1998). What would you infer about this person’s psychic ability?
3.102 Risk of a natural-gas pipeline accident. Process Safety Progress (Dec. 2004) published a risk analysis for a natural-gas pipeline between Bolivia and Brazil. The most likely scenario for an accident would be natural-gas leakage from a hole in the pipeline. The probability that the leak ignites immediately (causing a jet fire) is .01. If the leak does not immediately ignite, it may result in the delayed ignition of a gas cloud. Given no immediate ignition, the probability of delayed ignition (causing a flash fire) is .01. If there is no delayed ignition, the gas cloud will disperse harmlessly. Suppose a leak occurs in the natural-gas pipeline. Find the probability that either a jet fire or a flash fire will occur. Illustrate with a tree diagram.
3.103 Encryption systems with erroneous ciphertexts. In cryptography, ciphertext is encrypted or encoded text that is unreadable by a human or computer without the proper algorithm to decrypt it into plaintext. The impact of erroneous ciphertexts on the performance of an encryption system was investigated in IEEE Transactions on Information Forensics and Security (Apr. 2013). For one data encryption system, the probability of receiving an erroneous ciphertext is assumed to be $β$ $β$ , where $0 < β < 1$ $0 < β < 1$ . The researchers showed that if an erroneous ciphertext occurs, the probability of an error in restoring plaintext using the decryption system is .5. When no error occurs in the received ciphertext, the probability of an error in restoring plaintext using the decryption system is $α β$ $α β$ , where $0 < α < 1$ $0 < α < 1$ . Use this information to give an expression for the probability of an error in restoring plaintext using the decryption system.

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Table of Contents for 3.6 The Multiplicative Rule and Independent Events

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Table of Contents for
3.6 The Multiplicative Rule and Independent Events