3.4 The Additive Rule and Mutually Exclusive Events

In Section 3.2, we saw how to determine which sample points are contained in a union of two sets and how to calculate the probability of the union by summing the separate probabilities of the sample points in the union. It is also possible to obtain the probability of the union of two events by using the additive rule of probability.

By studying the Venn diagram in Figure 3.11, you can see that the probability of the union of two events A and B can be obtained by summing P(A) and P(B) and subtracting P(AB). We must subtract P(AB) because the sample point probabilities in AB have been included twice—once in P(A) and once in P(B).

Figure 3.11

Venn diagram of union

The formula for calculating the probability of the union of two events is given in the next box.

Additive Rule of Probability

The probability of the union of events A and B is the sum of the probability of event A and the probability of event B, minus the probability of the intersection of events A and B; that is,

[&P|pbo|A|Opunion|B|pbc||=|P|pbo|A|pbc||+|P|pbo|B|pbc||-|P|pbo|A|inter|B|pbc| &]

P(AB)=P(A)+P(B)P(AB)

Example 3.11 Applying the Additive Rule—Hospital Admissions Study

Problem

  1. Hospital records show that 12% of all patients are admitted for surgical treatment, 16% are admitted for obstetrics, and 2% receive both obstetrics and surgical treatment. If a new patient is admitted to the hospital, what is the probability that the patient will be admitted for surgery, for obstetrics, or for both?

Solution

  1. Consider the following events:

    [&*AS*A: *AP*|cbo|~rom~A patient admitted to the hospital receives surgical treatment~normal~.|cbc| &]

    [&*AS*B: *AP*|cbo|~rom~A patient admitted to the hospital receives obstetrics treatment~normal~.|cbc| &]

    A: {A patient admitted to the hospital receives surgical treatment.}B: {A patient admitted to the hospital receives obstetrics treatment.}

    Then, from the given information,

    [&P|pbo|A|pbc||=|.12 &]

    [&P|pbo|B|pbc||=|.16 &]

    P(A)=.12P(B)=.16

    and the probability of the event that a patient receives both obstetrics and surgical treatment is

    [&P|pbo|A|inter|B|pbc||=|.02 &]

    P(AB)=.02

    The event that a patient admitted to the hospital receives either surgical treatment, obstetrics treatment, or both is the union AB, the probability of which is given by the additive rule of probability:

    [&*AS*P|pbo|A|Opunion|B|pbc|*AP*|=|P|pbo|A|pbc||+|P|pbo|B|pbc||-|P|pbo|A|inter|B|pbc| &]

    [&*AS**AP*|=|.12|+|.16|-|.02|=|.26 &]

    P(AB)=P(A)+P(B)P(AB)=.12+.16.02=.26

    Thus, 26% of all patients admitted to the hospital receive either surgical treatment, obstetrics treatment, or both.

Look Back

From the information given, it is not possible to list and assign probabilities to all the sample points. Consequently, we cannot proceed through the five-step process (p. 125) for finding P(AB), and we must use the additive rule.

Now Work Exercise 3.45

A very special relationship exists between events A and B when AB contains no sample points. In this case, we call the events A and B mutually exclusive events.

Events A and B are mutually exclusive if AB contains no sample points—that is, if A and B have no sample points in common. For mutually exclusive events,

P(AB)=0

Figure 3.12 shows a Venn diagram of two mutually exclusive events. The events A and B have no sample points in common; that is, A and B cannot occur simultaneously, and P(AB)=0. Thus, we have the important relationship given in the next box.

Figure 3.12

Venn diagram of mutually exclusive events

Probability of Union of Two Mutually Exclusive Events

If two events A and B are mutually exclusive, the probability of the union of A and B equals the sum of the probability of A and the probability of B; that is, P(AB)=P(A)+P(B).

Caution

The preceding formula is false if the events are not mutually exclusive. In that case (i.e., two non-mutually exclusive events), you must apply the general additive rule of probability.

Example 3.12 Union of Two Mutually Exclusive Events— Coin-Tossing Experiment

Problem

  1. Consider the experiment of tossing two balanced coins. Find the probability of observing at least one head.

Solution

  1. Define the events

    [&*AS*A: *AP*|cbo|~rom~*N*[-1%0]Observe at least one head~normal~.*N*[-1.5%0]|cbc| &]

    [&*AS*B: *AP*|cbo|~rom~*N*[-1%0]Observe exactly one head~normal~.*N*[-1.5%0]|cbc| &]

    [&*AS*C: *AP*|cbo|~rom~*N*[-1.5%0]Observe exactly two heads~normal~.*N*[-1.5%0]|cbc| &]

    A: {Observe at least one head.}B: {Observe exactly one head.}C: {Observe exactly two heads.}

    Note that

    [&A|=|B|Opunion|C &]

    A=BC

    and that BC contains no sample points. (See Figure 3.13.) Thus, B and C are mutually exclusive, and it follows that

    Figure 3.13

    Venn diagram for coin-toss experiment

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

    P(A)=P(BC)=P(B)+P(C)=12+14=34

Look Back

Although this example is quite simple, it shows us that writing events with verbal descriptions that include the phrases “at least” or “at most” as unions of mutually exclusive events is very useful. This practice enables us to find the probability of the event by adding the probabilities of the mutually exclusive events.

Statistics in Action Revisited

The Probability of Winning Lotto with a Wheeling System

Refer to Florida’s Pick-6 Lotto game, in which you select six numbers of your choice from a field of numbers ranging from 1 to 53. In Section 3.1, we learned that the probability of winning Lotto on a single ticket is only 1 in approximately 23 million. The “experts” at Lottery Buster recommend many strategies for increasing the odds of winning the lottery. One strategy is to employ a wheeling system. In a complete wheeling system, you select more than six numbers—say, seven—and play every combination of six of those seven numbers.

For example, suppose you choose to “wheel” the following seven numbers: 2, 7, 18, 23, 30, 32, and 51. Every combination of six of these seven numbers is listed in Table SIA3.1. You can see that there are seven different possibilities. (Use the combinations rule with N=7 and n=6 to verify this.) Thus, we would purchase seven tickets (at a cost of $7) corresponding to these different combinations in a complete wheeling system.

Table SIA3.1 Wheeling the Seven Numbers 2, 7, 18, 23, 30, 32, and 51

Alternate View
Ticket #1 2  7 18 23 30 32
Ticket #2 2  7 18 23 30 51
Ticket #3 2  7 18 23 32 51
Ticket #4 2  7 18 30 32 51
Ticket #5 2  7 23 30 32 51
Ticket #6 2 18 23 30 32 51
Ticket #7 7 18 23 30 32 51

To determine whether this strategy does, in fact, increase our odds of winning, we need to find the probability that one of these seven combinations occurs during the 6/53 Lotto draw. That is, we need to find the probability that either Ticket #1 or Ticket #2 or Ticket #3 or Ticket #4 or Ticket #5 or Ticket #6 or Ticket #7 is the winning combination. Note that this probability is stated with the use of the word or, implying a union of seven events. Letting T1 represent the event that Ticket #1 wins, and defining T2, T3, … , T7 in a similar fashion, we want to find

[&P|pbo|T_{1} ~rom~or ~normal~T_{*N*[-0.95%0]2} ~rom~or~normal~ T_{*N*[-0.95%0]3} ~rom~or~normal~ T_{*N*[-0.95%0]4} ~rom~or~normal~ T_{*N*[-1.425%0]5} ~rom~or~normal~ T_{*N*[-0.475%0]6} ~rom~or~normal~ T_{7}|pbc|~norm~ &]

P(T1 or T2 or T3 or T4 or T5 or T6 or T7)

Recall (Section 3.1) that the 22,957,480 possible combinations in Pick-6 Lotto are mutually exclusive and equally likely to occur. Consequently, the probability of the union of the seven events is simply the sum of the probabilities of the individual events, where each event has probability 1/(22,957,480):

[&P|pbo|~rom~win Lotto with~normal~ 7 ~rom~Wheeled Numbers|pbc|~norm~ &]

[&|eq| P|pbo|T_{1} ~rom~or ~normal~T_{*N*[-0.95%0]2} ~rom~or~normal~ T_{*N*[-0.95%0]3} ~rom~or~normal~ T_{*N*[-0.95%0]4} ~rom~or~normal~ T_{*N*[-1.425%0]5} ~rom~or~normal~ T_{*N*[-0.475%0]6} ~rom~or~normal~ T_{7}|pbc|~norm~ &]

[&|=|7|sol||pbo|22,957,480|pbc||=|.0000003 &]

P(win Lotto with 7 Wheeled Numbers)=P(T1 or T2 or T3 or T4 or T5 or T6 or T7)=7/(22,957,480) = .0000003

In terms of odds, we now have 3 chances in 10 million of winning the Lotto with the complete wheeling system. Technically, the “experts” are correct: Our odds of winning Lotto have increased (from 1 in 23 million). However, the probability of winning is so close to 0 that we question whether the $7 spent on lottery tickets is worth the negligible increase in odds. In fact, it can be shown that to increase your chance of winning the 6/53 Lotto to 1 chance in 100 (i.e., .01) by means of a complete wheeling system, you would have to wheel 26 of your favorite numbers—a total of 230,230 combinations at a cost of $230,230!

Exercises 3.38–3.67

Understanding the Principles

  1. 3.38 Define in words mutually exclusive events.

  2. 3.39 Define in words the union of two events.

  3. 3.40 Define in words the intersection of two events.

  4. 3.41 Define in words the complement of an event.

  5. 3.42 State the rule of complements.

  6. 3.43 State the additive rule of probability for mutually exclusive events.

  7. 3.44 State the additive rule of probability for any two events.

Learning the Mechanics

  1. 3.45 A fair coin is tossed three times, and the events A and B are defined as follows:

    [&A: |cbo|~rom~At least one head is observed~normal~.*N*[-1%0]|cbc| &]

    [&B: |cbo|~rom~The number of heads observed is odd~normal~.*N*[-1%0]|cbc| &]

    A: {At least one head is observed.}B: {The number of heads observed is odd.}
    1. Identify the sample points in the events A,B,AB,Ac, and AB.

    2. Find P(A),P(B),P(AB),P(Ac), and P(AB) by summing the probabilities of the appropriate sample points.

    3. Use the additive rule to find P(AB). Compare your answer with the one you obtained in part b.

    4. Are the events A and B mutually exclusive? Why?

  2. 3.46 Suppose P(A)=.4,P(B)=.7, and P(AB)=.3. Find the following probabilities:

    1. P(Bc)

    2. P(Ac)

    3. P(AB)

  3. 3.47 Consider the following Venn diagram, where P(E2)=P(E3)=15,P(E4)=P(E5)= 120, P(E6)= 110,P(E7)= 15:

    Find each of the following probabilities:

    1. P(A)

    2. P(B)

    3. P(AB)

    4. P(AB)

    5. P(Ac)

    6. P(Bc)

    7. P(AAc)

    8. P(AcB)

  4. 3.48 Consider the following Venn diagram, where

    [&P|pbo|E_{1}|pbc||=|.10, P|pbo|E_{2}|pbc||=|.05, P|pbo|E_{3}|pbc||=|P|pbo|E_{4}|pbc||=|.2, &]

    [&P|pbo|E_{5}|pbc||=|.06, P|pbo|E_{6}|pbc||=|.3, P|pbo|E_{7}|pbc||=|.06, ~rom~and~normal~ &]

    [&P|pbo|E_{8}|pbc||=|.03: &]

    P(E1)=.10,P(E2)=.05,P(E3)=P(E4)=.2,P(E5)=.06,P(E6)=.3,P(E7)=.06,andP(E8)=.03:

    Find the following probabilities:

    1. P(Ac)

    2. P(Bc)

    3. P(AcB)

    4. P(AB)

    5. P(AB)

    6. P(AcBc)

    7. Are events A and B mutually exclusive? Why?

  5. 3.49 A pair of fair dice is tossed. Define the following events:

    [&A: |cbo|~rom~You will roll a~normal~ 7 |pbo|~rom~i.e., the sum of the numbers of~normal~ &]

    [&~rom~dots on the upper faces of the two dice is equal to~normal~ 7|pbc|.|cbc| &]

    [&B: |cbo|~rom~At least one of the two dice is showing a~normal~ 4.|cbc| &]

    A:{You will roll a 7 (i.e., the sum of the numbers ofdots on the upper faces of the two dice is equal to 7).}B:{At least one of the two dice is showing a 4.}
    1. Identify the sample points in the events A,B, AB,AB, and Ac.

    2. Find P(A),P(B),P(AB),P(AB), and P(Ac) and by summing the probabilities of the appropriate sample points.

    3. Use the additive rule to find P(AB). Compare your answer with that for the same event in part b.

    4. Are A and B mutually exclusive? Why?

  6. 3.50 Three fair coins are tossed. We wish to find the probability of the event A:{Observe at least one head.}

    1. Express A as the union of three mutually exclusive events. Using the expression you wrote, find the probability of A.

    2. Express A as the complement of an event. Using the expression you wrote, find the probability of A.

  7. 3.51 The outcomes of two variables are (Low, Medium, High) and (On, Off), respectively. An experiment is conducted in which the outcomes of each of the two variables are observed. The probabilities associated with each of the six possible outcome pairs are given in the following table:

    Alternate View
    Low Medium High
    On .50 .10 .05
    Off .25 .07 .03

    Consider the following events:

    [&A: |cbo|~rom~On|cbc|~normal~ &]

    [&B: |cbo|~rom~Medium or On|cbc|~normal~ &]

    [&C: |cbo|~rom~Off and Low|cbc|~normal~ &]

    [&D: |cbo|~rom~High|cbc|~normal~ &]

    A: {On}B: {Medium or On}C: {Off and Low}D: {High}
    1. Find P(A).

    2. Find P(B).

    3. Find P(C).

    4. Find P(D).

    5. Find P(Ac).

    6. Find P(AB).

    7. Find P(AB).

    8. Consider each possible pair of events taken from the events A, B, C, and D. List the pairs of events that are mutually exclusive. Justify your choices.

Applet Exercise 3.3

Use the applets entitled Simulating the Probability of Rolling a 6 and Simulating the Probability of Rolling a 3 or 4 to explore the additive rule of probability.

    1. Explain why the applet Simulating the Probability of Rolling a 6 can also be used to simulate the probability of rolling a 3. Then use the applet with n=1000 to simulate the probability of rolling a 3. Record the cumulative proportion. Repeat the process to simulate the probability of rolling a 4.

    2. Use the applet Simulating the Probability of Rolling a 3 or 4 with n=1000 to simulate the probability of rolling a 3 or 4. Record the cumulative proportion.

    3. Add the two cumulative proportions from part a. How does this sum compare with the cumulative proportion in part b? How does your answer illustrate the additive rule for probability?

Applet Exercise 3.4

Use the applets entitled Simulating the Probability of Rolling a 6 and Simulating the Probability of Rolling a 3 or 4 to simulate the probability of the complement of an event.

    1. Explain how the applet Simulating the Probability of Rolling a 6 can also be used to simulate the probability of the event rolling a 1, 2, 3, 4, or 5. Then use the applet with n=1000 to simulate this probability.

    2. Explain how the applet Simulating the Probability of Rolling a 3 or 4 can also be used to simulate the probability of the event rolling a 1, 2, 5, or 6. Then use the applet with n=1000 to simulate this probability.

    3. Which applet could be used to simulate the probability of the event rolling a 1, 2, 3, or 4? Explain.

Applying the Concepts—Basic

  1. 3.52 Social networking. According to the Pew Research Internet Project (Dec. 2013) survey, the two most popular social networking sites in the United States are Facebook and LinkedIn. Of all adult Internet users, 71% have a Facebook account, 22% have a LinkedIn account, and 18% visit both Facebook and LinkedIn.

    1. Draw a Venn diagram to illustrate the use of social networking sites in the United States.

    2. Find the probability that an adult Internet user visits either Facebook or LinkedIn.

    3. Use your answer to part b to find the probability that an adult Internet user does not have an account with either social networking site.

  2. 3.53 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. Use the rule of complements to find the probability that a randomly selected social robot is designed with either legs or wheels.

  3. 3.54 Study of analysts’ forecasts. The Journal of Accounting Research (Mar. 2008) published a study on relationship incentives and degree of optimism among analysts’ forecasts. Participants were analysts at either a large or small brokerage firm who made their forecasts either early or late in the quarter. Also, some analysts were only concerned with making an accurate forecast, while others were also interested in their relationship with management. Suppose one of these analysts is randomly selected. Consider the following events:

    [&~MAT~[1%2%L%120%C%A]*MAT*{A*N*[-0.5%0]|=|~rom~~norm~{~rom~The analyst is only concerned with making an~normal~}{ |em||em|*N*[-0.5%-2]|en|~rom~*N*[-2.5%0]accurate forecast~normal~.}~rom~}~norm~ &]

    [&B|=||cbo|~rom~The analyst makes the forecast early in the quarter.*N*[-1%0]|cbc|~normal~ &]

    [&C|=||cbo|~rom~The analyst is from a small brokerage firm.*N*[-1%0]|cbc|~norm~ &]

    A={The analyst is only concerned with making anaccurate forecast.}B={The analyst makes the forecast early in the quarter.}C={The analyst is from a small brokerage firm.}

    Describe each of the following events in terms of unions, intersections, and complements (e.g., AB,AB,AC, etc.).

    1. The analyst makes an early forecast and is only concerned with accuracy.

    2. The analyst is not only concerned with accuracy.

    3. The analyst is from a small brokerage firm or makes an early forecast.

    4. The analyst makes a late forecast and is not only concerned with accuracy.

  4. 3.55 Gene expression profiling. Gene expression profiling is a state-of-the-art method for determining the biology of cells. In Briefings in Functional Genomics and Proteomics (Dec. 2006), biologists reviewed several gene expression profiling methods. The biologists applied two of the methods (A and B) to data collected on proteins in human mammary cells. The probability that the protein is cross-referenced (i.e., identified) by method A is .41, the probability that the protein is cross-referenced by method B is .42, and the probability that the protein is cross-referenced by both methods is .40.

    1. Draw a Venn diagram to illustrate the results of the gene-profiling analysis.

    2. Find the probability that the protein is cross-referenced by either method A or method B.

    3. On the basis of your answer to part b, find the probability that the protein is not cross-referenced by either method.

  5. 3.56 Scanning errors at Wal-Mart. The National Institute of Standards and Technology (NIST) mandates that for every 100 items scanned through the electronic checkout scanner at a retail store, no more than 2 should have an inaccurate price. A study of the accuracy of checkout scanners at Wal-Mart stores in California (Tampa Tribune, Nov. 22, 2005) showed that, of the 60 Wal-Mart stores investigated, 52 violated the NIST scanner accuracy standard. If 1 of the 60 Wal-Mart stores is randomly selected, what is the probability that the store does not violate the NIST scanner accuracy standard?

  6. 3.57 Sleep apnea and sleep stage transitioning. Sleep apnea is a common sleep disorder characterized by collapses of the upper airway during sleep. Chance (Winter 2009) investigated the role of sleep apnea in how people transition from one sleep stage to another. The various stages of sleep for a large group of sleep apnea patients were monitored in 30-second intervals, or “epochs.” For each epoch, sleep stage was categorized as Wake, REM, or Non-REM. The table below provides a summary of the results. Each cell of the table gives the number of epochs that occurred when transitioning from the previous sleep stage to the current sleep stage. Consider the previous and current sleep stage of a randomly selected epoch from the study.

    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. List the sample points for this experiment.

    2. Assign reasonable probabilities to the sample points.

    3. What is the probability that the current sleep stage of the epoch is REM?

    4. What is the probability that the previous sleep stage of the epoch is Wake?

    5. What is the probability that the epoch transitioned from the REM sleep stage to the Non-REM sleep stage?

Applying the Concepts—Intermediate

  1. 3.58 Attempted suicide methods. A study of attempted suicide methods was published in the journal Death Studies (Vol. 38, 2014). Data for 59 suicide victims who had previously attempted suicide were collected. The table gives a breakdown of the number of suicide victims according to method used and gender. Suppose one of the suicide victims is selected at random.

    Alternate View
    Method Males Females Totals
    Self-poisoning 22 13 35
    Hanging  4  0  4
    Jumping from a height  7  4 11
    Cutting  6  3  9
    Totals 39 20 59
    1. Find P(A), where A={male}.

    2. Find P(B), where B={jumping from a height}.

    3. Are A and B mutually exclusive events?

    4. Find P(AC ).

    5. Find P(AB).

    6. Find P(AB).

  2. 3.59 Guilt in decision making. The effect of guilt emotion on how a decision maker focuses on the problem was investigated in the Jan. 2007 issue of the Journal of Behavioral Decision Making (see Exercise 1.31 , p. 22). A total of 171 volunteer students participated in the experiment, where each was randomly assigned to one of three emotional states (guilt, anger, or neutral) through a reading/writing task. Immediately after the task, the students were presented with a decision problem where the stated option has predominantly negative features (e.g., spending money on repairing a very old car). The results (number responding in each category) are summarized 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. Find the probability that the respondent is assigned to the guilty state.

    2. Find the probability that the respondent chooses the stated option (repair the car).

    3. Find the probability that the respondent is assigned to the guilty state and chooses the stated option.

    4. Find the probability that the respondent is assigned to the guilty state or chooses the stated option.

  3. 3.60 Abortion provider survey. The Alan Guttmacher Institute Abortion Provider Survey is a survey of all 358 known nonhospital abortion providers in the United States (Perspectives on Sexual and Reproductive Health, Jan./Feb. 2003). For one part of the survey, the 358 providers were classified according to case load (number of abortions performed per year) and whether they permit their patients to take the abortion drug misoprostol at home or require the patients to return to the abortion facility to receive the drug. The responses are summarized in the accompanying table. Suppose we select, at random, one of the 358 providers and observe the provider’s case load (fewer than 50, or 50 or more) and home use of the drug (yes or no).

    Alternate View
    Number of Abortions
    Permit Drug at Home Fewer than 50 50 or More Totals
    Yes 170 130 300
    No  48  10  58
    Totals 218 140 358

    Based on Henshaw, S. K., and Finer, L. B. “The accessibility of abortion services in the United States, 2001.” Perspectives on Sexual and Reproductive Health, Vol. 35, No. 1, Jan./Feb. 2003 (Table 4).

    1. Find the probability that the provider permits home use of the abortion drug.

    2. Find the probability that the provider permits home use of the drug or has a case load of fewer than 50 abortions.

    3. Find the probability that the provider permits home use of the drug and has a case load of fewer than 50 abortions.

  4. 3.61 Fighting probability of fallow deer bucks. In Aggressive Behavior (Jan./Feb. 2007), zoologists investigated the likelihood of fallow deer bucks fighting during the mating season. During the observation period, the researchers recorded 205 encounters between two bucks. Of these, 167 involved one buck clearly initiating the encounter with the other. In these 167 initiated encounters, the zoologists kept track of whether or not a physical contact fight occurred and whether the initiator ultimately won or lost the encounter. (The buck that is driven away by the other is considered the loser.) A summary of the 167 initiated encounters is provided in the table on p. 143. Suppose we select one of these 167 encounters and note the outcome (fight status and winner).

    1. What is the probability that a fight occurs and the initiator wins?

    2. What is the probability that no fight occurs?

    3. What is the probability that there is no clear winner?

    4. What is the probability that a fight occurs or the initiator loses?

    5. Are the events “No clear winner” and “Initiator loses” mutually exclusive?

    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.

  5. 3.62 Cell phone handoff behavior. A “handoff” is a term used in wireless communications to describe the process of a cell phone moving from a coverage area of one base station to another. Each base station has multiple channels (called color codes) that allow it to communicate with the cell phone. The Journal of Engineering, Computing and Architecture (Vol. 3., 2009) published a study of cell phone handoff behavior. During a sample driving trip that involved crossing from one base station to another, the different color codes accessed by the cell phone were monitored and recorded. The table below shows the number of times each color code was accessed for two identical driving trips, each using a different cell phone model. (Note: The table is similar to the one published in the article.) Suppose you randomly select one point during the combined driving trips.

    Color Code
    0 5 b c Totals
    Model 1 20 35 40 0  85
    Model 2 15 50  6 4  75
    Totals 35 85 46 4 160
    1. What is the probability that the cell phone is using color code 5?

    2. What is the probability that the cell phone is using color code 5 or color code 0?

    3. What is the probability that the cell phone used is Model 2 and the color code is 0?

  6. 3.63 Chemical signals of mice. The ability of a mouse to recognize the odor of a potential predator (e.g., a cat) is essential to the mouse’s survival. The chemical makeup of these odors—called kairomones—was the subject of a study published in Cell (May 14, 2010). Typically, the sources of these odors are major urinary proteins (Mups). Cells collected from lab mice were exposed to Mups from rodent species A, Mups from rodent species B, and kairomones (from a cat). The accompanying Venn diagram shows the proportion of cells that chemically responded to each of the three odors. (Note: A cell may respond to more than a single odor.)

    1. What is the probability that a lab mouse responds to all three source odors?

    2. What is the probability that a lab mouse responds to the kairomone?

    3. What is the probability that a lab mouse responds to Mups A and Mups B, but not the kairomone?

  7. 3.64 Employee behavior problems. The Organizational Development Journal (Summer 2006) reported on the results of a survey of human resource officers (HROs) at major firms. The focus of the study was employee behavior, namely absenteeism, promptness to work, and turnover. The study found that 55% of the HROs had problems with absenteeism. Also, 41% of the HROs had problems with turnover. Suppose that 22% of the HROs had problems with both absenteeism and turnover.

    1. Find the probability that a human resource officer selected from the group surveyed had problems with employee absenteeism or employee turnover.

    2. Find the probability that a human resource officer selected from the group surveyed did not have problems with employee absenteeism.

    3. Find the probability that a human resource officer selected from the group surveyed did not have problems with employee absenteeism nor with employee turnover.

Applying the Concepts—Advanced

  1. 3.65 Cloning credit or debit cards. Wireless identity theft is a technique of stealing an individual’s personal information from radio-frequency-enabled cards (e.g., credit or debit cards). Upon capturing this data, thieves are able to program their own cards to respond in an identical fashion via cloning. A method for detecting cloning attacks in radio-frequency identification (RFID) applications was explored in IEEE Transactions on Information Forensics and Security (Mar. 2013). The method was illustrated using a simple ball drawing game. Consider a group of 10 balls, 5 representing genuine RFID cards and 5 representing clones of one or more of these cards. A coloring system was used to distinguish among the different genuine cards. Because there were 5 genuine cards, 5 colors—yellow, blue, red, purple, and orange—were used. Balls of the same color represent either the genuine card or a clone of the card. Suppose the 10 balls are colored as follows: 3 yellow, 2 blue, 1 red, 3 purple, 1 orange. (See figure on p. 144.) Note that the singleton red and orange balls must represent the genuine cards (i.e., there are no clones of these cards). If two balls of the same color are drawn (without replacement) from the 10 balls, then a cloning attack is detected. For this example, find the probability of detecting a cloning attack.

  2. 3.66 Galileo’s passe-dix game. Passe-dix is a game of chance played with three fair dice. Players bet whether the sum of the faces shown on the dice will be above or below 10. During the late 16th century, the astronomer and mathematician Galileo Galilei was asked by the Grand Duke of Tuscany to explain why “the chance of throwing a total of 9 with three fair dice was less than that of throwing a total of 10.” (Interstat, Jan. 2004). The Grand Duke believed that the chance should be the same, since “there are an equal number of partitions of the numbers 9 and 10.” Find the flaw in the Grand Duke’s reasoning and answer the question posed to Galileo.

  3. 3.67 Encoding variability in software. At the 2012 Gulf Petrochemicals and Chemicals Association (GPCA) Forum, Oregon State University software engineers presented a paper on modeling and implementing variation in computer software. The researchers employed the compositional choice calculus (CCC)—a formal language for representing, generating, and organizing variation in tree-structured artifacts. The CCC language was compared to two other coding languages—the annotative choice calculus (ACC) and the computational feature algebra (CFA). Their research revealed the following: any type of expression (e.g., plain expressions, dimension declarations, or lambda abstractions) found in either ACC or CFA can be found in CCC; plain expressions exist in both ACC and CFA; dimension declarations exist in ACC but not CFA; lambda abstractions exist in CFA but not ACC. Based on this information, draw a Venn diagram that illustrates the relationships among the three languages. (Hint: An expression represents a sample point in the Venn diagram.)

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