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Chapter 4

Probability

4-2 Basic Concepts of Probability

1. To say that the probability of being injured while using recreation equipment in 1/500 meansthat approximately one injury occurs for every 500 times that recreation equipment is used.Note that this is an aggregate result, and not necessarily true for any one particular piece ofequipment since some pieces of equipment are more dangerous than others. Because theprobability 1/500 = 0.002 is small, such an injury is considered unusual.

2. While there are two possible outcomes (Republican, not Republican) for 2012, the probabilitythat a Republican will be elected that year is not 1/2 because the outcomes are not necessarilyequally likely. The probability will depend on which party has the incumbency, who theindividual candidates are, what the issues are, etc.

3. If A denotes the fact that some event occurs, then A denotes the fact that the event does not

occur. If P(A) = 0.995, then P( A ) = 1 0.995 = 0.005. If P(A) = 0.995, then A is very likely

to occur and it would be unusual for A to occur i.e., it would be unusual for A not to occur.

4. Answers will vary, but it seems very likely that such a delay will not occur. A reasonableprobability that one will not be delayed by a car crash blocking the road might be 0.999.

5. 4 in 21 = 4/21 = 0.190

6. 80% = 80/100 = 0.80

7. 5050 chance = 50% = 50/100 = 0.50

8. 5050 chance = 50% = 50/100 = 0.50

9. 6/36 = 0.167

10. 18/38 = 0.474

11. impossible = 0% = 0/100 = 0

12. certain = 100% = 100/100 = 1

13. All probability must be between 0 and 1 inclusive.The value 3:1 = 3/1 = 3 cannot be a probability because 3 > 1.The value 5/2 = 2.5 cannot be a probability because 2.5 > 1.

The value -0.5 cannot be a probability because -0.5 < 0.The value 321/123 = 2.610 cannot be a probability, because 2.610 > 1.

14. a. If event E is certain to occur, then P(E) = 1.b. If it is not possible for event E to occur, then P(E) = 0.c. If a sample space consists of 10 non-overlapping events are equally likely, then the

probability of any one of them is 1/10 = 0.10.d. If C represents answering correctly, then P(C) = 1/2 = 0/50.e. If C represents answering correctly, then P(C) = 1/5 = 0.20.

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Basic Concepts of Probability SECTION 4-2 73

15. a. Let having exactly one girl be the set with 3 simple events A = {bbg,bgb,gbb}.P(A) = 3/8 = 0.375

b. Let having exactly two girls be the set with 3 simple events B = {bgg,gbg,ggb}.P(B) = 3/8 = 0.375

c. Let having all three girls be the set with 1 simple event C = {ggg}.P(C) = 1/8 = 0.125

16. Let E = selecting the AA genotype. By the classical approach to probability, P(E) = 1/4 = 0.25.No, something that occurs 0.25 = 25% of the time is not an unusual event.

17. a. The total number of responses summarized in the table is 15 + 42 + 32 + 9 = 98.b. The number of negative responses (the test says the person did not lie) is 32 + 9 = 41.c. Let N = selecting one of the negative test responses. P(N) = 41/98.d. 41/98 = 0.418

18. The total number of responses summarized in the table is 15 + 42 + 32 + 9 = 98.a. The number of responses that were actually lies is 42 + 9 = 51.

b. Let L = selecting one of the lie responses. P(L) = 51/98.c. 51/98 = 0.520

19. Let F = selecting a false positive response. P(F) = 15/98 = 0.153.The probability of this type of error is high enough that such an occurrence would not beconsidered unusual; the test is not highly accurate.

20. Let F = selecting a false negative response. P(F) = 9/98 = 0.0918.The probability of this type of error is high enough that such an occurrence would not beconsidered unusual; the test is not highly accurate.

21. There are 84 + 16 = 100 total senators.Let W = selecting a woman. P(W) = 16/100 = 0.16.

No; this probability is too far below 0.50 to agree with the claim that men and women have

equal opportunities to become a senator.22. There were 428 + 152 = 580 total plants in the study.

Let G = getting an offspring that is green. By the relative frequency approximation ofprobability, the estimate is P(G) = 428/580 = 0.738. This result is very close to the 3/4 = 0.75expected by Mendel.

23. Let L = getting struck by lightning. By the relative frequency approximation of probability,the estimate is P(L) = 281/290,789,000 = 0.000000966. No; engaging in at risk behaviorincreases the probability of experiencing negative consequences.

24. Let G = getting a baby girl using the technique. By the relative frequency approximation ofprobability, P(G) = 668/726 = 0.920. Yes; since 0.920 is so far above 0.50 it appears that the

technique is effective.

25. a. Let B = the birthdate is correctly identified. P(B) = 1/365 = 0.003.

b. Yes. Since 0.003 0.05, it would be unusual to guess correctly on the first try.c. Most people would probably believe that Mikes correct answer was the result of having

inside information and not making a lucky guess.d. Fifteen years is a big error. If the guess were serious, Kelly would likely think that Mike

was not knowledgeable and/or be personally insulted and a second date is unlikely. If theguess were perceived as being made in jest, Kelly might well appreciate his sense of humorand/or handling of a potentially awkward situation and a second date is likely.

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74 CHAPTER 4 Probability

26. There were 117 + 617 = 734 total patients in the experiment.Let H = a patient experiences a headache. The estimate is P(H) = 117/734 = 0.159.No. Since 0.159 > 0.05, experiencing a headache from taking the drug is not an unusualoccurrence. Yes, the probability of experiencing a headache is high enough to be of concern especially since a headache is one symptom that can interfere with the intended purpose fortaking the drug in the first place.

27. Let F = a pacemaker malfunction is caused by firmware.Based on the given results, P(F) = 504/8834 = 0.0571.No. Since 0.0571 > 0.05, a firmware malfunction is not unusual among pacemakermalfunctions.

28. Let B = a Delta airline passenger is involuntarily bumped from a flight.Based on the given results, P(B) = 3/15378 = 0.000195.

Yes. Since 0.000195 0.05, such a bumping is unusual. Because the probability of such abumping is so low, it does not pose a serious problem for Delta passengers.

29. There were 795 + 10 = 805 total persons executed.

Let W = a randomly selected execution was that of a woman. P(W) = 10/805 = 0.0124.Yes. Since 0.0124 0.05, it is unusual for an executed person to be a woman. This is due tothe fact that more crimes worthy of the death penalty are committed by men than by women.

30. There were 481 + 401 + 120 = 1002 total persons in the survey.Let F = a person favors such use of federal tax dollars.Based on these results, P(F) = 481/1002 = 0.480.No. Since 0.480 > 0.05, it is not unusual for a person to favor such use of federal tax dollars.

31. There were 211+ 288 + 366 + 144 + 89 = 1098 total persons in the survey.Let F = a household has 4 or more cellphones in use.Based on these results, P(F) = 89/1098 = 0.0811.No. Since 0.0811 > 0.05, it is not unusual for a household to have 4 or more cellphones in use.NOTE: Technically, the given survey cannot be used to answer the question as posed. Asurvey made of persons selected at random cannot be used to answer questions abouthouseholds. If the goal is to make statements about households, then households (and notpersons) should be selected at random. Selecting persons at random means that householdswith more people have a higher probability of being in the survey than households with fewerpeople. Since households with more people are more likely to have more cellphones, thesurvey will tend to overestimate the number of cellphones per household.

32. There were 1065 + 240 + 14 + 66 = 1385 total persons in the survey.Let N = a worker does not make personal phone calls at work.Based on these results, P(N) = 66/1385 = 0.0477.

Yes. Since 0.0477 0.05, it is unusual for a worker not to make personal calls at work.

33. a. Listed by birth order, the sample space contains 4 simple events: bb bg gb gg.b. P(gg) = 1/4 = 0.25c. P(bg or gb) = 2/4 = 0.50

34. a. Listed by birth order, the sample space contains 16 simple events:bbbb bbbg bbgb bbgg bgbb bgbg bggb bggggbbb gbbg gbgb gbgg ggbb ggbg gggb gggg.

b. P(bbgg or bgbg or bggb or gbbg or gbgb or ggbb) = 6/16 = 0.375c. P(bbbb) = 1/16 = 0.0625

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Basic Concepts of Probability SECTION 4-2 75

35. a. Listed with the fathers contribution first, the sample space has 4 simple events:brown/brown brown/blue blue/brown blue/blue.

b. P(blue/blue) = 1/4 = 0.25c. P(brown eyes) = P(brown/brown or brown/blue or blue/brown) = 3/4 = 0.75

36. In each scenario. The contribution of the father is listed first.

a. For an xY father and an XX mother, there are 4 possibilities: xX xX YX YX.The sample space for a son has 2 simple events: YX YX.P(xY or Yx) = 0/2 = 0

b. For an xY father and an XX mother, there are 4 possibilities: xX xX YX YX.The sample space for a daughter has 2 simple events: xX xX.P(xx) = 0/2 = 0

c. For an XY father and an xX mother, there are 4 possibilities: Xx XX Yx YXThe sample space for a son has 2 simple events: Yx YX.P(xY or Yx) = 1/2 = 0.50

d. For an XY father and an xX mother, there are 4 possibilities: Xx XX Yx YXThe sample space for a daughter has 2 simple events: Xx XX.P(xx) = 0/2 = 0

37. The odd against winning are P(not winning)/P(winning) = (423/500)/(77/500) = 423/77,or 423:77 which is approximately 5.5:1, or 11:2.

38. Of the 38 slots, 18 are odd. Let W = winning because an odd number occursa. P(W) = 18/38 = 9/19 = 0.474b. odds against W = P(not W)/P(W) = (20/38)/(18/38) = 20/18 = 10/9, or 10:9c. If the payoff odds are 1:1, a win gets back your bet plus $1 for every $1 bet.

If you bet $18 and win, you get back 18 + 18 = $36 and your profit is $18.d. If the payoff odds are 10:9, a win gets get back your bet plus $10 for every $9 bet.

If you bet $18 and win, you get back 18 + 20 = $38 and your profit is $20.

39. a. If a $2 bet results in a net return of $14.20, the net profit is 14.20 2.00 = $12.20.b. If you get back your bet plus $12.20 for every $2 you bet, the payoff odds are 12.2:2

[typically expressed as either 6.1:1 or 61:10], or approximately 6:1.c. odds against W = P(not win)/P(W) = (443/500)/(57/500) = 443/57 or 443:57

[typically expressed as 7.77:1], or approximately 8:1.d. If the payoff odds are 7.77:1, a win gets back your bet plus $7.77 for every $1 bet.

If you bet $2 and win, you get back 2 + 15.54 = $17.54. The ticket would be worth $17.54and your profit would be 17.54 2.00 = $15.54.

40. In this problem the odds against winning are 97:3, and so a = 97 and b = 3.P(W) = a/(a+b) = 3/100 = 0.03

NOTE: If the odds against winning are a:b = a/b, then a/b = (a/x)/(b/x) = P( W )/P(W),

where x = a+b [the value necessary for P(W ) + P(W) = 1]. And so P(W) = b/(a+b).

41. Making a table like the one on the right organizesthe information and helps to understand the conceptsinvolved.pt = 26/2103 = 0.0124pc = 22/1671 = 0.0132

Headache?no yes

Nasonex 2077 26 2103

control 1649 22 1671

3276 48 3774

Treatment

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a. The relative risk is the P(risk) for the treatment as a percent of the P(risk) for the control.relative risk = pt/pc = 0.0124/0.0132 = 0.939 = 93.9%

b. The odds in favor of the risk are P(risk)/P(not risk).for the treatment: odds in favor= pt/(1-pt) = (26/2103)/(2077/2103) = 26/2077 = 0.012518/1

for the control: odds in favor= pc/(1-pc) = (22/1671)/(1649/1671) = 22/1649 = 0.013341/1The odds ratio is the odds in favor of the risk for the treatment as a percent of the odds in

favor of the risk for the control.odds ratio = [pt/(1-pt)/[pc/(1-pc)] = 0.012518/0.013314 = 0.938 = 93.8%

Both the relative risk and the odds ratio are less than 1.00, suggesting that the risk of headacheis actually smaller for Nasonex than it is for the control. Nasonex does not appear to pose arisk of headaches.

42. No matter where the two flies land, it is possible to cut the orange in half to include flies on thesame half. Since this is a certainty, the probability is 1. [Compare the orange to a globe.Suppose fly A lands on New York City, and consider all the circles of longitude. Wherever flyB lands, it is possible to slice the globe along some circle of longitude that places fly A and flyB on the same half.]NOTE: If the orange is marked into two predesignated halves before the flies land, theprobability is different once fly A lands, fly B has a chance of landing on the same half. Ifboth flies are to land on a specific one of the two predesignated halves, the probability isdifferent still fly A has a chance of landing on the specified half, and only the time willfly B pick the same half: the final answer would be of , or .

43. This difficult problem will be broken into two events and a conclusion. Let L denote thelength of the stick. Label the midpoint of the stick M, and label the first and second breakingpoints X and Y. Event A: X and Y must be on opposite sides of M. If X and Y are on the same side of M,

the side without X and Y would be longer than 0.5L and no triangle will be possible.Once X is set, P(Y on the same side as X) = P(Y on opposite side from X) = .

Event B: The distance |XY| must be less than 0.5L. Assuming X and Y are on opposite sidesof M, let Q denote the end closest to X and R denote the end closest to Y so that thefollowing sketch applies.

|--------|------------|----|----------------|Q X M Y R

In order to form a triangle, it must be true that |XY| = |XM| + |MY| < 0.5L. This happensonly when |MY| < |QX|. With all choices at random, there is no reason for |QX| to be largerthan |MY|, or vice-versa. This means that P(|MY|

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Addition Rule SECTION 4-3 77

4-3 Addition Rule

1. Two events in a single trial are disjoint if they are outcomes that cannot happen at the sametime.

2. Yes, two complementary events must also be disjoint. One event is the complement of another

if it includes exactly the outcomes not included in the other. If one event includes only theoutcomes not included in the other, then the events have no overlap and are disjoint.

3. In the context of the addition rule defined in this section, P(A and B) is the probability thatboth event A and event B occur at the same time in a single trial.

4. The addition rule as defined in this section refers to a single trial and the scenario givensuggests two separate trials: one involving the birth of a baby and one involving a coin toss.NOTE: The addition rule is a powerful tool, and there are two ways it could apply to the givenscenario. (1) Define a trial to be the birth of a baby and the tossing of a coin, with a samplespace consisting of 4 simple events: GH GT BH BT. (2) Apply the concepts of the additionrule to two trials. In either case the solution involves material not yet covered in this chapter.See exercise 37 in this section for further discussion of this exercise.

5. Not disjoint, since a physician can be a female surgeon.

6. Disjoint, since a subject cannot be both a Republican and a Democrat.

7. Disjoint, since a car that is free of defects cannot have a dead battery which is a defect.

8. Disjoint, since a fruit fly cannot have red eyes and dark brown eyes.

9. Not disjoint, since a subject who believes the evidence for global warming can also be opposedto stem cell research.

10. Disjoint, since a subject treated with the medication Lipitor cannot be a subject given nomedication.

11. Not disjoint, since an R-rated movie can receive a four-star review.

12. Not disjoint, since it is possible for a homeless person to be a college graduate.

13. P( M ) is the probability that a STATDISK user is not a Macintosh computer user

P( M ) = 1 P(M) = 1 0.05 = 0.95

14. Let C = a selected woman has red/green color blindness. P(C) = 0.25% = 0.0025.

P(C ) = 1 P(C) = 1 0.0025 = 0.9975, rounded to 0.998

15. Let M = a selected American believes it morally wrong not to report all income. P(M) = 0.79.

P( M ) = 1 P(M) = 1 0.79 = 0.21

16. P( I ) is the probability that a screened driver is not intoxicated

P( I ) = 1 P(I) = 1 0.00888 = 0.99112, rounded to 0.991

NOTE: For exercises 17-20, refer to the tableat the right and use the following notation.Let P = the polygraph test indicates a positiveLet L = the person actually lied.

Subject Lie?no yes

positive 15 42 57

negative 32 9 41

47 51 98

Test Result

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17. P(P or L ) = P(P) + P( L ) P(P and L )= 57/98 + 47/97 15/98= 89/98 = 0.908

18. P( L ) = 47/98 = 0.480

19. P( L and P ) = 32/98 = 0.327

20. P( P or L) = P( P ) + P(L) P( P and L)= 41/98 + 51/98 9/98= 83/98 = 0.847

NOTE: For exercises 21-26, refer to the tableat the right and use the following notation.Let M = the challenge was made by a male.Let S = the challenge was successful.

21. P(S ) = 512/839 = 0.610

22. P( M ) = 350/839 = 0.417

23. P(M or S) = P(M) + P(S) P(M and S)= 489/839 + 327/839 201/839= 615/839 = 0.733

24. P( M or S) = P( M ) + P(S) P( M and S)= 350/839 + 327/839 126/839= 551/839 = 0.657

25. P(M or S ) = P(M) + P( S ) P(M and S )

= 489/839 + 512/839 288/839

= 713/839 = 0.850

26. P(M or S ) = P(M ) + P(S ) P(M and S )

= 350/839 +512/839 224/839= 638/839 = 0.760

NOTE: For exercises 27-32, refer to the following table and notation.For a person selected at random, let the following designations apply.Y = there is a responseN = there is a refusalA = the age is 18-21

B = the age is 22-29C = the age is 30-39D = the age is 40-49E = the age is 50-59F = the age is 60+

27. P(N) = 156/1205 = 0.129Yes. Persons who refuse to answer form a subpopulation whose opinions will not be counted,thus preventing the survey from being representative of the entire population.

28. P(F and Y) = 202/1205 = 0.168

Successful?yes no

male 201 288 489

female 126 224 350

327 512 839

Gender

Age18-21 22-29 30-39 40-49 50-59 60+

yes 73 255 245 136 138 202 1049

no 11 20 33 16 27 49 156

84 275 278 152 165 251 1205

Response?

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29. P(Y or A) = P(Y) + P(A) P(Y and A)= 1049/1205 + 84/1205 73/1205= 1060/1205 = 0.880

30. P(N or F) = P(N) + P(F) P(N and F)= 156/1205 + 251/1205 49/1205

= 358/1205 = 0.29731. Use the intuitive approach rather than the formal addition rule.

P(Y or B or C) = P(Y) + P(B and N) + P(C and N)= 1049/1205 + 20/1205 + 33/1205= 1102/1205 = 0.915

32. Use the intuitive approach rather than the formal addition rule.P(N or A or F) = P(N) + P(A and Y) + P(F and Y)

= 156/1205 + 73/1205 + 202/1205= 431/1205 = 0.368

NOTE: For exercises 33-36 refer to the tableat the right and use the following notation.Let P = the drug test indicates a positiveLet M = the person actually uses marijuana

33. a. 300b. 178

c. P(M ) = 178/300 = 0.593

34. P(P or M) = P(P) + P(M) P(P and M)= 143/300 + 122/300 119/300= 146/300 = 0.487

35. P( P or M ) = P( P ) + P( M ) - P( P and M )= 157/300 + 178/300 154/300= 181/300 = 0.603

36. P(M) = 122/300 = 0.407No. The manner in which the subjects were chosen for this test is not stated, but there is noimplication that they were selected to be representative of the general population. Since thestudy seems to be aimed toward determining the accuracy of a drug test for marijuana usage, itis likely that participants were chosen from a pool which guaranteed a reasonable number ofappropriate subjects i.e., from a pool with a marijuana usage rate higher than that of thegeneral population.

37. P[(false positive) or (false negative)]= P[(P andM ) or (P and M)]

= P(P and M ) + ( P and M) P[(P and M ) and ( P and M)]= 24/300 + 3/300 0/300= 27/300 = 0.090

Using the 0.05 guideline, the occurrence of a false positive or a false negative is not unusual.It appears that the test is not highly accurate.

Use Marijuana?yes no

positive 119 24 143

negative 3 154 157

122 178 300

Test Result

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38. P[(true positive) or (true negative)]

= P[(P and M) or (P and M )]

= P(P and M) + ( P and M ) P[(P and M) and ( P and M )]= 119/300 + 154/300 0/300= 273/300 = 0.910

Exercise 37 asks for P(test result is in error), and this exercise asks for P(test result is correct).

Because these are complementary events, the above answer could have been found usingP(test result is correct) = 1 P(test result is in error) = 1 0.090 = 0.910.

39. Assuming girls and boys are equally likely, the birth-toss sample space consists of 4 equallylikely simple events: GH GT BH BT.Using the concepts of the formal addition rule,

P(G or H) = P(G) + P(H) P(G and H)= 2/4 + 2/4 1/4= 3/4 = 0.75

Using the concepts of the intuitive addition rule,P(G or H) = 3/4 = 0.75

40. No; the facts that A and B are disjoint and that B and C are disjoint doe not mean that A and Cmust be disjoint. Consider, for example, the following A, B and C.Let A = has prostate cancer.Let B = is a female.Let C = has testicular cancer.

41. If the exclusive oris used instead of the inclusive or, then the double-counted probability mustbe completely removed (i.e., it must be subtracted twice) and the formula becomes as follows.

P(A or B) = P(A) + P(B) 2P(A and B)

42. The expression can be derived using algebra as follows.P(A or B or C)

= P[(A or B) or C]= P(A or B) + P(C) P[(A or B) and C]= [P(A) + P(B) P(A and B)] + P(C) P[(A and C) or (B and C)]= P(A) + P(B) P(A and B) + P(C) [P(A and C) + P(B and C) P(A and B and C)]= P(A) + P(B) P(A and B) + P(C) P(A and C) P(B and C) + P(A and B and C)= P(A) + P(B) + P(C) P(A and B) P(A and C) P(B and C) + P(A and B and C)

43. The following two English/logic facts are used in this exercise. not (A or B) = not A and not B Is your sister either artistic or bright?

No? Then she is not artistic and she is not bright. not (A and B) = not A or not B Is your brother artistic and bright?

No? Then either he is not artistic or he is not bright.

a. P(A or B) = P( A and B ) from the first fact aboveor (from the rule for complementary events)

P(A or B) = 1 P(A or B) = 1 [P(A) + P(B) P(A and B)] = 1 P(A) P(B) + P(A and B

b. P( A or B ) = 1 P(A and B) from the second fact above

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c. They are different: part (a) gives the complement of A or Bpart (b) gives the complement of A and B

NOTE: The following example illustrates this concept. Consider the following 5 students andtheir majors: Rob (art), Steve (art), Tom (art and biology), Uriah (biology), Vern (math).Select one student at random and consider the major.P(A) = 3/5 P(B) = 2/5 P(A or B) = 4/5 P(A and B) = 1/5.

a. P(A or B) = 1 P(A or B)= 1 4/5= 1/5 [only Vern is neither an art major nor a biology major]

b. P( A or B ) = P( A ) + P( B ) P( A and B )= 2/5 + 3/5 1/5= 4/5 [everyone but Tom is either not an art major or not a biology major]

c. No, 1/5 4/5.

4-4 Multiplication Rule: Basics

1. Answers will vary, but the following are possibilities.Independent events: Getting an even number when a die is rolled and getting a head when a

coin is tossed.Dependent events: Selecting aces on two consecutive draws without replacement from an

ordinary deck of cards.

2. P(B|A) represents the probability that event B occurs after event A has already happened.

3. The events of selecting adults are dependent, because the adults are selected withoutreplacement. On any given pick, the probability of a particular person being selected dependsupon who and how many persons have already been selected.

4. Since the sample is 1068/477938 = 0.223% 5% of the population, the 5% guidelineindicates that the events in Exercise 3 can be treated as being independent.

5. Independent, since the population is so large that the 5% guideline applies.

6. Dependent, since the two items operate from the same power supply.

7. Dependent, if there is visual contact during the invitation, since visual impressions affectresponses.

Independent, if there is no visual contact during the invitation, since the invitee would have noknowledge of the inviters clothing.

8. Independent, since the two items have unrelated sources of power.

9. Dependent, since the two items operate from the same power supply.

10. Independent, since the two items have unrelated sources of power.

11. Independent, since the population is so large that the 5% guideline applies.

12. Dependent, since having access to a computer is necessary for using the Internet.

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13. Let F = a selected person had false positive results.P(F1 and F2) = P(F1)P(F2|F1)

= (15/98)(14/97) = 0.0221Yes; since 0.0221 0.05, getting two subjects who had false positives would be unusual.NOTE: Since n=2 is 2/98 = 0.0204 0.05 of the population, one could use the 5% guidelineand treat the selections as independent to get (15/98)2 = 0.0234. Typically the 5% guideline is

invoked only when the calculations for dependent events are extremely tedious.

14. Let F = a selected person had false positive results.P(F1 and F2 and F3) = P(F1)P(F2|F1)P(F3|F1 and F2)

= (15/98)(14/97)(13/96) = 0.00299Yes; since 0.00229 0.05, getting three subjects who had false positives would be unusual.

15. Let P = a selected person tested positiveP(P1 and P2 and P3 and P4) = P(P1)P(P2|P1)P(P3|P1 and P2)P(P4|P1 and P2 and P3)

= (57/98)(56/97)(55/96)(54/95) = 0.109No; since 0.109 > 0.05, getting four subjects who had positive results would not be unusual.

16. Let I = a selected person had incorrect results.P(I1) = 15/98 + 9/98 = 24/98P(I1 and I2 and I3 and I4) = P(I1)P(I2|I1)P(I3|I1 and I2)P(I4|I1 and I2 and I4)

= (24/98)(23/97)(22/96)(21/95) = 0.00294Yes; since 0.00295 0.05, getting four subjects who had incorrect results would be unusual.

NOTE: For exercises 17-20 refer to the table at the right.

17. P(O and Rh+) = 39/100a. (39/100)(39/100) = 0.152b. (39/100)(38/99) = 0.150

18. P(B and Rh-) = 2/100a. (2/100)(2/100)(2/100) = 0.00008b. (2/100)(1/99)(0/98) = 0

19. P(O and Rh-) = 6/100a. (6/100)(6/100)(6/100)(6/100) = 0.0000130b. (6/100)(5/99)(4/98)(3/97) = 0.00000383

20. P(AB and Rh+) = 4/100a. (4/100)(4/100)(4/100) = 0.000064b. (4/100)(3/99)(2/98) = 0.0000247

21. a. Yes; since the correctness of the first response has no effect on the correctness of the second,

the events are independent.b. P(C1 and C2) = P(C1)P(C2|C1)

= (1/2)(1/4) = 1/8 = 0.125c. No; since 0.125 > 0.05, getting both answers correct would not be unusual but it certainly

isnt likely enough to recommend such a practice.

22. Let P = a power supply unit is OKP(entire batch is accepted) = P(P1 and P2 and P3)

= P(P1)P(P2|P1)P(P3|P1 and P2)= (392/400)(391/399)(390/398) = 0.941

Blood GroupO A B AB

Rh+ 39 35 8 4 86

Rh- 6 5 2 1 1445 40 10 5 100

Type

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Multiplication Rule: Basics SECTION 4-4 83

23. Let A = a public opinion poll is accurate within its margin of error.P(A) = 0.95 for each polling organizationP(all 9 are accurate) = P(A1 and A2 andand A9)

= P(A1)P(A2)P(A9)= (0.95)

9= 0.630

No. With 9 independent polls the probability that at least one of them is not accurate within its

margin of error is 1 P(all accurate) = 1 0.630 = 0.370.

24. Let F = a choice is favorable to producing a total match.P(F1) = 5/5 = 1 NOTE: With nothing to match, it will be favorable no matter what.P(Fi) = 1/5, for i = 2 to 9 NOTE: Subsequent choices must match the first one.P(all the same choice) = P(F1 and F2 and F3 and F4 and F5 and F6 and F7 and F8 and F9)

= P(F1) P(F2) P(F3) P(F4) P(F5) P(F6) P(F7) P(F8) P(F9)= (5/5)(1/5)(1/5)(1/5)(1/5)(1/5)(1/5)(1/5)(1/5) = (1)(1/5)8 = 0.00000256

There is no reasonable doubt that the voice selected mostly closely matches the voice of thecriminal. But if the victims were under pressure to select one of the voices, and if this were theonly voice close to that of the criminal, a case could me made for reasonable doubt about theguilt of the selected person.

25. Let G = getting a girl.P(G) = 1/2 for each birthP(G1 and G2 and G3) = P(G1)P(G2)P(G3)

= (1/2)(1/2)(1/2) = 1/8 = 0.125No; since 0.125 > 0.05, getting 3 girls by chance alone is not an unusual event. The resultsare not necessarily evidence that the gender-selection method is effective.

26. Let G = getting a girl.P(G) = 1/2 for each birthP(G1 and G2 andand G10) = P(G1)P(G2)P(G10)

= (1/2)(1/2)(1/2) = (1/2)10 = 0.000977Yes; since 0.000977 0.05, getting 10 girls by chance alone is an unusual event. Since theprobability 0.000977 is so small, the results are evidence that the gender-selection method iseffective.

27. Let A = the alarm works.P(A) = 0.9 for each alarm

a. P(A ) = 1 P(A) = 1 0.9 = 0.1

b. P( 1 2A and A ) = P(A 1)P(A 2)

= (0.1)(0.1) = 0.01

c. P(being awakened) = P( 1 2A and A )

= 1 P( 1 2A and A )= 1 0.01 = 0.99

NOTE: The above parts (b) and (c) assume that the alarm clocks work independently of eachother. This would not be true if they are both electric alarm clocks.

28. Let F = the radio fails.P(F) = 0.002 for each radioP(F1 and F2) = P(F1)P(F2) = (0.002)(0.002) = 0.000004The second independent radio decreases the probability of radio failure from 0.002 to0.000004, which is a substantial increase in safety.

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29. Let G = getting a tire that is goodP(G) = 4800/5000 = 0.96a. P(G1 and G2 and G3 and G4) = P(G1)P(G2|G1)P(G3|G1 and G2)P(G4|G1 and G2 and G3)

= (4800/5000)(4799/4999)(4798/4998)(4797/4997) = 0.849b. Since n = 100 represents 100/5000 = 0.02 0.05 of the population, use the 5% guideline

and treat the repeated selections as being independent.

P(G1 and G2 andand G100) = P(G1)P(G2)P(G100)= (0.96)(0.96)(0.96) = (0.96)100 = 0.0169

Yes; since 0.0169 0.05, getting 100 good tires would be an unusual event and the outletshould plan on dealing with returns of defective tires.

30. Let G = a selected ignition system is good.P(G) = 195/200 = 0.975a. No; since the selections are made without replacement, the selection process does not

involve independent events.b. P(G1 and G2 and G3) = P(G1)P(G2|G1)P(G3|G1 and G2)

= (195/200)(194/199)(193/198) = 0.926c. Since n = 3 represents 3/200 = 0.015 0.05 of the population, use the 5% guideline and

treat the repeated selections as being independent.P(G1 and G2 and G3) = P(G1)P(G2)P(G3)

= (0.975)(0.975)(0.975) = 0.927d. The answer from part (b) is better because it is based on exact calculations using the

mathematical formula that accurately models the selection process.

31. Let W = the surge protector works correctlyP(W) = 0.99 for each of p and qa. The TV is protected if at least one of the surge protectors works correctly.

P(Wp or Wq) = P(Wp) + P(Wq) P(Wp and Wq)= 0.99 + 0.99 0.9801 [.9801 comes from part (b)]

= 0.9999b. The TV is protected only if both surge protectors work correctly.

P(Wp and Wq) = P(Wp)P(Wq) [since they operate independently]= (.99)(.99) = 0.9801

c. The series arrangement in part (a) provided the better coverage. The parallel arrangement inpart (b) actually provides poorer protection than a single protector alone.

32. Let D = a birthday is different from any yet selected.P(D1) = 365/365 = 1 NOTE: With nothing to match, it must be different.P(D2) = P(D2|D1) = 364/365P(D3) = P(D3|D1 and D2) = 363/365

P(D25) = P(D25|D1 and D2 and D3 andand D24) = 341/365P(all different) = P(D1 and D2 andand D25)

= P(D1)P(D2)P(D25)= (365/365)(364/365)(341/365) = 0.431

NOTE: An algorithm to perform this calculation can be constructed using a programminglanguage, spreadsheet, or statistical software package. In BASIC, for example, use

10 LET P=1 40 LET P=P*(365K)/365

15 PRINT HOW MANY PEOPLE? 50 NEXT K20 INPUT D 60 PRINT P(ALL DIFFERENT BIRTHDAYS) IS ;P

30 FOR K=1 TO D1 70 END

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Multiplication Rule: Basics SECTION 4-4 85

33. This problem can be done by two different methods. In either case, letA = getting an aceS = getting a spade. Consider the sample space

The first card could be any of 52 cards, and for each first card there are 51 possiblesecond cards. This makes a total of 5251 = 2652 equally likely outcomes in the sample

space. How many of them are A1 and S2?The aces of hearts, diamond and clubs can be paired with any of the 13 spades for a

total of 313 = 39 favorable possibilities. The ace of spades can only be paired with anyof the remaining 12 members of that suit for a total of 12 favorable possibilities. Sincethere are 39 + 12 = 51 total favorable possibilities among the equally likely outcomes,P(A1 and S2) = 51/2652 = 0.0192

Use the formulas (and express the event as two mutually exclusive possibilities)P(A1 and S2) = P([spadeA1 and S2]or [otherA1 and S2])

= P(spadeA1 and S2) + P(otherA1 and S2)= P(spadeA1)P(S2|spadeA1) + P(otherA1)P(S2|otherA1)= (1/52)(12/51) + (3/52)(13/51)

= 12/2652 + 39/2652 = 51/2652 = 0.0192

4-5 Multiplication Rule: Complements and Conditional Probability

1. If at least one of the 10 pacemakers is defective, the exact number of defective pacemakersmust be one of the following: 1,2,3,4,5,6,7,8,9,10.

2. P(B|A) denotes the probability that event B occurs once it is known that event A has alreadyoccurred.

3. The probability of a particular outcome is equal to one divided by the total number of

outcomes only when all the outcomes are equally likely and mutually exclusive. In this casethe two outcomes are not equally likely, and so that reasoning cannot be used. The actualsurvival rate was not considered, but it should have been.

4. In probability, confusion of the inverse is defined to be mistakenly believing that P(B|A) is thesame as P(A|B) or to incorrectly using the value of one for the other.

5. If it is not true that at least one of the 15 tests positive, then all 15 of them test negative.

6. If it is not true that all 6 of them are free of defects, than at least one of them is defective.

7. If it is not true that none of the 4 has the recessive gene, then at least one of the 4 has therecessive gene.

8. If it is not true that at least one of the 5 accepts the invitation, then all 5 of them rejected theinvitation.

9. P(at least one girl) = 1 P(all boys)= 1 P(B1 and B2 and B3 and B4 and B5 and B6)= 1 P(B1)P(B2)P(B3)P(B4)P(B5)P(B6)= 1 ()()()()()()= 1 1/64 = 63/64 = 0.984

Yes, that probability is high enough for the couple to be very confident that they will get atleast one girl in six children.

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10. P(at least one girl) = 1 P(all boys)= 1 P(B1 and B2 and B3 and B4 and B5 and B6 and B7 and B8)= 1 P(B1)P(B2)P(B3)P(B4)P(B5)P(B6)P(B7)P(B8)= 1 ()()()()()()()()= 1 1/256 = 255/256 = 0.996

If the couple has 8 boys, either a very rare event has occurred or there is some environmental

or genetic factor that makes boys more likely for this couple.NOTE: While the described event is certainly rare, it is expected to occur about 4 times inevery 1000 families with 8 children. This could simply be one of those 4 families.

11. P(at least one correct) = 1 (all wrong)= 1 P(W1 and W2 and W3 and W4)= 1 P(W1)P(W2)P(W3)P(W4)= 1 (4/5)(4/5)(4/5)(4/5)= 1 256/625 = 269/625 = 0.590

Since there is a greater chance of passing than of failing, the expectation is that such a strategywould lead to passing. In that sense, one can reasonably expect to pass by guess. But whilethe expectation for a single test may be to pass, such a strategy can be expected to lead tofailing about 4 times every 10 times it is applied.

12. P(at least one working calculator) = 1 P(all fail)= 1 P(F1 and F2)= 1 P(F1)P(F2)= 1 (0.04)(0.04)= 1 0.0016 = 0.9984, rounded to 0.998

Yes. The increase in the probability of being able to finish the exam with a working calculatorfrom 96% to 99.8% is probably significant enough to justify such action. In any event, theextra peace of mind such a student would gain is probably more than worth the extra effort ofcarrying an additional calculator.

13. Since each birth is an independent event, P(G4|G1 and G2 and G3) = P(G4) = P(G) = .One can also consider the 16-outcome sample space S for 4 births and use the formula forconditional probability as follows.

S = {BBBG, BBGG, BGBG, GBBG, BGGG, GBGG, GGBG, GGGG,BBBB, BBGB, BGBB, GBBB, BGGB, GBGB, GGBB, GGGB}

Let F = the first 3 children are girlsLet G4 = the fourth child is a girlP(F) = 2/16P(F and G4) = 1/16

44

P(G and F) 1/16P(G |F) = 1/ 2

P(F) 2 /16

= =

Note: Using the sample space S also verifies directly that P(G4) = 8/16 = 1/2 = P(G),independent of what occurred during the first three births.No. This probability is not the same as P(GGGG) = 1/16.

14. P(at least one delinquent) = 1 (all good)= 1 P(G1 and G2 and G3 and G4)= 1 P(G1)P(G2)P(G3)P(G4)= 1 (0.99)(0.99)(0.99)(0.99)= 1 0.9606 = 0.0394

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Multiplication Rule: Complements and Conditional Probability SECTION 4-5 87

15. Let P(C) = P(crash) = 0.0480, and P(N) = P(no crash) = 0.9520.P(at least one crash) = 1 P(no crashes)

= 1 P(N1 and N2 and N3 and N4)= 1 P(N1)P(N2)P(N3)P(N4)= 1 (0.9520)(0.9520)(0.9520)(9.8520)= 1 0.821 = 0.179

Yes, the above probability does not apply to the given story problem. The above solutionassumes that the cars are selected at random so that, for example, P(C2|C1) = P(C) = 0.0480.These cars were not selected at random, but they were all taken from the same family. Sincethe cars were driven by the same people, or at least by people likely to have similar drivinghabits, knowing that one of the cars has a crash would reasonably make a person think that itwas more likely that another of the cars would crash i.e., the events are not independent andP(C2|C1) > P(C) = 0.0480.

16. P(at least one girl) = 1 P(all boys)= 1 P(B1 and B2 and B3 and B4 and B5)= 1 P(B1)P(B2)P(B3)P(B4)P(B5)= 1 (0.5845)(0.5845)(0.5845)(0.5845)(0.5845)= 1 0.0682 = 0.9318, rounded to 0.932

Probably not. The system will eventually produce such a shortage of females that baby girlswill become a valuable asset and parents will take appropriate measures to change theprobabilities.

17. P(at least one with vestigial wings) = 1 P(all have normal wings)= 1 P(N1 and N2 and N3 and and N10)= 1 P(N1)P(N2)P(N3)P(N10)= 1 (3/4)(3/4)(3/4)(3/4)= 1 (3/4)10 = 1 0.0563 = 0.9437, rounded to 0.944

Yes, the researchers can be 94.4% certain of getting at least one such offspring.

18. Let P(C) = P(cleared with arrest) = 0.249, and P(N) = P(not cleared with arrest) = 0.751.a. P(at least one cleared) = 1 P(all not cleared)

= 1 P(N1 and N2 and N3 and and N10)= 1 P(N1)P(N2)P(N3)P(N10)= 1 (0.751)(0.751)(0.751)(0.751)= 1 (0.751)10 = 1 0.0571 = 0.9429, rounded to 0.943

b. P(clears all 10) = P(C1 and C2 and C3 and and C10)= P(C1)P(C2)P(C3)P(C10)= (0.249)(0.249)(0.249)(0.249) = (0.249)10 = 0.000000916

c. If the detective clears all 10 cases with arrests, we should conclude that the P(C) = 0.249 ratedoes not apply to this detective i.e., that the probability he clears a case with an arrest ismuch higher than 0.249.

NOTE: Technically, the probabilities in (a) and (b) do not apply to the given story problem.The above solution assumes that the cases are selected at random so that, for example,P(C2|C1) = P(C) = 0.249. These cases were not selected at random, but they were all takenfrom the same officers case load. Since the cases were handled by the same person, knowingthat one of the cases closes with an arrest would reasonably make a person think that it wasmore likely that other of the cases would close with an arrest i.e., the events are notindependent and P(C2|C1) > P(C) = 0.249. [See also exercise 3 in the Statistical Literacy andCritical Thinking exercises at the end of this chapter.] In addition, this is a new detective andthere is no reason to assume the usual rate would apply to a new recruit.

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NOTE: For exercises 19-22, refer to the tableat the right and use the following notation.Let P = the lie detector test indicates a positiveLet Y = the person actually lied

19. P(P| Y ) = 15/47 = 0.319

This case is problematic because it represents the testindicating incorrectly that a person lied when he was telling the truth.

20. P( P |Y) = 9/51 = 0.176This result suggests that the polygraph is not very reliable because 17.6% of the time it fails tocatch a person who really is lying.

21. P(Y|P ) = 9/41 = 0.220This result (the probability that a person is really lying when the test fails to accuse him) isdifferent from the one in Exercise 20 (the probability that the test fails to accuse a person whenhe really is lying).

22. a. P(P | Y ) = 32/47 = 0.681

b. P( Y | P ) = 32/41 = 0.780c. No. The first result (the probability that the test fails to accuse a person that is telling the

truth) is not the same as the second result (the probability that a person is telling the truthwhen the test does not accuse him of lying).

NOTE: For exercises 23-26, refer to thetable at the right.

23. a. P(identical) = 10/30 = 1/3 = 0.333b. P(identical|BB) = 5/10 = 1/2 = 0.5

24. a. P(fraternal) = 20/30 = 2/3 = 0.667b. P(fraternal|BG or GB) = 10/10 = 1

25. P(BG or GB|fraternal) = 10/20 = 1/2 = 0.5

26. P(GG|fraternal) = 5/20 = 1/4 = 0.25

27. P(at least one works) = 1 P(all fail)= 1 P(F1 and F2 and F3)= 1 P(F1)P(F2)P(F3)= 1 (0.10)(0.10)(0.10)= 1 0.001 = 0.999

Yes, the probability of a working clock rises from 90% with just one clock to 99.9% with 3

clocks. If the alarm clocks run on electricity instead of batteries, then the clocks do not operateindependently and the failure of one could be the result of a power failure or interruption andmay be related to the failure of another i.e., P(F2|F1) is no longer P(F) = 0.90.

28. a. P(at least one defective) = 1 (all good)= 1 P(G1 and G2)= 1 P(G1)P(G2|G1)= 1 (388/400)(387/399)= 1 0.9408 = 0.0592

NOTE: Even though the sample size is 2/400 = 0.5% 5% of the sample, the sampled

Actually Lie?no yes

positive 15 42 57

negative 32 9 41

47 51 98

Test Result

Twin GendersBB BG GB GG

identical 5 0 0 5 10

fraternal 5 5 5 5 20

10 5 5 10 30

type

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Multiplication Rule: Complements and Conditional Probability SECTION 4-5 89

items are not considered independent of each other. That guideline is employed as anapproximation only when it is cumbersome to use all the conditional probabilities.

b. P(at least one defective) = 1 P(all good)= 1 P(G1 and G2 and G3 and and G100)= 1 P(G1)P(G2)P(G3)P(G10)= 1 (0.97)(0.97)(0.97)(0.97)

= 1 (0.97)10 = 1 0.0476 = 0.9524, rounded to 0.952NOTE: Because the sample size is 100/4000 = 2.5% 5% of the sample, the sampled itemsmay be considered independent of each other.

29. P(HIV positive result) = P(at least one person is HIV positive)= 1 P(all persons are HIV negative)= 1 P(N1 and N2 and N3)= 1 P(N1) P(N2) P(N3)= 1 (0.9)(0.9)(0.9)= 1 0.729 = 0.271

NOTE: This plan is very efficient. Suppose, for example, there were 3000 people to be tested.Only in 0.271 = 27.1% of the groups would a retest need to be done for each of the threeindividuals. Those (0.271)(1000) = 271 groups would generate (271)(3) = 813 retests. Thetotal number of tests required would be 1813 (the 1000 original + the 813 retests), only 60% ofthe 3,000 tests that would have been required to test everyone individually.

30. P(the combined sample fails) = P(at least one pool fails)= 1 P(all pools pass)= 1 P(P1 and P2 andand P6)= 1 P(P1) P(P2) P(P6)= 1 (0.98)(0.98)(0.98)= 1 (0.98)6 = 1 0.886 = 0.114

NOTE: This plan is very efficient. Suppose, for example, there were 60 pools to be tested.

Only in 0.114 = 11.4% of the groups would a retest need to be done for each of the six pools.That (0.114)(10) = 1 case would generate (1)(6) = 6 retests. The total number of tests requiredwould be 16 (the 10 original + the 6 retests), only 27% of the 60 tests that would have beenrequired to test each pool individually.

31. a. Let D = a birthday is different from any yet selected.P(D1) = 365/365 = 1 NOTE: With nothing to match, it must be different.P(D2) = P(D2|D1) = 364/365P(D3) = P(D3|D1 and D2) = 363/365P(D25) = P(D25|D1 and D2 and D3 andand D24) = 341/365P(all different) = P(D1 and D2 andand D25)

= P(D1)P(D2)P(D25)= (365/365)(364/365)(341/365) = 0.431

NOTE: The above analysis ignores February 29. Including leap years (and countingFebruary 29 equal to the other days) changes the final calculation to(366/366)(365/366)(342/366) = 0.432. An algorithm to perform this calculation can beconstructed using a programming language, spreadsheet, or statistical software package.In BASIC, for example, use

10 LET P=1 40 LET P=P*(365K)/36515 PRINT HOW MANY PEOPLE? 50 NEXT K

20 INPUT D 60 PRINT P(ALL DIFFERENT BIRTHDAYS) IS ;P

30 FOR K=1 TO D1 70 END

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b. P(at least 2 share the same birthday) = 1 P(all different)= 1 0.431 = 0.569

NOTE: The above analysis ignores February 29. Including leap years (and countingFebruary 29 equal to the other days) changes the final calculation to 1 0.432 = 0.568.

32. Make a table like the one at the right.

Let D = the pacemaker is defective.Let A = the pacemaker came from Atlanta.P(A|D) = 3/5 = 0.600

33. Let F = getting a seat in the front row.P(F) = 2/24 = 1/12 = 0.0833 formulate the problem in terms of n rides

P(at least one F in n rides) = 1 P(no F in n rides)= 1 (11/12)n

try different values for n until P(at least one F) 0.95P(at least one F in 32 rides) = 1 (11/12)32 = 1 0.0616 = 0.9384P(at least one F in 33 rides) = 1 (11/12)33 = 1 0.0566 = 0.9434

P(at least one F in 34 rides) = 1 (11/12)34

= 1 0.0519 = 0.9481P(at least one F in 35 rides) = 1 (11/12)35 = 1 0.0476 = 0.9524

You must ride 35 times in order to have at least a 95% chance of getting a first row seat atleast once.

34. When 2 coins are tossed, there are four equally likely possibilities: HH HT TH TT.There are two possible approaches to this problem. use the sample space directly

If there is at least one H, that leaves 3 equally likely possibilities: HH HT TH.Since HH is 1 of those 3, P(HH|at least one H) = 1/3

apply an appropriate formula

Use P(B|A) = P(A and B)/P(A), where B = HH and A = at least one H.P(at least one H and HH) 1/ 4

P(HH|at least one H) = 1/ 3P(at least one H) 3/ 4

= =

4-6 Probabilities Through Simulations

1. No. Generating random numbers between 2 and 12 simulates a uniform distribution of equallylikely values, but the sums obtained by rolling two dice do not follow a uniform distribution.

2. Generate a random integer from 1 to 6 (inclusive), generate a second random number from 1 to

6 (inclusive), and then add the two numbers.3. Yes. Each birthday has the same chance of being chosen for each of the 25 selections.

4. A Simulation does not necessarily provide the exact probability, and each simulation generallyproduces a slightly different result.. The student should state that the probability of getting asequence of six 0s or six 1s is approximately 0.977.

PLANT

Atlan BaltoPRODUCTgood 397 798 1195

def. 3 2 5

400 800 1200

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Probabilities Through Simulations SECTION 4-6 91

NOTE: In exercises #5-8 we suggest using random two-digit numbers from 00 to 99 inclusive.For convenience of association with the natural counting numbers, consider 00 to represent 100 sothat the conceptual order of the digits is 01,02,03,98,99,00. Even though other schemes usingfewer than 100 possibilities could be used, this has the advantage of being able to read through aprinted list of random digits and associating them into pairs without having to discard any valuesas being outside the range of desired random numbers.

5. Generate 50 random two-digit integers as suggested in the NOTE before exercise #5.Let 01-95 represent an adult who recognizes the McDonalds brand name, and let 96-00represent an adult who does not recognize the McDonalds brand name.

6. Generate 15 random two-digit integers as suggested in the NOTE before exercise #5.Let 01-10 represent a person who is left-handed, and let 11-00 represent a person who is notleft-handed.

7. Generate a random three-digit integer from 001 to 000 as suggested in the NOTE beforeexercise #5.Let 001-528 represent a made free throw, and let 529-000 represent a missed free throw.

8. Generate 20 random two-digit integers as suggested in the NOTE before exercise #5.Let 01-75 represent a pea with a green pod, and let 76-00 represent a pea with a yellow pod.

NOTE: The answers for the remaining exercises will vary. The results shown are those obtained inthe preparation of this manual.

9. a. The 50 generated random numbers are:56 51 38 41 41 74 83 56 22 47 72 73 38 91 37 68 95 32 33 75 32 48 65 21 0986 33 27 46 80 18 37 72 81 45 30 61 08 27 28 66 35 52 71 19 98 65 41 47 06

Based on the scheme in Exercise 5, the corresponding Y-N brand recognition values are:Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y YY Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y N Y Y Y Y

The simulated proportion of adults who recognize the McDonald name is 49/50 = 0.98.This is close to the true value of 0.05

b. The results of the ten simulations are as follows.0.98 0.98 0.98 0.96 0.94 0.90 1.00 0.88 0.94 0.98

The results are rather variable, from 0.88 to 1.00 but while they are not narrowly centeredat 0.95, their mean value is 0.954. Based on these results, it would be extremely unusual fora sample of size 50 to yield a sample proportion near 0.50.

10. a. The 15 generated random numbers are:87 96 89 24 12 99 62 33 54 71 30 89 02 97 91

Based on the scheme in Exercise 6, the corresponding R-L values are:R R R R R R R R R R R R L R R

The simulated proportion of people who are left-handed is 1/15 = 0.067.This is close to the true value of 0.10.

b. The numbers of left-handed people in each of the ten simulations are as follows.1 2 0 1 1 2 0 2 1 0

Based on these results it appears not be unusual to find no left-handers in a random sampleof 15 people.

11. a. The 5 generated random numbers are: 497 424 938 011 655Based on the scheme in Exercise 7, the Y-N made free-throw values are: Y Y N Y NThe simulated proportion of made free-throws is 3/6 = 0.600.

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This is reasonably close to the true value of 0.528.b. The proportions of made free-throws in each of the ten simulations are as follows.

0.600 0.800 0.400 0.600 0.400 0.400 0.600 0.800 0.800 0.600Based on these results it appears that it would be unusual for ONeal to make all five free-throws at least, based on these results, the chances appear to be less than 1 in 10.

12. a. The 20 generated random numbers are:15 04 83 44 05 68 11 84 84 53 76 57 64 47 44 33 04 95 57 15Based on the scheme in Exercise 8, the G-Y values are:

G G Y G G G G Y Y G Y G G G G G G Y G GThe simulated proportion of peas with yellow pods is 5/20 = 0.25.This agrees with the true value of 0.25.

b. The numbers of peas with yellow pods in each of the ten simulations are as follows.5 3 5 3 4 5 3 4 3 3

The numbers of peas with yellow pods vary little. [These particular 10 simulations areunusual in that none of them yields a proportion of yellow pods greater than the expectedvalue of 0.25.] Based on these results it would be unusual to find that no pea plants out of asample of 20 have yellow pods.

13. Generate 5 random two-digit integers as suggested in theNOTE before exercise #5. Let 01-50 represent a boy andlet 51-00 represent a girl. Repeat this 20 times. Theresults are given in the box at the right.

The estimated probability of such an occurrence is4/20 = 0.25, and so such a run is not unusual.NOTE: The sample space for 5 births is the following32 equally like possibilities. Since 16 of the them include arun of three of the same gender, the true probability of suchan occurrence is 16/32 = 0.5 which is twice the estimate

produced by the simulation. In this instance the simulationdid not perform well.

bbbbb bbbbg bbbgb bbbgg bbgbb bbgbg bbggb bbgggbgbbb bgbbg bgbgb bgbgg bggbb bggbg bgggb bgggg.gbbbb gbbbg gbbgb gbbgg gbgbb gbgbg gbggb gbggg

ggbbb ggbbg ggbgb ggbgg gggbb gggbg ggggb ggggg.

14. Generate 6 random two-digit integers as suggested in theNOTE before exercise #5. Let 01-50 represent a boy andlet 51-00 represent a girl. Repeat this 20 times. Theresults are given in the box at the right.

The estimated probability of such an occurrence of runs

is 2/20 = 0.10, and so such a run is not unusual.NOTE: It can be shown that the true probability of such anoccurrence of runs is 0.21875 which is more than twicethe estimate produced by the simulation. In this instance the simulation did not perform well.

15. Generate 152 random 0s and 1s. Let the 0s represent a girl and let a 1 represent a boy. Sumthe values to determine the number of boys. Repeat this 20 times. The 20 numbers of boysthus obtained were: 71 91 72 77 72 78 74 77 76 82 83 69 78 79 75 85 76 79 71 72.It appears that getting 127 boys in 152 would be a very unusual event. It appears that theYSORT method is effective.

Exercise 13 Exercise 14result run? result run?

1 bbggb bgggbb2 bbgbg bbbgbb3 gbgbg ggbbbb4 bgbbg bggbgb5 gbbgg bgggbb6 bbgbb bggggb yes7 bbgbb gggbgb8 bggbb bbbgbb9 bbbbg yes gbbbgg

10 bbggb ggbggb11 bbgbb ggbggg12 bbbgg yes ggbbgb13 bgbgg bbgbbg14 bgbgg ggbbgb15 bbggb bbgbgg16 ggbgb gbbbgg17 gbbbg yes gbgbgb18 bgggg yes gggbbb19 ggbgg bggggb yes20 gbbgg bbgggb

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Probabilities Through Simulations SECTION 4-6 93

16. Generate 2103 random decimals from a uniform distribution from 0 to 1.Since 2/1671 = 0.001196888, let decimals below that value represent developing therespiratory infection and decimals above that value represent no such side effect. Repeat this20 times.Hint: Sorting the generated values for each simulation makes it easy to identify the ones below0.00119688 without having to look through all 2103 values. In Minitab the appropriate

commands are MTB > random 2103 c1-c20;SUBC> uniform 0 1.

MTB > sort c1 c1

MTB > sort c2 c2

...

MTB > sort c20 c20

The 20 numbers of respiratory infections thus obtained were:5 2 1 0 3 3 2 3 3 4 3 2 6 1 5 2 5 2 2 0

The simulation estimates that getting 6 respiratory infections by chance alone would occurabout 1/20 = 5% of the time. Since X is defined to be unusual whenever P(X) 0.05, theresults suggest that getting 6 respiratory infections by chance alone would be unusual and thatusing Nasonex may enhance such occurrences.

17. The probability of originally selecting the door with the car is 1/3.You can either switch or not switch.Let W = winning the car. If you make a selection and do not switch, P(W) = 1/3. If you make a selection and switch, there are only 2 possibilities.

A = you originally had the winning door and switched to a losing door.B = you originally had a losing door and switched to a winning door.

NOTE: You cannot switch from a losing door to another losing door. Monty Hall alwaysopens a losing door. If you have a losing door and Monty Hall opens a losing door, thenthe only possible door to switch to is the winning door. Since you win if B occurs,

P(W) = P(B) = P(originally had a losing door) = 2/3. Conclusion: the better strategy is to switch.While the preceding analysis makes it unnecessary, a simulation of n trails can be done asfollows.

1. Randomly select door 1,2 or 3 to hold the prize.2. Randomly select door 1,2 or 3 as your original choice.3. Determine m = the number of times the doors match.4. Estimated P(winning if you dont switch) = m/n. [m/n should be about 1/3]5. Estimated P(winning if you switch) = (n-m)/n. [(n-m)/n should be about 2/3]NOTE: If you switch, you lose m times and win (n-m) times if you originally selected alosing door, remember, a switch always wins the car.

The following Minitab commands produce the desired simulation, where column 1 indicatesthe winning door and column 2 indicates your choice.

MTB > RANDOM 100 C1 C2;

SUBC> INTEGER 1 3.

MTB > LET C3= C1-C2

MTB > TABLE C3

A zero in C3 indicates a match and that the original choice would win the car. The simulationproduced for this manual yielded 37 0s in C3.

Estimated P(winning if you dont switch) = 37/100 = 37%Estimated P(winning if you switch) = 63/100 = 63%

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18. a. Let A = at least two of 50 random people have the same birth date. Generate 50 randomintegers from 1 to 366. Order the numbers to see if there are any duplicates (or higher).Repeat this 100 times.P(A) (# of trials with duplicates)/100 [should be close to the true value of 0.970]

b. Let B = at three of 50 random people have the same birth date. Generate 50 random integersfrom 1 to 366. Order the numbers to see if there are any triplicates (or higher).

Repeat this 100 times.P(B) (# of trials with triplicates)/100 [should be close to the true value of 0.127]

NOTE: The above simulations include February 29 with the same frequency as the other dates.A slightly more accurate simulation could be obtained by giving special consideration to thenumber 60 (which represents February 29 in the ordered list) as follows: If 60 occurs, flip acoin twice if 2 heads occur [P(HH) = 1/4] use the 60, otherwise ignore the 60 and chooseanother number.

19. No, his reasoning is not correct. No, the proportion of girls will not increase. Each familywould consist of zero to several consecutive girls, followed by a boy. The types of possiblefamilies and their associated probabilities would be as follows.

P(B) = = 1/2 =16/32P(GB) = ()() = 1/4 = 8/32P(GGB) = ()()() = 1/8 = 4/32P(GGGB) = ()()()() = 1/16 = 2/32P(GGGGB) = ()()()()() = 1/32 = 1/32etc.

Each collection of 32 families would be expected to look like this, where * represents onefamily with 5 or more girls and a boy.

B B B B B B B BB B B B B B B BGB GB GB GB GB GB GB GBGGB GGB GGB GGB GGGB GGGB GGGGB *

A gender count reveals 32 boys and 31 or more girls, an approximately equal distribution ofgenders. In practice, however, families would likely stop after some maximum number ofchildren whether they had a boy or not. If that number were 5 for all families, then * =GGGGG and the expected gender count for the 32 families would be an exactly equaldistribution of 31 boys and 31 girls.NOTE: Since the stopping point would likely vary from family to family, the problem isactually more complicated. Suppose, for example, that 25% of the families decided in advanceto stop at two children regardless of gender. We expect those 1/4 = 8/32 of the families to bedistributed approximately evenly among the four rows in the above listing 2 in each row.The 6 such families in the first 3 rows would be stopped by the kings decree and remainunchanged, but the 2 such families in the last row would both become GG families. Countingthe * as GGGGG, this reduces the expected number of boys by 27/8 = 1.75 and the expectednumber of girls by 27/8 = 1.75 and there would still be an equal number of each gender.

4-7 Counting

1. The permutations rule applies when different orderings of the same collection of objectsrepresent a different solution. The combinations rule applies when all orderings of the samecollection of objects represent the same solution.

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Counting SECTION 4-7 95

2. No. Because order makes a difference, and the numbers must be used in the proper sequence,The device is actually a permutation lock and not a combination lock.

3. Permutations. Because order makes a difference, and a different ordering of the top 3 finishersrepresents a different race outcome, the trifecta involves permutations and not combinations.

4. Combinations. Because order makes no difference, and a different ordering of the top 2

finishers does not affect correct identification of the best 2 horses, the quinela involvescombinations and not permutations.

5. 5! = 54321 = 120

6. 9! = 987654321 = 362,880

7. 52C2 =52! 52 51

=50!2! 2!

= 1326

NOTE: This technique of reducing the problem by canceling out the 50! from both thenumerator and denominator can be used in most combinations and permutations problems. Ingeneral, a smaller factorial in the denominator can be completely divided into a larger factorial

in the numerator to leave only the excess factors not appearing in the denominator.Furthermore, nCrand nPrwill always be integers. More generally, a non-integer answer to anycounting problem (but not a probability problem) indicates that an error has been made.

8. 52C5 =52! 52 51 50 49 48

=47!5! 5!

= 2,598,960

9. 9P5 = 9!/4! = 98765 = 15,120

10. 50P3 = 50!/47! = 504948 = 117,600

11. 42C6 =42! 42 41 40 39 38 37

=36!6! 6!

= 5,245,786

12. 10P3 = 10!/7! = 1098 = 720

13. Let W = winning the described lottery with a single selection.The total number of possible combinations is

54C6 =54! 54 53 52 51 50 49

=48!6! 6!

= 25,827,165.

Since only one combination wins, P(W) = 1/25,827,165.


53C6 =53! 53 52 51 50 49 48

=47!6! 6!

= 22,957,480.

Since only one combination wins, P(W) = 1/22,957,480.


36C5 =36! 36 35 34 33 32

=31!5! 5!

= 376,992

Since only one combination wins, P(W) = 1/376,992.

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16. a. Let W = winning the described lottery with a single selection.The total number of possible combinations is

31C5 =31! 31 30 29 28 27

=26!5! 5!

= 169,911

Since only one combination wins, P(W) = 1/169,911.b. Let W = winning the described lottery with a single selection.

The total number of possible sequences is

31P5 = 31!/26! = 3130292827 = 20,389,320Since only one sequence wins, P(W) = 1/20,389,320.

17. a. Let G = generating a given social security number in a single trial.The total number of possible sequences is101010101010101010 = 109 = 1,000,000,000

Since only one sequence is correct, P(G) = 1/1,000,000,000.b. Let F = generating the first 5 digits of a given social security number in a single trial.

The total number of possible sequences is1010101010 = 105 = 100,000

Since only one sequence is correct, P(F) = 1/100,000. Since this probability is so small, oneneed not worry about the given scenario.

18. a. Let G = generating a given credit card number in a single trial.The total number of possible sequences is10101010101010101010101010101010 = 1016 = 10,000,000,000,000,000

Since only one sequence is correct, P(G) = 1/10,000,000,000,000,000.b. Let F = generating the first 12 digits of a given credit card number in a single trial.

The total number of possible sequences is101010101010101010101010 = 1012 = 1,000,000,000,000.

c. Let M = generating the middle digits of a given credit card number in a single trial.The total number of possible sequences is

1010101010101010= 108 = 100,000,000Since only one sequence is correct, P(M) = 1/100,000,000. This is not something to worryabout.

19. Since order makes a difference, use permutations.

20P6 = 20!/14! = 201918171615 = 27,907,200

20. There are 3 tasks to perform, and each task can be performed in any of 4 ways.The total number of possible sequences is 444 = 64.

21. Since the order of the two wires being tested is irrelevant, use combinations.

5C2 =5! 5 4

=

3!2! 2!

= 10

22. Once the 3 are selected for the United Way event, the other 2 are automatically determined.Since the order in which the 3 are picked makes no difference, use combinations.

5C3 =5! 5 4 3

=2!3! 3!

= 10

23. There are 8 tasks to perform, and each task can be performed in either of 2 ways.The total number of possible sequences is 22222222 = 28 = 256.Yes. The typical keyboard has approximately 50 keys, each with a lower an upper casepossibility, representing about 100 commonly used characters.

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24. Since the order in which the subjects are placed in the group is not relevant, use combinations.The total number of possible combinations is

20C5 =20! 20 19 18 17 16

=15!5! 5!

= 15,504

The probability of any one combination is 1/15,504.

25. The number of possible sequences of n different objects is n!, and 5! = 54321 = 120The unscrambled word is JUMBO. Since there is only one correct sequence, the probabilityof finding it with one random arrangement is 1/120.

26. The number of possible sequences of n objects is when some are alike is

1 2 k

n!

n !n ! n ! , and

5!

1!2!1!1!= 60

The unscrambled word is BAGGY. Since there is only one correct sequence, the probabilityof finding it with one random arrangement is 1/60.

27. a. Since order makes a difference, as there are 4 different offices, use permutations.

11P4 = 11!/7! = 111098 = 7920

b. Since the order in which the 4 are picked makes no difference, use combinations.

11C4 =11! 11 10 9 8

=7!4! 4!

= 330

28. There are 4 tasks to perform, and each task can be performed in any of 100 ways.The total number of possible sequences is 100100100100 = 1004 = 100,000,000.Since there is only one correct sequence, the probability of finding it in one random selectionis 1/100,000,000. Since there are so many possibilities, it would not be feasible to try openingthe safe by making random guesses.

29. a. There are 14 tasks to perform, and each task can be performed in either of 2 ways.The total number of possible sequences is 22222222222222= 214 = 16,384.

b. The number of possible sequences of n objects is when some are alike is

1 2 k

n!

n !n ! n ! , and

14!

13!1!= 14

c. P(13G,1B) = (# of ways to get 13G,1B)/(total number of ways to get 14 babies)= 14/16,384 = 0.000854

d. Yes. Since P(13G,1B) is so small, and since 13G,1B so far (only the 14G,0B result is moreextreme) from the expected 7G,7B result, the gender-selection method appears to yieldresults significantly different from those of chance alone.

30. Credit cards are rectangular. Even in the dark, one can distinguish the long side (which willnot fit into the slot in the ATM) from the short side. When one inserts the card, he has two

decisions to make and each decision has two options:(1) he must choose one of the short sides to insert, and there are 2 to chose from.(2) he must choose one of the faces of the card to insert next to the magnetic reader, and

there are 2 (front and back) to choose from.Since there are 2 tasks to perform, and each task can be performed in either of 2 ways, the totalnumber of possible insertion orientations is 22 = 4.a. Since only insertion orientation is correct, the probability that a random insertion will be

correct is 1/4 = 0.25b. Assuming the random process is started over each time, the probability of a failure followed

by a success is P(F1 and S2) = P(F1)P(S2) = (3/4)(1/4) = 3/16 = 0.188.

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c. Assuming the random process is started over each time, there is no finite number of tries thatwill guarantee a success i.e., he could keep choosing an incorrect position forever.

NOTE: Another valid interpretation of the problem to assume that each repeated selection ismade at random from the positions not already tried. In this case the answers to (b) and (c) areas follows.b. Assuming that each repeated selection is made at random from the positions not already

tried, the probability of a failure followed by a success is P(F1 and S2) = P(F1)P(S2|F1) =(3/4)(1/3) = 3/12 = 0.25.

c. Assuming that each repeated selection is made at random from the positions not alreadytried, the user is guaranteed a success by the fourth trial i.e., the worst he can do is tochoose the 3 incorrect orientation before finally being forced to chose to correct one.

31. Since order makes no difference when one is choosing groups, use combinations.

The number of possible treatment groups is 10C5 =10! 10 9 8 7 6

=5!5! 5!

= 252.

The number of ways to select a group of 5 men from the 5 males is 5C5 = 1.The number of ways to select a group of 5 women from the 5 females is 5C5 = 1.

Since the number of possible same sex groups is 1 + 1 = 2, the probability that randomselection determines a treatment group of all the same sex is 2/252 = 0.00794.Yes. Having a treatment group all of the same sex would create confounding the researchercould not say whether any difference in the treatment group was a treatment effect or a gendereffect.

32. a. There are 20 tasks to perform, and each task can be performed in either of 2 ways.The total number of possible sequences is 222 = 2

20= 1,048,576

b. The number of possible sequences of n objects is when some are alike is

1 2 k

n!

n !n ! n ! , and

20!

10!10!= 184,756.

c. P(10G,10B) = 184,756/1,048,576 = 0.176d. It is not unusual for an event with probability 0.176 to occur once, but repeated occurrencesshould be considered unusual as the probability of the event occurring twice in a row, forexample, is (0.176)(0.176) = 0.0310.

33. Let A be selecting the correct 5 numbers from 1 to 55. The number of possible selections is

55C5 =55! 55 54 53 52 51

=50!5! 5!

= 3,478,761.

Since there is only one winning selection, P(A) = 1/3,478,761.Let B be selecting the correct winning number from 1 to 42. There are 42 possible selections.Since there is only one winning selection, P(B) = 1/42.P(winning Powerball)

= P(A and B) = P(A)P(B) = (1/3,478,761)(1/42) = 1/146,107,962 = 0.00000000684

34. Let A be selecting the correct 5 numbers from 1 to 56. The number of possible selections is

56C5 =56! 56 55 54 53 52

=51!5! 5!

= 3,819,816.

Since there is only one winning selection, P(A) = 1/3,819,816.Let B be selecting the correct winning number from 1 to 46. There are 46 possible selections.Since there is only one winning selection, P(B) = 1/46.P(winning Mega Millions)

= P(A and B) = P(A)P(B) = (1/3,819,816)(1/46) = 1/175,711,536 = 0.00000000569

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35. The first digit could be any of the 8 numbers 2,3,4,5,6,7,8,9.The second digit could be either of the 2 numbers 0,1.The third digit could be any of 9 numbers as follows.

1,2,3,4,5,6,7,8,9 if the second digit is 0.0,2,3,4,5,6,7,8,9 if the second digit is 1.

By the fundamental counting rule, the number of possible 3-digit area codes is 829 = 144.

36. a. Since there are 64 teams at the start and only 1 left at the end, 63 teams must be eliminated.Since each game eliminates 1 team, it takes 63 games to eliminate 63 teams.

b. Let W = picking all 63 winners by random guessing.Since there are 2 possible outcomes for each game, there are 263 = 9.223 x 1018 possible setsof results. Since only one such result gives all the correct winners,

P(W) = 1/[9.223 x 1018

] = 1.084 x 10-19

.One could also reason: since each guess has a 50% chance of being correct,

P(W) = (0.5)63 = 1.084 x 10-19.c. Let E = picking all 63 winners by expert guessing.

Since each guess has a 70% chance of being correct, P(W) = (0.7)63 = 1.743 x 10-10.Since 1.743 x 10-10 = 1/5,738,831,575, the expert has a 1 in 5,738,831,575 chance [or abouta 1 in 5.7 billion chance] of selecting all 63 winners.

37. There are 26 possible first characters, and 36 possible characters for the other positions.Find the number of possible names using 1,2,3,,8 characters and then add to get the total.

characters possible names1 26 = 262 2636 = 9363 263636 = 33,6964 26363636 = 1,213,0565 2636363636 = 43,670,0166 263636363636 = 1,572,120,5767 26363636363636 = 56,596,340,736

8 2636363636363636 = 2,037,468,266,496total = 2,095,681,645,538

38. a. The number of handshakes is the number of ways 2 people can be chosen from 5,

5C2 =5! 5 4

=3!2! 2

= 10.

b. The number of handshakes is the number of ways 2 people can be chosen from n,

nC2 =n! n (n-1)

=(n-2)!2! 2

= n(n-1)/2.

c. Visualize the people entering one at a time, each person sitting to the right of the previousperson who entered. Who the first person is or where the first person sits is irrelevant i.e.,

it just establishes a reference point but does not affect the number of arrangements. Oncethe 1st person sits, there are 4 possibilities for the person to his right. Once the 2nd personsits, there are 3 possibilities for the person to his right. Once the 3rd person sits, there are 2possibilities for the person to his right. Once the 4

thperson sits, there is only 1 remaining

person to sit on his right. And so the number of possible arrangements is 4321 = 24.d. Use the same reasoning as in part (c). Once the 1st person sits, there are n-1 possibilities for

the person to his right. Once the 2nd person sits, there are n-2 possibilities for the person tohis right. And so on Once the next to the last person sits, there is only 1 possibility forthe person to his right. And so the number of possible arrangements is(n-1)(n-2)(1) = (n-1)!.

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39. a. The calculator factorial key gives 50! = 3.04140932 x 1064Using the approximation, K = (50.5)log(50) + 0.39908993 0.43429448(50)

= 85.79798522 + 0.39908993 21.71472400= 64.48235225

and then 50! = 10K

= 1064.48235225

= 3.036345215 x 1064NOTE: The two answers differ by 0.0051 x 1064 i.e., 5.1 x 1061, or 51 followed by 60zeros. While such an error of zillions and zillions may seem quite large, it is only an errorof (5.1 x 1061)/(3.04 x 1064) = 1.7%.

b. The number of possible routes is 300!Using the approximation, K = (300.5)log(300) + 0.39908993 0.43429448(300)

= 744.374937 + 0.39908993 130.288344= 614.485683

and then 300! = 10K

= 10614.485683

Since the number of digits in 10x is the next whole number above x, 300! has 615 digits.

40. a. Assuming the judge knows there are 4 computers and 4 humans and frames his guessesaccordingly, the number of ways he could guess is the number of ways 4 of the 8 could belabeled computer and is given by 8C4 = 8!/4!4! = 70. The probability of guessing thecorrect labels by chance alone is therefore 1/70.NOTE: If the judge guesses on each one individually, either not knowing or not caring thatthere are 4 computers and 4 humans, then the probability of guessing all 8 correctly is ()8 =1/256.

b. Under the original assumption in part (a), the probability that all 10 judges make all correctguesses is (1/70)

10= 3.54 x 10

-19.

41. This problem cannot be solved directly using permutations, combinations or other techniques

presented in this chapter. It is best solved by listing all the possible solutions in an orderlyfashion and then counting the number of solutions. Often this is the most reasonable approachto a counting problem.

While you are encouraged to develop your own systematic approach to the problem, thefollowing table represents one way to organize the solution. The table is organized by rows,according to the numbers of pennies in each way that change can be made. The numbers ineach row give the numbers of the other coins as explained in the footnotes below the table.

The three numbers in the 80 row, for example, indicate that there are three ways to makechange using 80 pennies. The 4 in the only 5 column represents one way (80 pennies, 4nickels). The 2 in the 10 column of the nickels and one other coin columns representsanother way (80 pennies, 2 dimes). The 1 in the 10 column of the nickels and one other

coin columns represents a third way (80 pennies, 1 dime, and [by default] 2 nickels).The bold-face 3 in the20 row and the 10/25 column of the nickels and two other coins

columns represents 20 pennies, 3 dimes, one quarter, and [by default] 5 nickels.Using the following table, or the system of your own design, you should be able to

(1) take a table entry and determine what way to make change it represents and(2) take a known way to make change and find its representation in the table.

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allc

nickels and two other coinsb 10only nickels and one other coina 10 10 25 25

1 5 10 25 50 25 50 50 50100 0 . . . . . . .

95 1 . . . . . . .90 2 1 . . . . . .

85 3 1 . . . . . .80 4 2,1 . . . . . .75 5 2,1 1 . . . . .70 6 3,2,1 1 . . . . .65 7 3,2,1 1 . 1 . . .60 8 4,3,2,1 1 . 1 . . .55 9 4,3,2,1 1 . 2,1 . . .50 10 5,4,3,2,1 2,1 1 2,1 . . .45 11 5,4,3,2,1 2,1 1 3,2,1 . . .40 12 6,5,4,3,2,1 2,1 1 3,2,1 1 . .35 13 6,5,4,3,2,1 2,1 1 4,3,2,1 1 . .30 14 7,6,5,4,3,2,1 2,1 1 4,3,2,1 2,1 . .25 15 7,6,5,4,3,2,1 3,2,1 1 5,4,3,2,1 2,1 1 .

20 16 8,7,6,5,4,3,2,1 3,2,1 1 5,4,3,2,1 3,2,1 1 .15 17 8,7,6,5,4,3,2,1 3,2,1 1 6,5,4,3,2,1 3,2,1 1 110 18 9,8,7,6,5,4,3,2,1 3,2,1 1 6,5,4,3,2,1 4,3,2,1 1 1

5 19 9,8,7,6,5,4,3,2,1 3,2,1 1 7,6,5,4,3,2,1 4,3,2,1 1 2,10 20 10,9,8,7,6,5,4,3,2,1 4,3,2,1 2,1 7,6,5,4,3,2,1 5,4,3,2,1 2,1 2,1

ways 21 100 34 12 56 25d 7 6e

aThe possible numbers of the non-nickel coin are given. The numbers of nickels are found by default.bThis is for a single occurrence of the second coin listed. The possible numbers of the leading non-

nickel coin are given. The numbers of nickels are found by default.cThis is for a single occurrence of the second and third coin listed. The possible numbers of the

leading non-nickel coin are given. The numbers of nickels are found by default.dThere are another 25 ways when the 50 is in the form of 2 quarters.eThere are another 6 ways when the 50 is in the form of 2 quarters.

The total number of ways identified is 21+100+34+12+56+25+25+7+6+6 = 292.If a one-dollar coin is also considered as change for a dollar, there are 293 ways.The one dollar bill itself should not be counted.

Statistical Literacy and Critical Thinking

1. An event that has probability 0.004 is unlikely to occur by chance. It would occur by chanceonly about 4 times in 1000 opportunities.

2. No. The detectors would be related in the sense that they would all be inoperable whenever

the home loses its electricity.3. No, her reasoning is incorrect for two reasons. (1) Since she is a new recruit she has no reason

to assume that the historical rate of 12.7% will apply to her. (2) Even if the 12.7% rate applies,the (0.127)(0.127) = 0.0161 calculation applies only to burglaries that are selected at randomand/or handled independently. Since the cases were handled by the same person, knowing thatone of the cases closes with an arrest would reasonably make a person think that it was morelikely the other case would close with an arrest i.e., the events are not independent andP(C2|C1) > P(C) = 0.127.

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4. No. Lottery numbers are selected so that past results have no effect on future drawings. Eachdrawing is statistically independent of the all the others.

Chapter Quick Quiz

1. P( C ) = 1 P(C)= 1 0.70 = 0.30

2. P(C1 and C2) = P(C1)P(C2)= (0.70)(0.70) = 0.49

3. Answers will vary, but such an occurrence is not very common. A reasonable guess for suchan interruption might be P(I) = 0.005.

4. No. A result with probability 0.342 could easily have occurred by chance.

5. P(A ) = 1 P(A) = 1 0.4 = 0.6

NOTE: For exercises 6-10, refer to the tableat the right and use A,B,P,F to identify in thenatural way the groups and qualifying exam results.

6. P(P) = 427/586 = 0.729

7. P(B or P) = P(B) + P(P) P(B and P)= 562/586 + 427/586 417/586= 572/586 = 0.976

8. P(A1 and A2) = P(A1)P(A2|A1)= (24/586)(23/585)= 0.00161

9. This problem may be worked either of two ways.directly from chart: P(A and P) = 10/586 = 0.0171by formula: P(A and P) = P(A)P(P|A)

= (24/586)(10/24) = 10/586 = 0.0171

10. P(P|A) = 10/24 = 0.417

Review Exercises

NOTE: For exercises 1-10, refer to the table

at the right and use the following notation.Let H = the rider wore a helmet.Let I = the rider experienced a head injury.

1. P(I) = 576/3562 = 0.162

2. P(H) = 752/3562 = 0.211

3. P(I or H) = P(I) + P(H) P(I and H)= 576/3562 + 752/3562 96/3562= 1232/3562 = 0.346

Resultpassed failed

A 10 14 24

B 417 145 562

427 159 586

Group

Head Injury?yes no

yes 96 656 752

no 480 2330 2810

576 2986 3562

Wore Helmet?

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Review Exercises 103

4. P( H or I ) = P(H ) + P( I ) P( H and I )= 2810/3562 + 2986/3562 2330/3562= 3466/3562 = 0.973

5. This problem may be worked either of two ways.directly from chart: P(H and I) = 96/3562 = 0.0270

by formula: P(H and I) = P(H)P(I|H)= (752/3562)(96/752) = 96/3562 = 0.0270

6. This problem may be worked either of two ways.

directly from chart: P( H and I ) = 2330/3562 = 0.654

by formula: P( H and I ) = P( H )P( I | H )= (2810/3562)(2330/2810) = 2330/3562 = 0.654

7. P(H1 and H2) = P(H1)P(H2|H1)= (752/3562)(751/3561) = 0.0445

8. P(I1 and I2) = P(I1)P(I2|I1)= (576/3562)575/3561) = 0.0261

9. P(H |I) = 480/576 = 0.833

10. P( I |H) = 656/752 = 0.872

11. Answers will vary. Black cars are not popular, but they are not rare enough to be consideredunusual. P(B) = 0.08 seems like a reasonable guess.

12. a. P(B ) = 1 P(B)= 1 0.35 = 0.65

b. P(B1 and B2 and B3 and B4) = P(B1) P(B2) P(B3) P(B4)= (0.35)(0.35)(0.35)(0.35) = (0.35)4 = 0.0150

c. Yes. Since 0.015

0.05, selecting four people at random and finding they all have blueeyes would be considered an unusual event.

13. a. P(O18) = 1/365 = 0.00274b. P(O) = 31/365 = 0.0849c. This would be a very rare event, probably occurring less than one time in a million tries. A

reasonable guess would be P(E) = 0.0000001d. Yes, based on the answer in (c). Since 0.0000001 0.05, randomly selecting an adult

American and fifing someone who knows that date would be considered an unusual event.

14. a. P(D) = 15.2/100,000 = 0.000152b. P(D1 and D2) = P(D1) P(D2)

= (0.000152)(0.000152) = (0.000152)

2

= 0.0000000231c. P(D ) = 1 P(D) = 1 0.000152 = 0.999848

P( 1D and 2D ) = P( 1D )P( 2D )

= (0.999848)(0.999848) = (0.999848)2 = 0.999696

15. a. P(A ) = 1 P(A)= 1 0.40 = 0.60

b. No, for two reasons. First, the sample was limited to America OnLine subscribers who arenot necessarily representative of the general population. Second, the poll was based on avoluntary response sample and not on people selected at random. People who respond in a

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voluntary response sample are typically those with strong feelings one way or the other (i.e.,either people who really like Sudoku, or those who cant stand it) and are not necessarilyrepresentative of the general population.

16. For each person, P(negative) = 1 P(positive) = 1 0.00320 = 0.99680.P(positive group result) = P(at least one person is positive)

= 1 P(all persons are negative)= 1 P(N1 and N2 andand N10)= 1 P(N1) P(N2) P(N10)= 1 (0.99680)(0.99680)(0.99680)= 1 (0.99680)10 = 1 0.9685 = 0.0315

No. Since 0.0315 0.05, a positive group result is not considered likely.NOTE: This plan is very efficient. Suppose, for example, there were 3000 people to be tested.Only in 0.0315 = 3.15% of the 300 groups would a retest need to be done for each of the tenindividuals. Those (0.0315)(300) 9 groups would generate (9)(10) = 90 retests. Thetotal number of tests required would be 390 (the 300 original + the 90 retests), only about 13%of the 3,000 tests that would have been required to test everyone individually.

17. Let D = getting a Democrat.P(D) = 0.85 for each selectionP(D1 and D2 andand D30) = P(D1)P(D2)P(D30)

= (0.85)(0.85)(0.85) = (0.85)30 = 0.00763Yes; since 0.00763 0.05, getting 30 Democrats by chance alone is an unusual event. Sincethe probability 0.00763 is so small, the results are evidence that the pollster is not telling thetruth.

18. Let M = a male in that age bracket not surviving. P( M ) = 114.4/100,000 = 0.001144

Let F = a female in that age bracket not surviving. P( F) = 44/100,000 = 0.00044

a. P(M) = 1 P(M )

= 1 0.001144 = 0.998856b. P(M1 and M2) = P(M1)P(M2)

= (.998856)(0.998856) = (0.998856)2 = 0.998

c. P(F) = 1 P(F) = 1 0.00044 = 0.99956P(F1 and F2) = P(F1)P(F2)

= (.99956)(0.99956) = (0.99956)2

= 0.999d. Males are more likely than females to be involved in situations where death is a possible

result (e.g., military combat, violent crimes, hazardous occupations, etc.).

19. The number of possible selections is 38C5 =38! 38 37 36 35 34

=33!5! 5!

= 501,942.

Since there is only one winning selection, P(W) = 1/501,942 = 0.00000199.Since 0.00000199 0.05, winning the jackpot with a single selection is an unusual event. Butsince hundreds of thousands of tickets are purchased, it is not unusual for there to be a winner.

20 There are 10 possibilities for each digit: 0,1,2,3,4,5,6,7,8,9.By the fundamental counting rule, the number of possibilities is now

101010 = 1013

= 10,000,000,000,000 = 10 trillion

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Cumulative Review Exercises 105

Cumulative Review Exercises

1. values in order are: 17 18 18 18 18 19 19 19 19 19 20 20 20 20 20 20 20 20 21 21The summary statistics are: n = 20 x = 386 x2 = 7472

a. x = (x)/n = 386/20 = 19.3 oz

b. x = (19+20)/2 = 19.5 oz

c. s2

= [n(x2) (x)

2]/[n(n-1)]

= [20(7472) (386)2]/[20(19)]= 444/380 = 1.168

s = 1.081, rounded to 1.1 oz

d. s2 = 1.168, rounded to 1.2 oz2 [from part (c) above]

e.

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