B. If the Sample Points Are Equally Likely, Then

Probability Chapter 3

3.2 a. This is a Venn Diagram.

b. If the sample points are equally likely, then

1 P(1) = P(2) = P(3) =  = P(10) = 10

Therefore,

1 1 1 3 P(A) = P(4) + P(5) + P(6) = + + = = .3 10 10 10 10 1 1 2 P(B) = P(6) + P(7) = + = = .2 10 10 10

1 1 3 5 c. P(A) = P(4) + P(5) + P(6) = + + = = .25 20 20 20 20 3 3 6 P(B) = P(6) + P(7) = + = = .3 20 20 20

骣9 9! 9鬃 8 7 鬃 6 5 鬃 4 3 鬃 2 1 3.4 a. 琪 = = = 126 桫4 4!(9 - 4)! 4鬃 3 2 鬃 1 5 鬃 4 3 鬃 2 1

骣7 7! 7鬃 6 5 鬃 4 3 鬃 2 1 b. 琪 = = = 21 桫2 2!(7 - 2)! 2鬃 1 5 鬃 4 3 鬃 2 1

骣4 4! 4鬃 3 2 1 c. 琪 = = = 1 桫4 4!(4 - 4)! 4鬃 3 2 鬃 1 1

骣5 5! 5鬃 4 3 鬃 2 1 d. 琪 = = = 1 桫0 0!(5 - 0)! 1鬃 5 4 鬃 3 2 1

骣6 6! 6鬃 5 4 鬃 3 2 1 e. 琪 = = = 6 桫5 5!(6 - 5)! 5鬃 4 3 鬃 2 1 1

56 Chapter 3 3.6 a. The 36 sample points are:

1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6

b. If the dice are fair, then each of the sample points is equally likely. Each would have a probability of 1/36 of occurring.

1 c. There is one sample point in A: 3,3. Thus, P(A) = . 36

6 1 There are 6 sample points in B: 1,6 2,5 3,4 4,3 5,2 and 6,1. Thus, P(B) = = . 36 6

There are 18 sample points in C: 1,1 1,3 1,5 2,2 2,4 2,6 3,1 3,3 3,5 4,2 4,4 18 1 4,6 5,1 5,3 5,5 6,2 6,4 and 6,6. Thus, P(C) = = . 36 2

3.8 Each student will obtain slightly different proportions. However, the proportions should be close to P(A) = 1/10, P(B) = 6/10 and P(C) = 3/10.

3.10 Define the following event:

B: {Postal worker was assaulted on the job in the past year}

600 P(B) = = .05 12,000

3.12 a. The 5 sample points are:

Total population, Agricultural change, Presence of industry, Growth, and Population concentration.

b. The probabilities are best estimated with the sample proportions. Thus,

P(Total population) = .18 P(Agricultural change) = .05 P(Presence of industry) = .27 P(Growth) = .05 P(Population concentration) = .45

c. Define the following event:

A: {Factor specified is population-related}

P(A) = P(Total population) + P(Growth) + P(Population concentration) = .18 + .05 + .45 = .68.

57 Chapter 3 3.14 a. The sample points of this experiment correspond to each of the 8 possible types of commodities. Suppose we introduce notation to make the listing of the sample points easier.

A: {carload contains agricultural products} CH: {carload contains chemicals} CO: {carload contains coal} F: {carload contains forest products} MO: {carload contains metallic ores and minerals} MV: {carload contains motor vehicles and equipment} N: {carload contains nonmetallic minerals and products} O: {carload contains other}

The eight sample points are: A CH CO F MO MV N O

b. The probability of each sample point is found by dividing the number of carloads for each sample point by the total number of carloads. The probabilities are:

P(A) = 41,690 / 335,770 = .124

P(CH) = 38,331 / 335,770 = .114

P(CO) = 124,595 / 335,770 = .371

P(F) = 21,929 / 335,770 = .065

P(MO) = 34,521 / 335,770 = .103

P(MV) = 22,906 / 335,770 = .068

P(N) = 37,416 / 335,770 = .111

P(O) = 14,382 / 335,770 = .043

c. P(MV) = .068

P(nonagricultural products) = P(CH) + P(CO) + P(F) + P(MO) + P(MV) + P(N) + P(O) = .114 + .371 + .065 + .103 + .068 + .111 + .043 = .875

d. P(CH) + P(CO) = .114 + .371 = .485

e. Since there were 335,770 carloads that week, the probability of selecting any one in particular would be 1 / 335,770 = .00000298. Thus, the probability of selecting the carload with the serial number 1003642 is .00000298.

Probability 58 3.16 a. Since order does not matter, the number of different bets would be a combination of 8 things taken 2 at a time.

The number of ways would be

骣8 8! 8鬃 7 6 鬃 5 4 鬃 3 2 1 40,320 琪 = = = = 28 桫2 2!(8- 2)! 2鬃 1 6 鬃 5 4 鬃 3 2 1 1440

b. If all players are of equal ability, then each of the 28 sample points would be equally likely. Each would have a probability of occurring of 1/28. There is only one sample point with values 2 and 7. Thus, the probability of winning with a bet of 2-7 would by 1/28 or .0357.

3.18 a. Let I = Infiniti, St = Saturn, and Sb = Saab. All possible rankings are as follows, where the first dealer listed is ranked first, the second dealer listed is ranked second, and the third dealer listed is ranked third:

I,St,Sb I,Sb,St St,Sb,I St,I,Sb Sb,I,St Sb,St,I

b. If each set of rankings is equally likely, then each has a probability of 1/6.

The probability that Saturn is ranked first = P(St,Sb,I) + P(St,I,Sb) = 1/6 + 1/6 = 2/6 = 1/3.

The probability that Saturn is ranked third = P(I,Sb,St) + P(Sb,I,St) = 1/6 + 1/6 = 2/6 = 1/3.

The probability that Saturn is ranked first and Infiniti is second = P(St,I,Sb) = 1/6.

3.20 We will denote the five successful utility companies as S1, S2, S3, S4, and S5 and the two failing companies as F1 and F2. There are

骣7 7! 7 鬃 6 5 鬃 4 3 鬃 2 1 琪 = = = 35 桫3 3!4! 3 鬃 2 1 鬃 4 3 鬃 2 1

possible ways to choose three companies from the seven, as shown below:

(S1, S2, S3) (S1, S3, S4) (S1, S4, S5) (S1, S5, F1) (S1, S2, S4) (S1, S3, S5) (S1, S4, F1) (S1, S5, F2) (S1, S2, S5) (S1, S3, F1) (S1, S4, F2) (S1, S2, F1) (S1, S3, F2) (S1, F1, F2) (S1, S2, F2)

(S2, S3, S4) (S2, S4, S5) (S2, S5, F1) (S2, F1, F2) (S2, S3, S5) (S2, S4, F1) (S2, S5, F2) (S2, S3, F1) (S2, S4, F2) (S2, S3, F2)

59 Chapter 3 (S3, S4, S5) (S3, S5, F1) (S3, F1, F2) (S3, S4, F1) (S3, S5, F2) (S3, S4, F2)

(S4, S5, F1) (S5, F1, F2) (S4, S5, F2) (S4, F1, F2)

a. Each outcome is equally likely, so each sample point has probability 1/35. From the 35

events listed, 10 do not contain F1 or F2. Therefore, P(selecting none) = 10/35.

b. From the 35 events listed, 20 contain either F1 or F2, but not both. Therefore, P(selecting one) = 20/35.

c. From the 35 events listed, 5 contain both F1 and F2. Therefore, P(selecting both) = 5/35.

3.22 a. P( Bc )= 1 - P ( B ) = 1 - .7 = .3

b. P( Ac )= 1 - P ( A ) = 1 - .4 = .6

c. P( A� B )+ P ( A ) - P ( B ) � P ( A + B ) - .4 = .7 .3 .8

3.24 The experiment consists of rolling a pair of fair dice. The sample points are:

1, 1 2, 1 3, 1 4, 1 5, 1 6, 1 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6

Since each die is fair, each sample point is equally likely. The probability of each sample point is 1/36.

a. A: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

B: {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}

A  B: {(3, 4), (4, 3)}

A  B: {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (1, 6), (2, 5), (5, 2), (6, 1)}

Ac: {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (4, 1), (4, 2), (4, 4), (4, 5), (4, 6), (5, 1), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Probability 60 骣1 6 1 b. P(A) = 6琪 = = 桫36 36 6 骣1 11 P(B) = 11琪 = 桫36 36 骣1 2 1 P(A  B) = 2琪 = = 桫36 36 18 骣1 15 5 P(A  B) = 15琪 = = 桫36 36 12 骣1 30 5 P(Ac) = 30琪 = = 桫36 36 6

1 11 1 6 + 11 - 2 15 5 c. P(A  B) = P(A) + P(B)  P(A  B) = + - = = = 6 36 18 36 36 12

d. A and B are not mutually exclusive. To be mutually exclusive, P(A  B) must be 0. Here, 1 P(A  B) = . 18

c 3.26 a. P(A ) = P(E3) + P(E6) = .2 + .3 = .5

c b. P(B ) = P(E1) + P(E7) = .10 + .06 = .16

c c. P(A  B) = P(E3) + P(E6) = .2 + .3 = .5

d. P(A  B) = P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6) + P(E7) = .10 + .05 + .20 + .20 + .06 + .30 + .06 = .97

e. P(A  B) = P(E2) + P(E4) + P(E5) = .05 + .20 + .06 = .31

c c f. P(A  B ) = P(E1) + P(E7) + P(E3) + P(E6) = .10 + .06 + .20 + .30 = .66

g. No. A and B are mutually exclusive if P(A  B) = 0. Here, P(A  B) = .31.

3.28 a. The outcome "On" and "High" is A  D.

b. The outcome "Low" or "Medium" is Dc.

3.30 a. The event A  B is the event the outcome is black and odd. The event is A  B: {11, 13, 15, 17, 29, 31, 33, 35}

b. The event A  B is the event the outcome is black or odd or both. The event A  B is {2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, 35, 1, 3, 5, 7, 9, 19, 21, 23, 25, 27}

61 Chapter 3 c. Assuming all events are equally likely, each has a probability of 1/38.

骣1 18 9 P(A) = 18琪 = = 桫38 38 19 骣1 18 9 P(B) = 18琪 = = 桫38 38 19 骣1 8 4 P(A  B) = 8琪 = = 桫38 38 19 骣1 28 14 P(A  B) = 28琪 = = 桫38 38 19 骣1 18 9 P(C) = 18琪 = = 桫38 38 19

d. The event A  B  C is the event the outcome is odd and black and low. The event A  B  C is {11, 13, 15, 17}.

9 9 4 14 e. P(A  B) = P(A) + P(B)  P(A  B) = + = 19 19 19 19

骣1 4 2 f. P(A  B  C) = 4琪 = = 桫38 38 19

g. The event A  B  C is the event the outcome is odd or black or low. The event A  B  C is:

{1, 2, 3, ... , 29, 31, 33, 35} or {All sample points except 00, 0, 30, 32, 34, 36}

骣1 32 16 h. P(A  B  C) = 32琪 = = 桫38 38 19

3.32 a. P  S  A

Products 6 and 7 are contained in this intersection.

b. P(possess all the desired characteristics) = P(P  S  A) 1 1 1 = P(6) + P(7) = + = 10 10 5

c. A  S

P(A  S) = P(2) + P(3) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) 1 1 1 1 1 1 1 1 8 4 = + + + + + + + = = 10 10 10 10 10 10 10 10 10 5

Probability 62 d. P  S

1 1 1 3 P(P  S) = P(2) + P(6) + P(7) = + + = 10 10 10 10

3.34 First, convert the percentages in the table to probabilities by dividing the percent by 100%.

a. P(A) = .259 + .169 + .115 = .543 P(B) = .003 P(C) = .037 + .078 + .016 + .002 + .047 + .027 = .207 P(D) = .414

b. P(A  D) = .156 + .094 + .043 = .293 P(A  D) = P(A) + P(B)  P(A  D) = .543 + .414  .293 = .664

c. Ac: {The worker is under 40} Bc: {The worker is 20 or older or is not part-time} Dc: {The worker is not part-time}

d. P(Ac) = 1  P(A) = 1  .543 = .457 P(Bc) = 1  P(B) = 1  .003 = .997 P(Dc) = 1  P(D) = 1  .414 = .586

3.36 Define the following events:

A: {Wheelchair user had an injurious fall} B: {Wheelchair user had all five features installed in the home} C: {Wheelchair user had no falls} D: {Wheelchair user had none of the features installed in the home}

48 a. P( A )= = .157 306

9 b. P( B )= = .029 306

89 c. P( C� D )= .291 306

3.38 a. Yes, the probabilities in the table sum to 1.

.05 + .16 + .05 + .19 + .32 + .05 + .11 + .05 + .02 = 1

b. P(A) = .05 + .16 + .05 = .26 P(B) = .05 + .19 + .11 = .35 P(C) = .05 + .16 + .19 + .32 = .72 P(D) = .05 + .05 + .11 + .05 + .02 = .28 P(E) = .05

63 Chapter 3 c. P(A  B) = .05 + .16 + .05 + .19 + .11 = .56 P(A  B) = .05 P(A  C) = .05 + .16 + .05 + .19 + .32 = .77

d. P(Ac) = 1  P(A) = 1  .26 = .74

The probability that a managerial prospect is not highly motivated is .74. Only about 1/4 of the prospects are highly motivated.

e. The pairs of events that are mutually exclusive have no sample points in common.

A  B contains the event "Prospect places in the high motivation category and in the high talent category." Therefore, A and B are not mutually exclusive.

A  C contains the event "Prospect places in the high motivation category and in the medium or high talent category." Therefore, A and C are not mutually exclusive.

A  D contains the event "Prospect places in the high motivation category and in the low talent category." Therefore, A and D are not mutually exclusive.

A  E contains the event "Prospect places in the high motivation category and in the high talent category." Therefore, A and E are not mutually exclusive.

B  C contains the event "Prospect places in the high talent category and in the medium or high motivation category." Therefore, B and C are not mutually exclusive.

B  D contains the event "Prospect places in the high talent category and in the low motivation category." Therefore, B and D are not mutually exclusive.

B  E contains the event "Prospect places in the high talent category and in the high motivation category." Therefore, B and E are not mutually exclusive.

C  D contains no events. Therefore, C and D are mutually exclusive.

C  E contains the event "Prospect places in the high talent category and in the high motivation category." Therefore, C and E are not mutually exclusive.

D  E contains no events. Therefore, D and E are mutually exclusive.

3.40 a. P( A� B )= P ( A | B ) P ( B ) = .6(.2) .12

P( A B ) .12 b. P( B | A )= = = .3 P( A ) .4

3.42 a. Since A and B are mutually exclusive events, P(A  B) = P(A) + P(B) = .30 + .55 = .85

b. Since A and C are mutually exclusive events, P(A  C) = 0

Probability 64 P( A B ) 0 c. P(A│B) = = = 0 P( B ) .55

d. Since B and C are mutually exclusive events, P(B  C) = P(B) + P(C) = .55 + .15 = .70

e. No, B and C cannot be independent events because they are mutually exclusive events.

3.44 a. If two fair coins are tossed, there are 4 possible outcomes or simple events. They are:

(1) HH (2) HT (3) TH (4) TT

Event A contains the simple events (2), (3), and (4). Event B contains the simple events (2) and (3).

A Venn diagram of this would be:

A B 2

3 4

Since the coins are fair, each of the sample points is equally likely. Each would have probabilities of ¼.

骣1 3 b. P( A )= 3琪 = = .75 桫4 4

骣1 2 1 P( B )= 2琪 = = = .5 桫4 4 2

1 1 2 1 P( A� B )+ P (2) = P (3) + = = = .5 4 4 4 2

65 Chapter 3 P( A B ) .5 c. P( A | B )= = = 1 P( B ) .5

P( A B ) .5 P( B | A )= = = .667 P( A ) .75

3.46 The 36 possible outcomes obtained when tossing two dice are listed below:

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

A: {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}

B: {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5), (6, 6)}

A  B: {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5)}

If A and B are independent, then P(A)P(B) = P(A  B).

18 1 7 6 1 P(A) = = P(B) = P(A  B) = = 36 2 36 36 6 1 7 7 1 P(A)P(B) = � � P( A B ) . Thus, A and B are not independent. 2 36 72 6

3.48 Define the following events:

C: {Executive admitted cheating at golf} D: {Executive lied in business}

From the problem, we know P(C) = .55 and P( C� D ) .20 .

P( C D ) .20 P( D | C )= = = .364 P( C ) .55

3.50 Define the following events:

A: {Winner is from the American League} B: {Winner is from the National League} C: {Winner is from the Eastern Division} D {Winner is from the Central Division} E: {Winner is from the Western Division}

Probability 66 6 P( A C ) 6 a. P( C | A )= =12 = = .75 P( A )8 8 12

1 P( B D ) 1 b. P( B | D )= =12 = = .5 P( D )2 2 12

2 P(( D惹 E ) B ) 2 c. P( D� E | B )= =12 = .5 P( B )4 4 12

3.52 Define the following events:

C: {Public school building has inadequate plumbing} D: {Public school has plans for repairing building}

From the problem, we know P(C) = .25 and P(D|C) = .38.

P( C� D )= P ( D | C ) P ( C ) = .38(.25) .095

3.54 Define the following events:

Ai : {ith CEO has bachelors degree}

6 a. P( A )= = .24 1 25

2 b. P( A | A乔 A A� A ) = .095 5 1 2 3 4 21

3.56 If A and B are independent, then P( A� B ) P ( A ) P ( B ) . For this Exercise, 1385+ 786 2171 1385+ 1175 2560 P( A )= = = .552 , P( B )= = = .651, and 3934 3934 3934 3934

1385 P( A� B )= .352 . 3934

P( A ) P ( B )= .552(.651) = .359� .352 P ( A B ) . Thus, A and B are not independent.

360 3.58 a. P( A )= = .36 1000

b. P( B | A )= .75

c. P( B | Ac )= .20

67 Chapter 3 d. P( A B ) is the probability that a student fails both the HSD exam and the written paper.

e. P( A� B )= P ( B | A ) P ( A ) = .75(.36) .27

3.60 First, define the following event:

A: {CVSA correctly determines the veracity of a suspect} P(A) = .98 (from claim)

a.The event that the CVSA is correct for all four suspects is the event A  A  A  A. P(A  A  A  A) = .98(.98)(.98)(.98) = .9224

b. The event that the CVSA is incorrect for at least one of the four suspects is the event (A  A  A  A)c. P(A  A  A  A)c = 1  P(A  A  A  A) = 1  .9224 = .0776

3.62 a. Define the following events:

W: {Player wins the game Go} F: {Player plays first (black stones)}

P(W  F) = 319/577 = .553

b. P(W  F│CA) = 34/34 = 1 P(W  F│CB) = 69/79 = .873 P(W  F│CC) = 66/118 = .559 P(W  F│BA) = 40/54 = .741 P(W  F│BB) = 52/95 = .547 P(W  F│BC) = 27/79 = .342 P(W  F│AA) = 15/28 = .536 P(W  F│AB) = 11/51 = .216 P(W  F│AC) = 3/39 = .077

c. There are three combinations where the player with the black stones (first) is ranked higher than the player with the white stones: CA, CB, and BA.

P(W  F│CA  CB  BA) = (34 + 69 + 40)/(34 + 79 + 54) = 143/167 = .856

d. There are three combinations where the players are of the same level: CC, BB, and AA.

P(W  F│CC  BB  AA) = (66 + 52 + 15)/(118 + 95 + 28) = 133/241 = .552

3.64 Define the following event:

E: {Error produced when dividing}

From the problem, P(E) = 1/9,000,000,000

a. P(Ec) = 1  P(E) = 8,999,999,999/9,000,000,000 = .999999999  1.0000

Probability 68 b. P(Ec  Ec) = .999999999(.999999999) = .999999999  1 (assuming the events are independent)

c. Assuming that all 1,000,000,000 divides are independent, the probability that there will be no errors in 1,000,000,000 divides is:

P(Ec  Ec  Ec   Ec) = [P(Ec)]1,000,000,000 = (8,999,999,999/9,000,000,000)1,000,000,000 = .9048

d. The event "at least one error in 1 billion divisions" is the complement of the event described in part c. Thus, the probability of at least one error in 1 billion divisions is:

1  .9048 = .0952

3.66 a. Suppose the elements of the population are:

1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

The possible samples of size 2 are:

(1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) (1, 9) (1, 10) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8) (2, 9) (2, 10) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) (3, 9) (3, 10) (4, 5) (4, 6) (4, 7) (4, 8) (4, 9) (4, 10) (5, 6) (5, 7) (5, 8) (5, 9) (5, 10) (6, 7) (6, 8) (6, 9) (6, 10) (7, 8) (7, 9) (7, 10) (8, 9) (8, 10) (9, 10)

Since there are N = 10 elements in the population, the number of samples of size n = 2 is a combination of 10 things taken 2 at a time or

骣10 10! 10 鬃 9 8 鬃 7 6 鬃 5 4 鬃 3 2 1 琪 = = 1 = 45 桫2 2!8! (2 鬃 1)(8 7 鬃 6 5 鬃 4 3 鬃 2 1)

Therefore, there are 45 different samples of size n = 2 that can be selected from a population of N = 10.

b. If random sampling is employed, every pair of elements has an equal probability of being selected. Therefore, the probability of drawing a particular pair is 1/45.

c. To draw a random sample of 2 elements from 10, we will number the elements from 0 to 9. Then, starting in an arbitrary position in Table I, Appendix B, we will select two numbers by going either down a column or across a row. Suppose that we start in the third position of column 6 and row 9. We will proceed down the column. The first sample drawn will be 1 and 5. The second sample drawn will be 9 and 4. The 20 samples selected are:

69 Chapter 3 Sample Number Items Selected Sample Number Items Selected 1 1, 5 11 0, 9 2 9, 4 12 1, 0 3 4, 2 13 3, 7 4 9, 3 14 3, 9 5 8, 1 15 0, 8 6 5, 6 16 3, 4 7 1, 3 17 0, 4 8 0, 2 18 9, 7 9 4, 6 19 8, 4 10 8, 0 20 0, 5

There are actually two pairs of samples that match: Samples 10 and 15, and samples 4 and 14. Given the low probability of each pair occurring, it is not that likely to have two pairs of samples that match.

3.68 First, number the elements of the population from 1 to 200,000. Starting in row 10, column 1, of Table I of Appendix B and reading down, take the first ten 6-digit numbers. Eliminate any duplicates, the number 000000, and all numbers greater than 200,000.

The 10 numbers selected for the random sample are:

094299 103656 071199 023682 010115 070569 024883 007425 053660 005820

Elements with the above numbers are selected for the sample.

3.70 To draw a random sample of 1,000 households from 534,322, we will number the households from 1 to 534,322. Then, starting in an arbitrary position in Table I, Appendix B, we will select 6-digit numbers by proceeding down a column. We will continue selecting numbers until we have 1,000 different 6-digit numbers, eliminating 000000 and any numbers between 534,323 and 999,999.

3.72 a. Give each stock in the NYSE-Composite Transactions table of the Wall Street Journal a number (1 to m). Using Table I of Appendix B, pick a starting point and read down using the same number of digits as in m until you have n different numbers between 1 and m, inclusive.

3.74 a. P( B1� A )= P ( A | B 1 ) P ( B 1 ) = .3(.75) .225

b. P( B2� A )= P ( A | B 2 ) P ( B 2 ) = .5(.25) .125

c. P( A )= P ( B1� A ) � P ( B 2 + A ) = .225 .125 .35

Probability 70 P( B1 A ) .225 d. P( B1 | A )= = = .643 P( A ) .35

P( B2 A ) .125 e. P( B2 | A )= = = .357 P( A ) .35

3.76 If A is independent of B1, B2, and B3, then P( A | B1 )= P ( A ) = .4 .

P( A | B1 ) P ( B 1 ) .4(.2) Then P( B1 | A )= = = .2 P( A ) .4

d. If there was a serious error, the probability that the error was made by engineer 3 is .526. This probability is higher than for any of the other engineers. Thus engineer #3 is most likely responsible for the error.

3.80 Define the following events:

S: {System shuts down}

F1: {Hardware failure} F2: {Software failure} F3: {Power failure}

From the Exercise, we know:

P(F1) = .01, P(F2) = .05, and P(F3) = .02. Also, P(S|F1) = .73, P(S|F2) = .12, and P(S|F3) = .88.

The probability that the current shutdown is due to a hardware failure is:

The probability that the current shutdown is due to a software failure is:

The probability that the current shutdown is due to a power failure is:

3.82 a. The two probability rules for a sample space are that the probability for any sample point

Probability 72 is between 0 and 1 and that the sum of the probabilities of all the sample points is 1.

For this Exercise, all the probabilities of the sample points are between 0 and 1 and

P( Si )= P ( S1 ) + P ( S 2 ) + P ( S 3 ) + P ( S 4 ) = .2 + .1 + .3 + .4 = 1.0 i=1

b. P( A )= P ( S1 ) + P ( S 4 ) = .2 + .4 = .6

3.84 P( A� B )+ P ( A ) - P ( B ) � P ( A + B ) - .7 = .5 .4 .8

3.86 a. If the Dow Jones Industrial Average increases, a large New York bank would tend to decrease the prime interest rate. Therefore, the two events are not mutually exclusive since they could occur simultaneously.

b. The next sale by a PC retailer could not be both a laptop and a desktop computer. Since the two events cannot occur simultaneously, the events are mutually exclusive.

c. Since both events cannot occur simultaneously, the events are mutually exclusive.

3.88 a. Because events A and B are independent, we have:

P(A  B) = P(A)P(B) = (.3)(.1) = .03

Thus, P(A  B)  0, and the two events cannot be mutually exclusive.

P( A B ) .03 P( A B ) .03 b. P(A│B) = = = .3 P(B│A) = = = .1 P( B ) .1 P( A ) .3

c. P(A  B) = P(A) + P(B)  P(A  B) = .3 + .1  .03 = .37

3.90 Mutually exclusive events are also dependent events since the assumption that one event occurs alters the probability of the occurrence of the other one. If we assume that one event has occurred, it is impossible for the other one to occur simultaneously since they are mutually exclusive. In other words, if A and B are mutually exclusive, P(A  B) = 0. P(A│B) = P( A B ) 0 = = 0. Since P(A)  0, A and B are dependent. P( B ) P ( B )

3.92 a. The event {The manager was involved in the ISO 9000 registration} contains the sample points {The manager was very involved}, {The manager had moderate involvement}, and {The manager had minimal involvement}. Thus, P(A) is:

9 16 12 37 P(A) = + + = = .925 40 40 40 40

b. The event {The length of time to achieve ISO 9000 registration was more than 2 years} contains the sample points {The length of time to achieve ISO 9000 registration was between 2.1 and 2.5 years} and {The length of time to achieve ISO 9000 registration was greater than 2.5 years}. Thus, P(B) is:

73 Chapter 3 2 3 5 P(B) = + = = .125 40 40 40

c. We cannot determine if events A and B are independent from the data given because there is no way of finding the P(A  B). In order to find P(A  B), the 40 individuals would have to be classified on both variables at the same time. In the data provided, the individuals are first classified on the first variable and then classified on the second variable.

3.94 Define the following events:

A: {male worker} B: {female worker} C: {service worker} D: {managerial/professional worker} E: {operator/fabricator/laborer} F: {technical/sales/administrative worker}

a. P(A  C) = .05

b. P(D) = P(A  D) + P(B  D) = .16 + .16 = .32

c. P[(B  D)  (B  E)] = .16 + .03 = .19

d. P(Fc) = 1  P(F) = 1  [P(A  F) + P(B  F)] = 1  (.10 + .18) = 1  .28 = .72

3.96 a. We will define the events the same as in Exercise 3.17.

There are a total of 9  2 = 18 sample points for this experiment. There are 9 sources of CO poisoning, and each source of poisoning has 2 possible outcomes, fatal or nonfatal. Suppose we introduce some notation to make it easier to write down the sample points. Let FI = Fire, AU = Auto exhaust, FU = Furnace, K = Kerosene or spaceheater, AP = Appliance, OG = Other gas-powered motors, FP = Fireplace, O = Other, and U = Unknown. Also, let F = Fatal and N = Nonfatal. The 18 sample points are:

FI, F AU, F FU, F K, F AP, F OG, F FP, F O, F U, F FI, N AU, N FU, N K, N AP, N OG, N FP, N O, N U, N

P(F│FI) = 63/116 = .543

b. P(AU│N) = 178/807 = .221

c. P(U│F) = 9/174 = .052

d. The event "not fire or fireplace" would include AU, FU, K, AP, OG, O, and U.

P(AU  FU  K  AP  OG  O  U│N) = (178 + 345 + 18 + 63 + 73 + 19 + 42)/807 = 738/807 = .914

Probability 74 3.98 Since there are 11 individuals who are willing to serve on the panel, the number of different panels of 5 experts is a combination of 11 things taken 5 at a time or

骣11 11! 11 鬃 10 9 鬃 8 7 鬃 6 5 鬃 4 3 鬃 2 1 琪 = = = 462 桫5 5!6! (5 鬃 4 3 鬃 2 1)(6 鬃 5 4 鬃 3 2 1)

3.100 The possible ways of ranking the blades are:

GSW SGW WGS GWS SWG WSG

If the consumer had no preference but still ranked the blades, then the 6 possibilities are equally likely. Therefore, each of the 6 possibilities has a probability of 1/6 of occurring.

1 1 2 1 a. P(Ranks G first) = P(GSW) + P(GWS) = + = = 6 6 6 3

1 1 2 1 b. P(Ranks G last) = P(SWG) + P(WSG) = + = = 6 6 6 3

1 c. P(ranks G last and W second) = P(SWG) = 6

1 d. P(WGS) = 6

3.102 a. Consecutive tosses of a coin are independent events since what occurs one time would not affect the next outcome.

b. If the individuals are randomly selected, then what one individual says should not affect what the next person says. They are independent events.

c. The results in two consecutive at-bats are probably not independent. The player may have faced the same pitcher both times which may affect the outcome.

d. The amount of gain and loss for two different stocks bought and sold on the same day are probably not independent. The market might be way up or down on a certain day so that all stocks are affected.

e. The amount of gain or loss for two different stocks that are bought and sold in different time periods are independent. What happens to one stock should not affect what happens to the other.

f. The prices bid by two different development firms in response to the same building construction proposal would probably not be independent. The same variables would be present for both firms to consider in their bids (materials, labor, etc.).

3.104 a. We will define the following events:

75 Chapter 3 A:{The first activation device works properly; i.e., activates the sprinkler when it should} B:{The second activation device works properly}

From the statement of the problem, we know

P(A) = .91 and P(B) = .87

Furthermore, since the activation devices work independently, we conclude that

P(A  B) = P(A)P(B) = (.91)(.87) = .7917

Now, if a fire starts near a sprinkler head, the sprinkler will be activated if either the first activation device or the second activation device, or both, operates properly. Thus,

P(Sprinkler head will be activated) = P(A  B) = P(A) + P(B)  P(A  B) = .91 + .87  .7917 = .9883

b. The event that the sprinkler head will not be activated is the complement of the event that the sprinkler will be activated. Thus,

P(Sprinkler head will not be activated) = 1  P(Sprinkler head will be activated) = 1  .9883 = .0117

c. From part a, P(A  B) = P(A)P(B) = .7917

d. In terms of the events we have defined, we wish to determine

P(A  Bc) = P(A)P(Bc) (by independence) = .91(1  .87) = .91(.13) = .1183

3.106 Define the following events:

A: (ATV driver is under age 12} B: (ATV driver is aged 12–15} C: (ATV driver is under age 25}

a. P(ATV driver is 15 years old or younger) = P(A  B) = P(A) + P(B) = .14 + .13 = .27

b. P(ATV driver is 25 years old or older) = P(C) = 1  P(C) = 1  .48 = .52

P( A C ) .14 c. P(A│C) = = = .2917 P( C ) .48

d. No, the events A and C are not mutually exclusive. P(A  C) = .14  0

e. No, the events A and C are not independent. P(A)P(C) = .14(.48) = .0672  P(A  C) = .14

3.108 Define the following events:

Probability 76 C: {Committee judges joint acceptable} I: {Inspector judges joint acceptable}

The sample points of this experiment are:

C  I C  Ic Cc  I Cc  Ic

a. The probability the inspector judges the joint to be acceptable is:

101 23 124 P(I) = P(C  I) + P(Cc  I) = + =  .810 153 153 153

The probability the committee judges the joint to be acceptable is:

101 10 111 P(C) = P(C  I) + P(C  Ic) = + =  .725 153 153 153

b. The probability that both the committee and the inspector judge the joint to be acceptable is:

101 P(C  I) =  .660 153

The probability that neither judge the joint to be acceptable is:

19 P(Cc  Ic) =  .124 153

c. The probability the inspector and committee disagree is:

10 23 33 P(C  Ic) + P(Cc  I) = + =  .216 153 153 153

The probability the inspector and committee agree is:

101 19 120 P(C  I) + P(Cc  Ic) = + =  .784 153 153 153

77 Chapter 3 3.110 a. Define the following events:

A1: {Component 1 works properly} A2: {Component 2 works properly} B3: {Component 3 works properly} B4: {Component 4 works properly} A: {Subsystem A works properly} B: {Subsystem B works properly}

The probability a component fails is .1, so the probability a component works properly is 1  .1 = .9.

Subsystem A works properly if both components 1 and 2 work properly.

P(A) = P(A1  A2) = P(A1)P(A2) = .9(.9) = .81 (since the components operate independently)

Similarly, P(B) = P(B1  B2) = P(B1)P(B2) = .9(.9) = .81

The system operates properly if either subsystem A or B operates properly.

The probability the system operates properly is:

P(A  B) = P(A) + P(B) - P(A  B) = P(A) + P(B)  P(A)P(B) = .81 + .81  .81(.81) = .9639

b. The probability exactly one subsystem fails is:

P(A  Bc) + P(Ac  B) = P(A)P(Bc) + P(Ac)P(B) = .81(1  .81) + (1  .81).81 = .1539 + .1539 = .3078

c. The probability the system fails is the probability that both subsystems fail or:

P(Ac  Bc) = P(Ac)P(Bc) = (1  .81)(1  .81) = .0361

d. The system operates correctly 99% of the time means it fails 1% of the time. The probability one subsystem fails is .19. The probability n subsystems fail is .19n. Thus, we must find n such that

.19n  .01

Thus, n = 3.

3.112 Define the events:

A: {A bottle comes from machine A} B: {A bottle comes from machine B} R: {A bottle is rejected}.

Probability 78 Then the given probabilities are:

1 1 P(A) = .75, P(B) = .25, P(R│A) = , P(R│B) = 20 30

The proportion of rejected bottles is:

P(R) = P(A  R) + P(B  R) = P(RA)P(A) + P(R│A)P(B) 1 1 = (.75) + (.25)  .0458 20 30

The probability that a bottle comes from machine A, given that it is accepted is:

c P( A亲 Rc )P( R A ) P ( A ) (19/ 20) (.75) P(A│Rc) = = =  .7467 R( Rc ) 1- P ( R ) 1 - .0458

3.114 A person will acquire pneumonia only if exposed. Therefore, to calculate the probability that someone (vaccinated or not) will contract pneumonia, it is necessary to calculate the probability that the person is exposed and comes down with the disease. This is the intersection of two events, exposed and contracts pneumonia. We let:

A: {Person is exposed to bacterial pneumonia} B: {Vaccinated person contracts pneumonia} BC: {Vaccinated person does not contract pneumonia} C: {Non-vaccinated person contracts pneumonia} CC: {Non-vaccinated person does not contract pneumonia}

From the Exercise, the given probabilities are:

P(A) = .40 P(BC|A) = .90 Thus, P(B|A) = 1 – P(BC|A) = 1  .90 = .10 P(C|A) = .95

The probability that an elderly person inoculated with the new vaccine acquires pneumonia is:

P( A� B )= P ( B | A ) P ( A ) = .10(.40) .04

The probability that an elderly person not inoculated with the new vaccine acquires pneumonia is:

P( A� C )= P ( C | A ) P ( A ) = .95(.40) .38

79 Chapter 3 3.116 There are a total of 6  6 = 36 outcomes when rolling 2 dice. If we let the first number in the pair represent the outcome of die number 1 and the second number in the pair represent the outcome of die number 2, then the possible outcomes are:

1,1 2,1 3,1 4,1 5,1 6,1 1,2 2,2 3,2 4,2 5,2 6,2 1,3 2,3 3,3 4,3 5,3 6,3 1,4 2,4 3,4 4,4 5,4 6,4 1,5 2,5 3,5 4,5 5,5 6,5 1,6 2,6 3,6 4,6 5,6 6,6

If both dice are fair, then each of these outcomes are equally like and have a probability of 1/36.

a. To win on the first roll, a player must roll a 7 or 11. There are 6 ways to roll a 7 and 2 ways to roll an 11. Thus the probability of winning on the first roll is:

8 P(7 or 11)= = .2222 36

b. To lose on the first roll, a player must roll a 2 or 3. There is 1 way to roll a 2 and 2 ways to roll a 3. Thus the probability of losing on the first roll is:

3 P(2 or 3)= = .0833 36

c. If a player rolls a 4 on the first roll, the game will end on the next roll if the player rolls the original roll again (player wins) or if the player rolls a seven (player loses). Now, there are 3 ways of getting a 4 on the first roll: 1,3, 2,2, or 3,1.

If the first roll was 2,2, then the game would end on the next roll if the player threw a 2,2, 1,6, 2,5, 3,4, 4,3, 5,2, or 6,1 on the next roll. The probability of the game ending on the next roll would be:

7 P(2,2 or 7 on second toss | 2,2 on first)= = .1944 36

Now, suppose the first roll ended with a 1 and a 3. Since the dice are not marked, this result could have happened two ways: 1, 3 or 3,1. Regardless of how the original 1 and 3 were obtained, the player would have 2 ways of winning on the next roll: 1,3 or 3,1. For the game to end on the next roll, the player could throw 1,3, 3,1, 1,6, 2,5, 3,4, 4,3, 5,2, or 6,1. The probability of the game ending on the next roll would be:

8 P(1,3 or 3,1 or 7 on second toss |1 and 3 on first)= = .2222 36

Since there were 3 ways to get a 4 on the first roll, and each were equally likely, P(2,2) = 1/3 and P[1 and 3 (any order)] = 2/3.

Probability 80 The probability that the game ends on the second roll is P(2,2 or 7 on second toss | 2,2 on first) P (2,2 on first) +P(1,3 or 3,1 or 7 on second toss |1 and 3 on first) P (1 and 3 on first) 骣1 骣 2 = .1944琪 + .2222 琪 = .0648 + .1481 = .2129 桫3 桫 3

81 Chapter 3