Probability Chapter 3 - Mihaylo Faculty...

Probability Chapter 3

3.2 a. This is a Venn Diagram. b. If the sample points are equally likely, then

P(1) = P(2) = P(3) = ⋅⋅⋅ = P(10) = 110

Therefore,

P(A) = P(4) + P(5) + P(6) = 1 1 1 310 10 10 10

+ + = = .3

P(B) = P(6) + P(7) = 1 1 210 10 10

+ = = .2

c. P(A) = P(4) + P(5) + P(6) = 1 1 3 520 20 20 20

+ + = = .25

P(B) = P(6) + P(7) = 3 3 620 20 20

+ = = .3

3.4 a. 9 9! 9 8 7 6 5 4 3 2 14 4!(9 4)! 4 3 2 1 5 4 3 2 1

⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⎝ ⎠ ⋅

= 126

b. 7 7! 7 6 5 4 3 2 12 2!(7 2)! 2 1 5 4 3 2 1

⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ − ⋅ ⋅ ⋅ ⋅ ⋅⎝ ⎠ ⋅

= 21

c. 4 4! 4 3 2 14 4!(4 4)! 4 3 2 1 1

⎛ ⎞ ⋅ ⋅ ⋅= =⎜ ⎟ − ⋅ ⋅ ⋅⎝ ⎠ ⋅

= 1

d. 5 5! 5 4 3 2 10 0!(5 0)! 1 5 4 3 2 1

⎛ ⎞ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ − ⋅ ⋅ ⋅ ⋅⎝ ⎠ ⋅

= 1

e. 6 6! 6 5 4 3 2 15 5!(6 5)! 5 4 3 2 1 1

⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ − ⋅ ⋅ ⋅ ⋅⎝ ⎠ ⋅

= 6

Probability 55

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.

c. There is one sample point in A: 3,3. Thus, P(A) = 136

.

There are 6 sample points in B: 1,6 2,5 3,4 4,3 5,2 and 6,1. Thus, P(B) = 6 136 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

4,6 5,1 5,3 5,5 6,2 6,4 and 6,6. Thus, P(C) = 18 136 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}

P(B) = 600 .0512,000

=

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.

56 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 57

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

8 8! 8 7 6 5 4 3 2 1 40,320 282 2!(8 2)! 2 1 6 5 4 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 1435, TP = Toyota Prius, and C = Chevrolet Corvette. 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,TP,C I,C,TP C,I,TP C,TP,I TP,I,C TP,C, I

b. If each set of rankings is equally likely, then each has a probability of 1/6. The probability that the Toyota Prius is ranked first = P(TP,I,C) + P(TP,C, I) =1/6 + 1/6 = 2/6 = 1/3.

The probability that the Infinity 1435 is ranked third = P(C,TP,I) + P(TP,C, I) =1/6 + 1/6 = 2/6 = 1/3.

The probability that the Toyota Prius is ranked first and the Chevrolet Corvette is ranked

second = P(TP,C, I) =1/6. 3.20 First, we need to compute the total number of ways we can select 2 bullets (pair) from 1,837

bullets. This is a combination of 1,837 things taken 2 at a time. The number of pairs is:

366,686,1218361837

11834183512118361837

)!2837,1(!2!837,1

2 837,1

=⋅

=⋅⋅⋅⋅⋅⋅

⋅⋅⋅⋅⋅=

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

The probability of a false positive is the number of false positives divided by the number of pairs and is: P(false positive) = # false positives / # pairs = 693 / 1,686,366 = .0004 This probability is very small. There would be only about 4 false positives out of every 10,000. I would have confidence in the FBI’s forensic evidence.

58 Chapter 3

3.22 a. ( ) 1 ( ) 1 .7 .3cP B P B= − = − =

b. ( ) 1 ( ) 1 .4 .6cP A P A= − = − = c. ( ) ( ) ( ) ( ) .4 .7 .3 .8P A B P A P B P A B∪ = + − ∩ = + − = 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)}

b. P(A) = 1 6636 36 6

⎛ ⎞ = =⎜ ⎟⎝ ⎠

1

P(B) = 1 11136 36

⎛ ⎞ =⎜ ⎟⎝ ⎠

1

P(A ∩ B) = 1 2236 36 18

⎛ ⎞ = =⎜ ⎟⎝ ⎠

1

P(A ∪ B) = 1 15 51536 36 12

⎛ ⎞ = =⎜ ⎟⎝ ⎠

P(Ac) = 1 303036 36 6

⎛ ⎞ = =⎜ ⎟⎝ ⎠

5

c. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 1 11 1 6 11 2 15 56 36 18 36 36 12

+ −+ − = = =

d. A and B are not mutually exclusive. To be mutually exclusive, P(A ∩ B) must be 0. Here,

P(A ∩ B) = 118

.

Probability 59

3.26 a. P(Ac) = P(E3) + P(E6) = .2 + .3 = .5 b. P(Bc) = P(E1) + P(E7) = .10 + .06 = .16 c. P(Ac ∩ 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 f. P(Ac ∪ Bc) = 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 Define the following events: A: {problems with absenteeism} T: {problems with turnover} From the problem, P(A) = .55, P(T) = .41, and P(A ∩ T) = .22 P(problems with either absenteeism or turnover) = P(A ∪ T) = P(A) + P(T) − P(A ∩ T) = .55 + .41 − .22 = .74 3.32 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}

60 Chapter 3

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

P(A) = 1 18 91838 38 19

⎛ ⎞ = =⎜ ⎟⎝ ⎠

P(B) = 1 18 91838 38 19

⎛ ⎞ = =⎜ ⎟⎝ ⎠

P(A ∩ B) = 1 8838 38 19

⎛ ⎞ 4= =⎜ ⎟

⎝ ⎠

P(A ∪ B) = 1 28 12838 38 19

⎛ ⎞ = =⎜ ⎟⎝ ⎠

4

P(C) = 1 18 91838 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}.

e. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 9 9 4 1419 19 19 19

+ − =

f. P(A ∩ B ∩ C) = 1 4438 38 19

⎛ ⎞ 2= =⎜ ⎟

⎝ ⎠

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}

h. P(A ∪ B ∪ C) = 1 32 13238 38 19

⎛ ⎞ 6= =⎜ ⎟

⎝ ⎠

3.34 a. P ∩ S ∩ A Products 6 and 7 are contained in this intersection. b. P(possess all the desired characteristics) = P(P ∩ S ∩ A)

= P(6) + P(7) = 1 110 10 5

1+ =

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 8 410 10 10 10 10 10 10 10 10 5

+ + + + + + + = =

Probability 61

d. P ∩ S

P(P ∩ S) = P(2) + P(6) + P(7) = 1 1 1 310 10 10 10

+ + =

3.36 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 3.38 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}

a. 48( ) .157306

P A = =

b. 9( ) .029306

P B = =

c. 89( ) .2306

P C D∩ = = 91

3.40 There are a total of 6 x 6 x 6 = 216 possible outcomes from throwing 3 fair dice. To help

demonstrate this, suppose the three dice are different colors – red, blue and green. When we roll these dice, we will record the outcome of the red die first, the blue die second, and the green die third. Thus, there are 6 possible outcomes for the first position, 6 for the second, and 6 for the third. This leads to the 216 possible outcomes.

62 Chapter 3

The Grand Duke argued that the chance of getting a sum of 9 and the chance of getting a sum of 10 should be the same since the number of partitions for 9 and 10 are the same. These partitions are:

9 10 126 136 135 145 144 226 225 235 234 244 333 334 In each case, there are 6 partitions. However, if we take into account the three colors of the

dice, then there are various ways to get each partition. For instance, to get a partition of 126, we could get 126, 162, 216, 261, 612, and 621 (again, think of the red die first, the blue die second, and the green die third). However, to get a partition of 333, there is only 1 way. To get a partition of 144, there are 3 ways: 144, 414, and 441. The numbers of ways to get each of the above partitions are:

9 # ways 10 # ways 126 6 136 6 135 6 145 6 144 3 226 3 225 3 235 6 234 6 244 3 333 _ 1 334 _3 25 27 Thus, there are a total of 25 ways to get a sum of 9 and 27 ways to get a sum of 10. The chance of throwing a sum of 9 (25 chances out of 216 possibilities) is less than the chance of throwing a 10 (27 chances out of 216 possibilities). 3.42 a. ( ) ( | ) ( ) .6(.2) .12P A B P A B P B∩ = = =

b. ( ) .12( | ) .3( ) .4

P A BP B AP A∩

= = =

3.44 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

c. P(A│B) = ( )( ) .55

P A B P B∩

=0 = 0

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.

Probability 63

3.46 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 B2

34

1

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

b. 1 3( ) 3 .754 4

P A ⎛ ⎞= = =⎜ ⎟⎝ ⎠

1 2 1( ) 2 .54 4 2

P B ⎛ ⎞= = = =⎜ ⎟⎝ ⎠

1 1 2 1( ) (2) (3)4 4 4 2

.5B P P∩ = + = + = = =P A

c. ( ) .5( | ) 1( ) .5

P A BP A BP B∩

= = =

( ) .5( | ) .667( ) .75

P A BP B AP A∩

= = =

64 Chapter 3

3.48 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).

P(A) = 18 136 2

= P(B) = 736

P(A ∩ B) = 6 136 6

=

P(A)P(B) = 1 7 7 1 (2 36 72 6

P A B⋅ = ≠ = ∩ ) . Thus, A and B are not independent.

3.50 Define the following events: S: {cause of fatal crash is speeding} C: {cause of fatal crash is missing a curve} From the problem, we know P(S) = .3 and P(S ∩ C) = .12.

( ) .12( | ) .4( ) .3

P C SP C SP S∩

= = =

3.52 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}

a. 7( ) 715( | ) .710( ) 1015

P A CP C AP A∩

= = = =

Probability 65

b. 1( ) 115( | ) .3333( ) 315

P B DP B DP D∩

= = = =

c. 2(( ) ) 215( | ) 5( ) 515

P D E BP D E BP B∪ ∩

∪ = = = = .4

3.54 Define the following events: A: {electrical switch monitors quality of power} B: {electrical switch not wired properly} From the problem, P(A) = .90 and P(B | A) = .90. P(A ∩ B) = P(B | A) P(A) = .90(.90) = .81. 3.56 Define the following events:

iA : {ith CEO has bachelors degree}

a. 18( ) .2040

P A = =

b. If the first 4 CEO’s have just bachelor’s degree, then on the next pick there are only 4 left

to choose from. Similarly, after picking 4 CEO’s, there are only 36 observations left to choose from.

5 1 2 3 44( | ) .11

36P A A A A A∩ ∩ ∩ = = 1

3.58 If A and B are independent, then ( ) ( ) ( )P A B P A P B∩ = . For this Exercise,

1385 786 2171( ) .5523934 3934

P A += = = , 1385 1175 2560( ) .651

3934 3934P B +

= = = , and

1385( ) .33934

P A B∩ = = 52

∩

.

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

3.60 The probability of a false positive is P(A | B).

66 Chapter 3

3.62 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 3.64 Define the following events: I: {Leak ignites immediately (jet fire)} D: {Leak has delayed ignition (flash fire)} From the problem, P(I) = .01 and P(D | Ic) = .01 The probability of a jet fire or a flash fire = P(I ∪ D) = P(I) + P(D) – P(I ∩ D) = P(I) + P(D | Ic)P(Ic) − P(I ∩ D) = .01 + .01(1 − .01) – 0 = .01 + .0099 = .0199 A tree diagram of this problem is:

.01 I

Ic.99

D(.01)

Dc

(.99)

Ic∩D .99(.01)=.0099

Ic∩Dc .99(.99)=.9801

I .01

3.66 a. Define the following events: W: {Player wins the game Go} F: {Player plays first (black stones)} P(W ∩ F) = 319/577 = .553

Probability 67

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 3.68 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

10 10! 10 9 8 7 6 5 4 3 2 12 2!8! (2 1)(8 7 6 5 4 3 2 1)

⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⎝ ⎠

1 = 45

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.

68 Chapter 3

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:

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.

3.70 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. 3.72 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.

Probability 69

3.74 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.

3.76 a. 1 1 1( ) ( | ) ( ) .3(.75) .225P B A P A B P B∩ = = = b. 2 2 2( ) ( | ) ( ) .5(.25) .125P B A P A B P B∩ = = = c. 1 2( ) ( ) ( ) .225 .125 .35P A P B A P B A= ∩ + ∩ = + =

d. 11

( ) .225( | ) .643( ) .35

P B AP B AP A∩

= = =

e. 22

( ) .125( | ) .357( ) .35

P B AP B AP A

∩= = =

3.78 If A is independent of B1, B2, and B3, then 1( | ) ( ) .4P A B P A= = .

Then 1 11

( | ) ( ) .4(.2)( | ) .2( ) .4

P A B P BP B AP A

= = =

3.80 a.

11

1 1

1 1 2 2 3 3

( )( | )( )

( | ) ( )( | ) ( ) ( | ) ( ) ( | ) (

.01(.30) .003 .003 .158.01(.30) .03(.20) .02(.50) .003 .006 .01 .019

P E errorP E errorP error

P error E P EP error E P E P error E P E P error E P E

∩=

=+ +

= =+ + + +

)

= =

b.

22

2 2

1 1 2 2 3 3

( )( | )( )

( | ) ( )( | ) ( ) ( | ) ( ) ( | ) (

.03(.20) .006 .006 .316.01(.30) .03(.20) .02(.50) .003 .006 .01 .019



∩=

=+ +

= =+ + + +

)

= =

70 Chapter 3

c. 33

3 3

1 1 2 2 3 3

( )( | )( )

( | ) ( )( | ) ( ) ( | ) ( ) ( | ) (

.02(.50) .01 .01 .526.01(.30) .03(.20) .02(.50) .003 .006 .01 .019



∩=

=+ +

= =+ + + +

)

= =

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.

3.82 Define the following events: D: {Defect in steel casting} H: {NDE detects ‘Hit” or defect in steel casting} From the problem, P(H | D) = .97, P(H | Dc) = .005, and P(D) = .01. P(H) = P(H | D)P(D) + P(H | Dc)P(Dc) = .97(.01) + .005(.99) = .0097 + .00495 = .01465

( ) ( | ) ( ) .97(.01) .0097( | ) .6621( ) ( ) .01465 .01465

P D H P H D P DP D HP H P H∩

= = = = =

3.84 Define the following events: A: {Alarm A sounds alarm} B: {Alarm B sounds alarm} I: {Intruder}

From the problem: P(A | I ) = .9 P(B | I ) = .95 P(A | Ic ) = .2 P(B | Ic ) = .1 P( I ) = .4

Since the two systems are operating independently of each other, P(A ∩ B | I ) = P(A | I ) P(B | I ) = .9 (.95) = .855 P(A ∩ B ∩ I ) = P(A ∩ B | I ) P( I ) = .855(.4) = .342 P(A ∩ B | Ic ) = P(A | Ic ) P(B | Ic ) = .2 (.1) = .02

Probability 71

P(A ∩ B ∩ Ic ) = P(A ∩ B | Ic ) P( Ic ) = .02(.6) = .012

Thus, P(A ∩ B) = P(A ∩ B ∩ I ) + P(A ∩ B ∩ Ic ) = .342 + .012 = .354 Finally, P(I | A ∩ B ) = P(A ∩ B ∩ I ) / P(A ∩ B) = .342 / .354 = .966

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

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

4

1 2 3 41

( ) ( ) ( ) ( ) ( ) .2 .1 .3 .4 1.0ii

P S P S P S P S P S=

= + + + = + + + =∑ b. 1 4( ) ( ) ( ) .2 .4 .6P A P S P S= + = + =

3.88 ( ) ( ) ( ) ( ) .7 .5 .4 .8P A B P A P B P A B∪ = + − ∩ = + − = 3.90 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. 3.92 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.

b. P(A│B) = ( ) .0

( ) .1P A B

P B∩

=3

= .3 P(B│A) = ( ) .0

( ) .3P A B

P A∩

=3

= .1

c. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = .3 + .1 − .03 = .37 3.94 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) =

( ) 0( ) ( )

P A B P B P B∩

= = 0. Since P(A) ≠ 0, A and B are dependent.

72 Chapter 3

3.96 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. ( ) ( | ) ( ) .38(.25) .095P C D P D C P C∩ = = = 3.98 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:

P(A) = 9 16 1240 40 40

+ + = 3740

= .925

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:

P(B) = 2 340 40

+ = 540

= .125

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.

3.100 a. The experiment consists of selecting 159 employees and asking each to indicate how

strongly he/she agreed or disagreed with the statement "I believe that management is committed to CQI." There are five sample points: "Strongly agree," "Agree," "Neither agree nor disagree," "Disagree," and "Strongly disagree."

b. Since we have frequencies for each of the sample points, good estimates of the

probabilities are the relative frequencies. To find the relative frequencies, divide all of the frequencies by the sample size of 159. The estimates of the probabilities are:

Strongly Agree

Agree Neither Agree Nor Disagree

Disagree Strongly Disagree

.189 .403 .258 .113 .038 c. The probability that an employee agrees or strongly agrees with the statement is .189 + .403 = .592.

Probability 73

d. The probability that an employee does not strongly agree with the statement is equal to the sum of all the probabilities except that for "strongly agree" = .403 + .258 + .113 + .038 = .812.

3.102 a. 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

b. The set of all sample points is called the sample space. c. The event A is made up of the following sample points: FI, F and FI, N Then, P(A) = P(FI, F) + P(FI, N) = 63/981 + 53/981 = 116/981 = .118 d. The event B is made up of the following sample points: (FI, F); (AU, F); (FU, F); (K, F); (AP, F); (OG, F); (FP, F); (O, F); (U, F) Then, P(B) = P(FI, F) + P(AU, F) + P(FU, F) + P(K, F) + P(AP, F) + P(OG, F) + P(FP, F) + P(O, F) + P(U, F) = 63/981 + 60/981 + 18/891 + 9/981 + 9/981 + 3/981 + 0/981 + 3/981 + 9/981 = 174/981 = .177 e. The event C is made up of the following sample points: (AU, F) and (AU, N) Then, P(C) = P(AU, F) + P(AU, N) = 60/981 + 178/981 = 238/981 = .243 f. The event D is made up of the following sample point: AU, F Then, P(D) = P(AU, F) = 60/981 = .061 g. The event E is made up of the following sample point: FI, N Then, P(E) = P(FI, N) = 53/981 = .054 3.104 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

11 11! 11 10 9 8 7 6 5 4 3 2 15 5!6! (5 4 3 2 1)(6 5 4 3 2 1)

⎛ ⎞ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= =⎜ ⎟ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⎝ ⎠

= 462

74 Chapter 3

3.106 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.

a. P(Ranks G first) = P(GSW) + P(GWS) = 1 16 6

+ = 26

= 13

b. P(Ranks G last) = P(SWG) + P(WSG) = 1 16 6

+ = 26

= 13

c. P(ranks G last and W second) = P(SWG) = 16

d. P(WGS) = 16

3.108 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.).

Probability 75

3.110 a. We will define the following events: 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 3.112 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.

76 Chapter 3

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

1 1 11

1 1 2 2 3 3

( ) ( | ) ( )( | )( ) ( | ) ( ) ( | ) ( ) ( | ) ( )

.73(.01) .0073 .0073 .2362.73(.01) .12(.05) .88(.02) .0073 .006 .0176 .0309

P F S P S F P FP F SP S P S F P F P S F P F P S F P F∩

= =+ +

= = =+ + + +

=

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

2 2 22

1 1 2 2 3 3

( ) ( | ) ( )( | )( ) ( | ) ( ) ( | ) ( ) ( | ) ( )

.12(.05) .006 .006 .1942.73(.01) .12(.05) .88(.02) .0073 .006 .0176 .0309


= =+ +

= = =+ + + +

=

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

3 3 33

1 1 2 2 3 3

( ) ( | ) ( )( | )( ) ( | ) ( ) ( | ) ( ) ( | ) ( )

.88(.02) .0176 .0176 .5696.73(.01) .12(.05) .88(.02) .0073 .006 .0176 .0309


= =+ +

= = =+ + + +

=

3.114 Define the following events: C: {Committee judges joint acceptable} I: {Inspector judges joint acceptable} The sample points of this experiment are: C ∩ I C ∩ I c

C c ∩ I Cc ∩ I c

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

P(I) = P(C ∩ I) + P(C c ∩ I) = 101 23 124153 153 153

+ = = .810

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

P(C) = P(C ∩ I) + P(C ∩ I c) = 101 10 111153 153 153

+ = = .725

Probability 77

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

P(C ∩ I) = 101153

= .660

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

P(C c ∩ I c) = 19153

= .124

c. The probability the inspector and committee disagree is:

P(C ∩ I c) + P(C c ∩ I) = 10 23 33153 153 153

+ = = .216

The probability the inspector and committee agree is:

P(C ∩ I) + P(C c ∩ I c) = 101 19 120153 153 153

+ = = .784

3.116 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(BB1 ∩ B2B ) = P(BB1)P(B2B ) = .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

78 Chapter 3

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. 3.118 Define the events: A: {A bottle comes from machine A} B: {A bottle comes from machine B} R: {A bottle is rejected}. Then the given probabilities are:

P(A) = .75, P(B) = .25, P(R│A) = 120

, P(R│B) = 130

The proportion of rejected bottles is: P(R) = P(A ∩ R) + P(B ∩ R) = P(R⏐A)P(A) + P(R│A)P(B)

= 120

(.75) + 130

(.25) = .0458

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

P(A│Rc) = ( ) ( )( ) (19 / 20) (.7

( ) 1 ( ) 1 .0458

cc

c

P R A P AP A RR R P R

⋅∩ ⋅= =

− −5) = .7467

Probability 79

3.120 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(7 11) .222236

P or = =

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(2 3) .083336

P or = =

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(2,2 or 7 on second toss | 2,2 on first) .194436

P = =

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(1,3 or 3,1 or 7 on second toss |1 and 3 on first) .222236

P = =

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.

80 Chapter 3

The probability that the game ends on the second roll is

(2,2 or 7 on second toss | 2,2 on first) (2,2 on first)(1,3 or 3,1 or 7 on second toss |1 and 3 on first) (1 and 3 on first)

1 2 .1944 .2222 .0648 .1481 .21293 3

P PP P+

⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3.122 Suppose we define the following event: E: {Error produced when dividing} From the problem, we know that P(E) = 1 / 9,000,000,000 The probability of no error produced when dividing is P(Ec) = 1 – P(E) = 1 – 1 / 9,000,000,000 = 8,999,999,999 / 9,000,000,000 = .999999999 ≈ 1.0000 Suppose we want to find the probability of no errors in 2 divisions (assuming each division is independent): P(Ec ∩ Ec) = .999999999(.999999999) = .999999999 ≈ 1.0000 Thus, in general, the probability of no errors in k divisions would be: P(Ec ∩ Ec ∩ Ec ∩ …∩ Ec) = P(Ec)k = [8,999,999,999 / 9,000,000,000]k

k times Suppose a user ran a program that performed 1 billion divisions. The probability of no errors in these 1 billion divisions would be: P(Ec)1,000,000,000 = [8,999,999,999 / 9,000,000,000]1,000,000,000 = .9048 Thus, the probability of at least 1 error in 1 billion divisions would be 1 − P(Ec)1,000,000,000 = 1 - [8,999,999,999 / 9,000,000,000]1,000,000,000 = 1 − .9048 = .0852 For a heavy MINITAB user, this flawed chip would be a problem because the above

probability is not that small.

Probability 81

Probability Chapter 3 - Mihaylo Faculty...

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