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Students’ Solutions Manual

for

Carlton and Devore’s

Probability

with Applications in Engineering, Science, and Technology

Matthew A. Carlton

Cal Poly State University

CONTENTS

Chapter 1 Probability 1

Chapter 2 Discrete Random Variables and Probability

Distributions

Chapter 3 Continuous Random Variables and

Probability Distributions

Chapter 4 Joint Probability Distributions and Their

Applications

Chapter 5 The Basics of Statistical Inference 118

Chapter 6 Markov Chains 137

Chapter 7 Random Processes 156

Chapter 8 Introduction to Signal Processing 180

a. The 3 3 = 27 possible outcomes are numbered below for later reference.

Outcome Outcome Number Outcome Number Outcome 1 111 15 223 2 112 16 231 3 113 17 232 4 121 18 233 5 122 19 311 6 123 20 312 7 131 21 313 8 132 22 321 9 133 23 322 10 211 24 323 11 212 25 331 12 213 26 332 13 221 27 333 14 222

b. Outcome numbers 1, 14, 27 above.

c. Outcome numbers 6, 8, 12, 16, 20, 22 above.

d. Outcome numbers 1, 3, 7, 9, 19, 21, 25, 27 above.

a. S = { BBBAAAA, BBABAAA, BBAABAA, BBAAABA, BBAAAAB, BABBAAA, BABABAA, BABAABA, BABAAAB, BAABBAA, BAABABA, BAABAAB, BAAABBA, BAAABAB, BAAAABB, ABBBAAA, ABBABAA, ABBAABA, ABBAAAB, ABABBAA, ABABABA, ABABAAB, ABAABBA, ABAABAB, ABAAABB, AABBBAA, AABBABA, AABBAAB, AABABBA, AABABAB, AABAABB, AAABBBA, AAABBAB, AAABABB, AAAABBB }.

b. AAAABBB, AAABABB, AAABBAB, AABAABB, AABABAB.

a. In the diagram on the left, the shaded area is ( A ∪ B )′. On the right, the shaded area is A ′, the striped area is B ′, and the intersection A ′∩ B ′ occurs where there is both shading and stripes. These two diagrams display the same area.

b. In the diagram below, the shaded area represents ( A ∩ B )′. Using the right-hand diagram from (a), the union of A ′ and B ′ is represented by the areas that have either shading or stripes (or both). Both of the diagrams display the same area.

Section 1.

a. .07.

b. .15 + .10 + .05 = .30.

c. Let A = the selected individual owns shares in a stock fund. Then P ( A ) = .18 + .25 = .43. The desired probability, that a selected customer does not shares in a stock fund, equals P ( A′ ) = 1 – P ( A ) = 1 – .43 = .57. This could also be calculated by adding the probabilities for all the funds that are not stocks.

a. A 1 (^) ∪ A 2 = “awarded either #1 or #2 (or both)”: from the addition rule, P ( A 1 ∪ A 2 ) = P ( A 1) + P ( A 2) – P ( A 1 ∩ A 2) = .22 + .25 – .11 = .36.

b. A 1 ′ ∩ A 2 ′= “awarded neither #1 or #2”: using the hint and part (a), P A ( 1 ′ ∩ A 2 ′ ) = P (( A 1 ∪ A 2 (^) ) )′ = 1 − P A ( 1 (^) ∪ A 2 ) = 1 – .36 = .64.

c. (^) A 1 (^) ∪ A 2 (^) ∪ A 3 = “awarded at least one of these three projects”: using the addition rule for 3 events, P A ( 1 (^) ∪ A 2 ∪ A 3 )= P A ( 1 (^) ) + P A ( 2 (^) ) + P A ( 3 (^) ) − P A ( 1 (^) ∩ A 2 (^) ) − P A ( 1 (^) ∩ A 3 (^) ) − P A ( 2 (^) ∩ A 3 (^) ) + P A ( 1 (^) ∩ A 2 (^) ∩ A 3 )= .22 +.25 + .28 – .11 – .05 – .07 + .01 = .53.

d. (^) A 1 ′ (^) ∩ A 2 ′ (^) ∩ A 3 ′= “awarded none of the three projects”: P A ( 1 ′ ∩ A 2 ′ ∩ A 3 ′)= 1 –^ P (awarded at least one) = 1 – .53 = .47.

a. Let E be the event that at most one purchases an electric dryer. Then E ′ is the event that at least two purchase electric dryers, and P ( E ′) = 1 – P ( E ) = 1 – .428 = .572.

b. Let A be the event that all five purchase gas, and let B be the event that all five purchase electric. All other possible outcomes are those in which at least one of each type of clothes dryer is purchased. Thus, the desired probability is 1 – [ P ( A ) – P ( B )] = 1 – [.116 + .005] = .879.

a. The probabilities do not add to 1 because there are other software packages besides SPSS and SAS for which requests could be made.

b. P ( A ′) = 1 – P ( A ) = 1 – .30 = .70.

c. Since A and B are mutually exclusive events, P ( A ∪ B ) = P ( A ) + P ( B ) = .30 + .50 = .80.

d. By deMorgan’s law, P ( A ′ ∩ B ′) = P (( A ∪ B )′) = 1 – P ( A ∪ B ) = 1 – .80 = .20. In this example, deMorgan’s law says the event “neither A nor B ” is the complement of the event “either A or B .” (That’s true regardless of whether they’re mutually exclusive.)

21. Let A be that the selected joint was found defective by inspector A , so P ( A ) = (^10724) , 000. Let B be analogous for

inspector B , so P ( B ) = (^10751) , 000. The event “at least one of the inspectors judged a joint to be defective is A ∪ B ,

so P ( A ∪ B ) = (^101159) , 000.

a. By deMorgan’s law, P (neither A nor B ) = P ( A ′^ ∩ B ′)= 1 – P(A∪B) = 1 – (^101159) , 000 = (^108841) , 000 = .8841.

b. The desired event is B ∩ A ′. From a Venn diagram, we see that P ( B ∩ A ′) = P ( B ) – P ( A ∩ B ). From the addition rule, P ( A ∪ B ) = P ( A ) + P ( B ) – P ( A ∩ B ) gives P ( A ∩ B ) = .0724 + .0751 – .1159 = .0316. Finally, P ( B ∩ A ′) = P ( B ) – P ( A ∩ B ) = .0751 – .0316 = .0435.

23. In what follows, the first letter refers to the auto deductible and the second letter refers to the homeowner’s deductible. a. P ( MH ) = .10.

b. P (low auto deductible) = P ({ LN, LL, LM, LH }) = .04 + .06 + .05 + .03 = .18. Following a similar pattern, P (low homeowner’s deductible) = .06 + .10 + .03 = .19.

c. P (same deductible for both) = P ({ LL, MM, HH }) = .06 + .20 + .15 = .41.

d. P (deductibles are different) = 1 – P (same deductible for both) = 1 – .41 = .59.

e. P (at least one low deductible) = P ({ LN, LL, LM, LH, ML, HL }) = .04 + .06 + .05 + .03 + .10 + .03 = .31.

f. P (neither deductible is low) = 1 – P(at least one low deductible) = 1 – .31 = .69.

25. Assume that the computers are numbered 1-6 as described and that computers 1 and 2 are the two laptops. There are 15 possible outcomes: (1,2) (1,3) (1,4) (1,5) (1,6) (2,3) (2,4) (2,5) (2,6) (3,4) (3,5) (3,6) (4,5) (4,6) and (5,6).

a. P (both are laptops) = P ({(1,2)}) = 151 =.067.

b. P (both are desktops) = P ({(3,4) (3,5) (3,6) (4,5) (4,6) (5,6)}) = 156 = .40.

c. P (at least one desktop) = 1 – P (no desktops) = 1 – P (both are laptops) = 1 – .067 = .933.

d. P (at least one of each type) = 1 – P (both are the same) = 1 – [ P (both are laptops) + P (both are desktops)] = 1 – [.067 + .40] = .533.

27. By rearranging the addition rule, P ( A ∩ B ) = P ( A ) + P ( B ) – P ( A ∪ B ) = .40 + .55 – .63 = .32. By the same

method, P ( A ∩ C ) = .40 + .70 – .77 = .33 and P ( B ∩ C ) = .55 + .70 – .80 = .45. Finally, rearranging the

addition rule for 3 events gives P ( A ∩ B ∩ C ) = P ( A ∪ B ∪ C ) – P ( A ) – P ( B ) – P ( C ) + P ( A ∩ B ) + P ( A ∩ C ) + P ( B ∩ C ) = .85 – .40 – .55 – .70 + .32 + .33 + .45 = .30.

These probabilities are reflected in the Venn diagram below.

a. P ( A ∪ B ∪ C ) = .85, as given.

b. P (none selected) = 1 – P (at least one selected) = 1 – P ( A ∪ B ∪ C ) = 1 – .85 = .15.

c. From the Venn diagram, P (only automatic transmission selected) = .22.

d. From the Venn diagram, P (exactly one of the three) = .05 + .08 + .22 = .35.

29. Recall there are 27 equally likely outcomes.

a. P (all the same station) = P ((1,1,1) or (2,2,2) or (3,3,3)) = 273 = 91.

b. P (at most 2 are assigned to the same station) = 1 – P (all 3 are the same) = 1 – 19 = 89.

c. P (all different stations) = P ((1,2,3) or (1,3,2) or (2,1,3) or (2,3,1) or (3,1,2) or (3,2,1))

A B

C

a. Since there are 5 receivers, 4 CD players, 3 speakers, and 4 turntables, the total number of possible selections is (5)(4)(3)(4) = 240.

b. We now only have 1 choice for the receiver and CD player: (1)(1)(3)(4) = 12.

c. Eliminating Sony leaves 4, 3, 3, and 3 choices for the four pieces of equipment, respectively: (4)(3)(3)(3) = 108.

d. From a , there are 240 possible configurations. From c , 108 of them involve zero Sony products. So, the number of configurations with at least one Sony product is 240 – 108 = 132.

e. Assuming all 240 arrangements are equally likely, P (at least one Sony) =

Next, P (exactly one component Sony) = P (only the receiver is Sony) + P (only the CD player is Sony) + P (only the turntable is Sony). Counting from the available options gives

P (exactly one component Sony) =

a. There are 12 American beers and 8 + 9 = 17 imported beers, for a total of 29 beers. The total number of

five-beer samplers is

= 118,755. Among them, the number with at least 4 American beers, i.e.

exactly 4 or exactly 5, is

= 9,207. So, the probability that you get at least 4

American beers is

b. The number of five-beer samplers consisting of only American beers is

= 792. Similarly, the

number of all-Mexican and all-German beer samplers are

= 56 and

= 126, respectively.

Therefore, there are 792 + 56 + 126 = 974 samplers with all beers from the same country, and the

probability of randomly receiving such a sampler is 118

a. Since order doesn’t matter, the number of possible rosters is

b. The number of ways to select 2 women from among 5 is

= 10, and the number of ways to select 4

men from among 11 is

= 330. By the Fundamental Counting Principle, the total number of (2-

woman, 4-man) teams is (10)(330) = 3300.

c. Using the same idea as in part b , the count is

d. P (exactly 2 women) =

= .4121; P (at least 2 women) =

41. There are (^)  

= 10 possible ways to select the positions for B ’s votes: BBAAA, BABAA, BAABA, BAAAB,

ABBAA, ABABA, ABAAB, AABBA, AABAB , and AAABB. Only the last two have A ahead of B throughout the vote count. Since the outcomes are equally likely, the desired probability is 2/10 = .20.

a. There are 6 75W bulbs and 9 other bulbs. So, P (select exactly 2 75W bulbs) = P (select exactly 2 75W

bulbs and 1 other bulb) =

b. P (all three are the same rating) = P (all 3 are 40W or all 3 are 60W or all 3 are 75W) =

c. P (one of each type is selected) =. 2637 455

d. It is necessary to examine at least six bulbs if and only if the first five light bulbs were all of the 40W or 60W variety. Since there are 9 such bulbs, the chance of this event is 9 5 126 . 15 3003 5

n k

n n k k

n k n k

n k

n ( )!!

The number of subsets of size k equals the number of subsets of size n – k , because to each subset of size k there corresponds exactly one subset of size n – k : the n – k objects not in the subset of size k. The combinations formula counts the number of ways to split n objects into two subsets: one of size k , and one of size n – k.

Section 1.

51. Let A be that the individual is more than 6 feet tall. Let B be that the individual is a professional basketball player. Then P ( A|B ) = the probability of the individual being more than 6 feet tall, knowing that the individual is a professional basketball player, while P ( B|A ) = the probability of the individual being a professional basketball player, knowing that the individual is more than 6 feet tall. P ( A|B ) will be larger. Most professional basketball players are tall, so the probability of an individual in that reduced sample space being more than 6 feet tall is very large. On the other hand, the number of individuals that are pro basketball players is small in relation to the number of males more than 6 feet tall.

a. (^) 2 1 2 1 1

P A A

A

A

P

P A

= = = .50. The numerator comes from Exercise 28.

b. (^) 1 2 3 1 1 2 3 1 1 2 3 1 1

([ ]

P A A A A P ( A A

A A A

P

A

P A

A P A

∩ ∩ = = = = .0833. The numerator simplifies

because A 1 (^) ∩ A 2 (^) ∩ A 3 is a subset of A 1 , so their intersection is just the smaller event.

c. For this example, you definitely need a Venn diagram. The seven pieces of the partition inside the three circles have probabilities .04, .05, .00, .02, .01, .01, and .01. Those add to .14 (so the chance of no defects is .86). Let E = “exactly one defect.” From the Venn diagram, P ( E ) = .04 + .00 + .01 = .05. From the addition above, P (at least one defect) = P A ( 1 (^) ∪ A 2 ∪ A 3 )= .14. Finally, the answer to the question is

1 2 3 1 2 3 1 2 3 1 2 3

[ ) ).

( ] (

P E A P E

P E A

P A P A

A A

A A

A A A A

= .3571. The numerator simplifies

because E is a subset of A 1 (^) ∪ A 2 (^) ∪ A 3.

d. (^) 3 1 2 3 1 2 1 2

[ ) .

( ]

P A A A

P A A

P A

A

A

′ (^) ∩ = = .8333. The numerator is Exercise 28(c), while the

denominator is Exercise 28(b).

a. P (M ∩ LS ∩ PR) = .05, directly from the table of probabilities.

b. P (M ∩ Pr) = P (M ∩ LS ∩ PR) + P (M ∩ SS ∩ PR) = .05 + .07 = .12.

c. P (SS) = sum of 9 probabilities in the SS table = .56. P (LS) = 1 – .56 = .44.

d. From the two tables, P (M) = .08 + .07 + .12 + .10 + .05 + .07 = .49. P (Pr) = .02 + .07 + .07 + .02 + .05 + .02 = .25.

e. P (M|SS ∩ Pl) =

(M SS Pl). . (SS Pl) .04 .08.

P

P

f. P (SS|M ∩ Pl) =

(SS M Pl). . (M Pl) .08.

P

P

. P (LS|M ∩ Pl) = 1 – P (SS|M ∩ Pl) = 1 – .444 = .556. 57. We know that P ( A 1 ∪ A 2) = .07 and P ( A 1 ∩ A 2) = .01, and that P ( A 1 ) = P ( A 2 ) because the pumps are identical. There are two solution methods. The first doesn’t require explicit reference to q or r : Let A 1 be the event that #1 fails and A 2 be the event that #2 fails. Apply the addition rule: P ( A 1 ∪ A 2 ) = P ( A 1) + P ( A 2 ) – P ( A 1 ∩ A 2 ) ⇒ .07 = 2 P ( A 1 ) – .01 ⇒ P ( A 1) = .04.

Otherwise, we assume that P ( A 1) = P ( A 2 ) = q and that P ( A 1 | A 2) = P ( A 2 | A 1) = r (the goal is to find q ). Proceed as follows: .01 = P ( A 1 ∩ A 2) = P ( A 1) P ( A 2 | A 1) = qr and .07 = P ( A 1 ∪ A 2) = P A ( 1 (^) ∩ A 2 (^) ) + P A ( 1 ′ ∩ A 2 (^) )+ P A ( 1 (^) ∩ A 2 ′)= .01 + q (1 – r ) + q (1 – r ) ⇒ q (1 – r ) = .03. These two equations give 2 q – .01 = .07, from which q = .04 (and r = .25).

a. P ( A 2 | A 1 ) =. 50

. 22

1

P A

P A A

. If the firm is awarded project 1, there is a 50% chance they will also

be awarded project 2.

b. P ( A 2 ∩ A 3 | A 1) =. 0455

. 22

1

P A

P A A A

. If the firm is awarded project 1, there is a 4.55%

chance they will also be awarded projects 2 and 3.

c. ( )

[( ) ( )]

[ ( )]

1

1 2 1 3 1

1 2 3 (^2 31) PA

P A A A A

PA

PA A A

P A A A

1

= 1 ∩^2 +^1 ∩^3 −^1 ∩^2 ∩^3 = =

PA

P A A PA A PA A A

. If the firm is awarded project 1, there is a

68.2% chance they will also be awarded at least one of the other two projects.

d.. 0189

. 53

1 2 3

1 2 3 (^1 23123) ∪ ∪ = =

PA A A

PA A A

P A A A A A A. If the firm is awarded at least one of

the projects, there is a 1.89% chance they will be awarded all three projects.

71. First, partition the sample space into statisticians with both life and major medical insurance, just life insurance, just major medical insurance, and neither. We know that P (both) = .20; subtracting them out, P (life only) = P (life) – P (both) = .75 – .20 = .55; similarly, P (medical only) = P (medical) – P (both) = .45 – .20 = .25. a. Apply the Law of Total Probability: P (renew) = P (life only) P (renew | life only) + P (medical only) P (renew | medical only) + P (both) P (renew | both) = (.55)(.70) + (.25)(.80) + (.20)(.90) = .765.

b. Apply Bayes’ Rule: P (both | renew) =

(both) (renew | both) (

re

new)

P P

P

73. A tree diagram can help. We know that P (day) = .2, P (1-night) = .5, P (2-night) = .3; also, P (purchase | day) = .1, P (purchase | 1-night) = .3, and P (purchase | 2-night) = .2.

Apply Bayes’ rule: e.g., P (day | purchase) =

(day purchase) (.2)(.1) (purchase) (.2)(.1) (.5)(.3) (.3)(.2)

P

P

Similarly, P (1-night | purchase) =

= .652 and P (2-night | purchase) = .261.

75. Let T denote the event that a randomly selected person is, in fact, a terrorist. Apply Bayes’ theorem, using P ( T ) = 1,000/300,000,000 = .0000033:

P ( T | +) =

P T P T

P T P T P T P T

+ + ′^ + ′

= .003289. That is to say,

roughly 0.3% of all people “flagged” as terrorists would be actual terrorists in this scenario.

77. Let’s see how we can implement the hint. If she’s flying airline #1, the chance of 2 late flights is (30%)(10%) = 3%; the two flights being “unaffected” by each other means we can multiply their probabilities. Similarly, the chance of 0 late flights on airline #1 is (70%)(90%) = 63%. Since percents add to 100%, the chance of exactly 1 late flight on airline #1 is 100% – (3% + 63%) = 34%. A similar approach works for the other two airlines: the probability of exactly 1 late flight on airline #2 is 35%, and the chance of exactly 1 late flight on airline #3 is 45%.

The initial (“prior”) probabilities for the three airlines are P ( A 1) = 50%, P ( A 2 ) = 30%, and P ( A 3) = 20%. Given that she had exactly 1 late flight (call that event B ), the conditional (“posterior”) probabilities of the three airlines can be calculated using Bayes’ Rule:

2 2 3

1 1 1 1 1 3

P A P B A

B

P A P B A P A P B A P A P

P A

B A

2 2 2 1 1 2 2 3 3

P A P B A

B

P A P B A P A P B A P A P

P A

B A

= .2877; and

3 3 3 1 1 2 2 3 3

P A P B A

B

P A P B A P A P B A P A P

P A

B A

Notice that, except for rounding error, these three posterior probabilities add to 1.

The tree diagram below shows these probabilities.

Section 1.

79. Using the definition, two events A and B are independent if P ( A | B ) = P ( A );

P ( A | B ) = .6125; P ( A ) = .50; .6125 ≠ .50, so A and B are not independent. Using the multiplication rule, the events are independent if P ( A ∩ B )= P ( A ) P ( B ); P ( A ∩ B ) = .25; P ( A ) P ( B ) = (.5)(.4) = .2. .25 ≠ .2, so A and B are not independent.

81. P ( A 1 ∩ A 2 ) = .11 while P ( A 1) P ( A 2 ) = .055, so A 1 and A 2 are not independent.

P ( A 1 ∩ A 3 ) = .05 while P ( A 1) P ( A 3 ) = .0616, so A 1 and A 3 are not independent. P ( A 2 ∩ A 3 ) = .07 and P ( A 2 ) P ( A 3) = .07, so A 2 and A 3 are independent.

83. Using subscripts to differentiate between the selected individuals,

P (O 1 ∩ O (^) 2) = P (O 1 ) P (O 2 ) = (.45)(.45) = .2025. P (two individuals match) = P (A 1 ∩ A2) + P (B 1 ∩ B 2 ) + P (AB 1 ∩ AB (^) 2) + P (O 1 ∩O (^) 2) = .40 2 + .11 2 + .04 2 + .45 2 = .3762.

85. Follow the same logic as in the previous exercise. With p = 1/9,000,000,000 and n = 1,000,000,000, the probability of at least one error is 1 – (1 – p ) n^ ≈ 1 – .8948 = .1052.

Note: For extremely small values of p , (1 – p ) n^ ≈ 1 – np. So, the probability of at least one occurrence under these assumptions is roughly 1 – (1 – np ) = np. Here, that would equal 1/9 ≈ .1111.

87. P (at least one opens) = 1 – P (none open) = 1 – (.05) 5 = .99999969.

P (at least one fails to open) = 1 – P (all open) = 1 – (.95) 5 = .2262.

P A ( ∩ B = P A × P B A = = .039984. Events A and B are not independent, since P ( B ) =.

while

P B A = = , and these are not equal.

b. If A and B were independent, we’d have P A ( ∩ B ) = P A ( ) × P B ( ) = (.2)(.2) = .0 4. This is very close to the answer .039984 from part a. This suggests that, for most practical purposes, we could treat events A and B in this example as if they were independent.

c. Repeating the steps in part a , you again get P ( A ) = P ( B ) = .2. However, using the multiplication rule, 2 1 ) ( ) ( | ) 10

P A ∩ B = P A × P B A = × =.0222. This is very different from the value of .04 that we’d get if A

and B were independent!

The critical difference is that the population size in parts a-b is huge, and so the probability a second board is green almost equals .2 (i.e., 1,999/9,999 = .19992 ≈ .2). But in part c , the conditional probability of a green board shifts a lot: 2/10 = .2, but 1/9 = .1111.

a. For route #1, P (late) = P (stopped at 2 or 3 or 4 crossings) = 1 – P (stopped at 0 or 1) = 1 – [.9 4 + 4(.9) 3 (.1)] = .0523. For route #2, P (late) = P (stopped at 1 or 2 crossings) = 1 – P (stopped at none) = 1 – .81 = .19. Thus route #1 should be taken.

b. Apply Bayes’ Rule: P (4-crossing route | late) =

(4-crossing late) (late)

P

P

a. Similar to several previous exercises, P (win at least once) = 1 – P (lose all n times) = 1 – P (lose) n^ = 1 – (1 – 1 /N ) n. b. The probability of rolling at least one 4 in n rolls is 1 – (5/6) n , while the “ n/N ” answer is n /6. The correct answers are .4212 ( n = 3), .6651 ( n = 6), and .8385 ( n = 10). The wrong answer 3/6 = .5 is not very accurate for n = 3, 6/6 = 1 is absurd for n = 6, and 10/6 for n = 10 is impossible since probability cannot exceed 1! Clearly, n/N is a terrible approximation to the correct probability. c. From Exercise 85, the probability of at least one error was .1052, compared to n/N = 10 9 /(9×10 9 ) = 1/9 ≈ .1111. Those aren’t too far off, so arguably n/N might not be a bad approximation to 1 – (1 – 1 /N ) n^ when N is very large (equivalently, p = 1/ N is small).

d. The binomial theorem states that 2 0

) n^ n^ k^ k^ n^ n^ n

n

k

n n a b a b a na b a b k

− − −

∑ ^. Applying the

binomial theorem with a = 1 and b = –1/ N gives (1 1/ ) 1 1 1 ( 1/ ) 2 ( 1 )^2 2

n n n (^) 1 n / n N n −^ N − N

n n n N N

− + +. But if n << N , then the third term in that expression and all those thereafter will be

negligible compared to the first two terms. Thus, (1 – 1 /N ) n^ ≈ 1 – n/N when N is very large. It follows that the probability of at least one success in the given setting, whose exact probability equals 1 – (1 – 1 /N ) n , can be approximated by 1 – (1 – n/N ) = n/N when N is very large.

Section 1.

a. Let A = exactly one of B 1 or B 2 occurs = ( B 1 ∩ B 2 ′) ∪ ( B 2 ∩ B 1 ′). The Matlab and R code below has been modified from Example 1.40 to count how often, out of 10,000 independent runs event A occurs.

A=0;

for i=1: u1=rand; u2=rand; if (u1<.6 && u2>=.7)|| (u1>=.6 && u2<.7) A=A+1; end end

A<-

for(i in 1:10000){ u1<-runif(1); u2<-runif(1) if((u1<.6 && u2>=.7)|| (u1>=.6 && u2<.7)){ A<-A+ } }

Executing the Matlab code returned A=4588, so

ˆ^4588

P A = = .4588.

The exact probability is P ( A ) = P ( B 1 ∩ B 2 ′) + P ( B 2 ∩ B 1 ′) = P ( B 1) – P ( B 1 ∩ B 2) + P ( B 2) – P ( B 1 ∩ B 2 ) = P ( B 1) + P ( B 2 ) – 2 P ( B 1 ∩ B 2) = P ( B 1) + P ( B 2 ) – 2 P ( B 1) P ( B 2) = .6 + .7 – 2(.6)(.7) = .46.

Note: In both programs, the code (u1<.6 && u2>=.7)||(u1>=.6 && u2<.7) can be replaced by a single “exclusive or” command: xor(u1<.6,u2<.7).

b. The estimated standard error of P A ˆ( ) is

ˆ ( )[1 ˆ( )] (.4588)(1 .4588)

P A P A

n

103. In the code below, seven random numbers are generated, one for each of the seven components. The sequence of and/or conjunctions matches the series and parallel ties in the system design.

A=0;

for i=1: u=rand(7,1); if (u(1)<.9 | u(2)<.9) & ((u(3)<.8 & u(4)<.8) | (u(5)<.8 & u(6)<.8)) & u(7)<. A=A+1; end end

A<-

for(i in 1:10000){ u<-runif(7) if((u[1]<.9 | u[2]<.9) & ((u[3]<.8 & u[4]<.8) | (u[5]<.8 & u[6]<.8)) & u[7]<.95){ A<-A+ } }

Executing the Matlab code gave A=8159, so

ˆ^8159

P A = = .8159.