# Solution Manual for Complete Business Statistics 7th Edition Aczel

August 24, 2017 | Author: hanand6 | Category: Median, Probability And Statistics, Statistics, Mathematics

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Solution Manual for Complete Business Statistics 7th Edition Aczel...

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CHAPTER 2

PROBABILITY 2-1.

Objective and subjective.

2-2.

An event is a set of basic outcomes of an experiment. The union of two events is the set containing all basic outcomes that are either in one event or in the other, or in both. The intersection of two events is the set of basic outcomes that are members of both events.

2-3.

The sample space is the universal set pertinent to a given experiment. It is the set of all possible outcomes of an experiment.

2-4.

The probability of an event is a measure of the likelihood of the occurrence of the event. When sample points are equally likely, the probability of the event is the relative size of the set comprising the event within the sample space.

2-5.

The union G  F is the event that the baby is either a girl, or is over 5 pounds (of either sex). The intersection G  F is the event that the baby is a girl over 5 pounds.

2-6.

The union is the event that the player scores in the game with A, or in the game with B, or in both. The intersection is the event that the player scores in both games.

2-7. Sample Space

first toss second toss > first

There are 36 possible outcomes tossing two dice. There are 15 possible outcomes where the second toss is greater than the first. P(Second Toss > First) = 15/36 = 0.417

First Toss Second Toss > First 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6 none

2-8.

R  T is the event that a randomly chosen person is exposed to the ad on the radio or the ad on television, or both. R  T is the event that a randomly chosen person is exposed to the ad on the radio and the ad on television.

2-9.

S  B : purchase stock or bonds, or both. S  B : purchase stock and bonds.

2-10.

Since there are only 37 possible outcomes, the probability of winning is 1/37 = 0.027, better odds (for the player) than the American version. Because of this, the house admission fee makes sense.

2.11.

P(Public Sector) + P(Corporate Sector) – P(both sectors) = 0.08 + 0.06 - 0.02 = 0.12

2-12.

P(nonfunctioning data device) = 5/18 = 0.2778

2.13.P(nonfunctioning cell phone) = 3/18 = 0.1667 P(nonfunctioning data device and nonfunctioning cell phone) = 1/18 = 0.0556 P(nonfunctioning data device or nonfunctioning cell phone) = 0.2778 + 0.1667 – 0.0556 = 0.3889 2-14.

Based on murder statistics for a given time period: total murders per 100,000 population.

2-15.

0.85, for example, is a typical "very likely" probability.

2-16.

More likely to occur than not to occur.

2-17.

The team is very likely to win.

2-18.

P(first shopper detected) + P(second detected) - P(both detected) = 0.98 + 0.94 - 0.93 = 0.99

2-19.

a. The two events are mutually exclusive. b. Let 0, D be the events: machine is out-of-control, down (respectively). Then we need P(O  D)  P(O)  P(D)  P(O  D) = 0.02 + 0.015 - 0 = 0.035 c.

P(D )  I  P(D)  I  0.015  0.985 . This event and D are mutually exclusive.

2-20.

P(F) + P(>50) – P(F & >50) = 12/20 + 2/20 – 2/20 = 0.6 P(< 30) = 2/20 = 0.1

2-21.

P (T  R )  P (T )  P ( R )  P (T  R ) = 0.25 + 0.34 – 0.10 = 0.49

P( S  B )  P( S )  P( B)  P( S  B ) = 0.85 + 0.33 - 0.28 = 0.90

2-23.

P(VT  CE) = 380/550 + 412/550 - 357/550 = 0.7909

2.24.

From (2-12): P(nonfunctioning data device) = 0.2778

P(functioning data device) = 1 – P(nonfunctioning data device) = 1 – 0.2778 = 0.7222 2-25.

P(M  H)  P(M)  P(H)  P(M  H) = (11 + 8 – 5) / 28 = 1 / 2 = 0.50

2-26.

25%: Reading the market shares as 2 %, 2%, and 4%, the answer is 2(2+2+4) = 25%

2-27.

0.34

2-28.

P(M | R) = 0.32 Given that P(M  R) = 0.80 and P(R) = 0.4,

Given that P(R | B) = 0.85 and P(B) = 0.40, P(R  B) = P(R | B) P(B) = (.85)(.40) = 0.34

P(M | R) = P(M / R) P(R) = (0.80)(0.40) = 0.32 2-29.

P(D | L) = 0.60

Given that P(L  D) = 0.12 and P(L) = 0.20,

P(D | L) = P(L  D) / P(L) = .12 / .20 = 0.60 2-30.

2.5% of the packages are late. Given that P(N | D) = 0.25 and P(D) = 0.10, P(N  D) = P(N | D) P(D) = (.25)(.10) = 0.025

2-31.

a. b. c. d. e.

P(M) = 198 / 1976 = 0.1002 P(E) = 408/1976 = 0.2065 Mutually exclusive events: P(M  S) = P(M) + P(S) = (198 + 968) / 1976 = 0.59 P(H) = 284/1976 = 0.144 P(S | H) = P(S  H) / P(H) = (128 / 1976) / (284 / 1976) = 0.451 f. P(P | E) = P(P  E) / P(E) = (233 / 1976) / (408 / 1976) = 0.571 g. P(W | O) = P(W  O) / P(O) = (99 / 1976) / (590 / 1976) = 0.168 h. P(E  O) = P(E) + P(O) – P(E  O) = (408 + 590 – 100) / 1976 = 0.454 i. P(H  S) = P(H) + P(S) – P ( H  S ) = (284 + 968 – 128) / 1976 = 0.569

2-32.

61.1% successfully completed. Given that P(A | H) = 0.94 and P(H) = 0.65, P(A  H) = P(A | H) P(H) = (.94)(.65) = 0.611

a. P(I) = 119 / 246 = 0.484 b. P(D) = 112/ 246 = 0.455 c. P(I  D) = 34 / 246 = 0.138 d. P(I  D) = 49/ 246 = 0.199 e. P(D | I) = P(D  I) / P(I) = 0.138 / 0.484 = 0.285 f. P(I | D) = P(I  D)/P(D) = (85 / 246) / (134 / 246) = 0.634 g. P(D  I) = P(D) + P(I) – P ( D  I ) = 0.455 + 0.484 – 0.138 = 0.801

2-34.

Let E,S denote the events: top Executive made over \$1M, Shareholders made money, respectively. Then: a. P(E) = 3 / 10 = 0.30 b. P( S ) = 3 / 10 = 0.30 c. P(E | S ) = P(I  S )/P( S ) = (2 / 10) / (3 / 10) = 2/3 = 0.667 d. P(S | E) = P(S  E)/P(E) = (1/10)/ (3/10) = 1/3 = 0.333

2.35.

P(Critical Danger | Dependence on Forested Areas) = P(CD ∩ DFA) / P(DFA) = 0.30 / 0.90 = 0.333

2-36.

(Use template: Probability of at least 1.xls) For any single person, the probability that the person is obese is 65% and the probability that the person is not obese is 35%. The probability that none of the five people chosen is obese would be (0.35)5 = 0.00525. The probability that at least one of the people chosen is overweight and obese is: P(≥ 1 overweight & obese) = 1 – P(none are overweight and obese) = 1 – 0.00525 = 0.99475. Probability of at least one success from many independent trials. Success Probs 1 2 3 4 5

2-37.

0.65 0.65 0.65 0.65 0.65

Prob. of at least one success

0.9947

  2   2   2   5   5   19   19   P(at least one job) = 1 – P(no jobs) = 1 –                (by independence) =   3   3   3   6   6   20   20   0.8143

download full file at http://testbankcafe.com (Use template: Probability of at least 1.xls) Probability of at least one success from many independent trials. Success Probs 1 2 3 4 5 6 7

0.333 0.333 0.333 0.167 0.167 0.05 0.05

Prob. of at least one success

0.8142

2-38.

Assume independence: P(at least one arrives on time) = 1 – P(all three fail to arrive) = 1 – (1 – .90)(1 – .88)(1 – .91) = 0.99892

2-39.

P(increase sales all three countries) = P(increase sales in US) x P(increase sales Australia) x P(increase sales in Japan) P(increase sales all three countries) = (0.95)(0.90)(0.85) = 0.72675

2-40.

P(device works satisfactorily) = 1 – P(both components fail) = 1 – (0.02)(0.1) = 1 – 0.002 = 0.998

2-41.

P(at least one benefit) = 1 – P(no benefits) = 1 – (1 – 0.93)(1 – 0.55)(1 – 0.70) = 1 - 0.00945 = 0.99055

2-42.

(Use template: Probability of at least 1.xls) Probability of at least one success from many independent trials. Success Probs 1 2 3

0.9 0.75 0.6

Prob. of at least one success

0.9900

(Use template: Probability of at least 1.xls) Probability of at least one success from many independent trials. Success Probs 1 2 3 4

2-44.

0.7 0.82 0.8 0.9

Prob. of at least one success

0.9989

P(H)P(M) = (284/1976)(198/1976) = 0.01440 P(H  M) = 29/1976 = 0.01468  0.01440 The two events are not independent (but they are close)

2-45.

P(D)P(I) = (112/246)(119/246) = 0.2202 P(D  I) = 34/246 = 0.1382  0.2202 The two events are not independent.

2-46.

P(E)P( S ) = (3/10)(3/10) = 0.09 P ( E  S ) = 2/10 = 0.20  0.09

The two events are not independent. There may be some sort of relationship between them as a general rule, if seen in all firms. 2-47.

P(getting at least one disease) = 1 – P(getting none of the three) = 1 – P(not getting Mal.)P(not getting schist.)P(not getting s.s.) = 1 – (1 – P(getting mal.))(1 – P(getting schist.))(1 – P(getting s.s.)) = 1 – (1 – (110/2100))(1 – (200/600))(1 – (.025/50)) = 1 – (0.9476)(0.6667)(0.9995) = 0.3686

2-48.

The device works if at least one out of three works. P(device works) = 1 – P(all components fail) = 1 – (1 – 0.96)(1 – 0.91)(1 – 0.80) = 0.99928

2-49.

(Use template: Probability of at least 1.xls) The probability of dying in a car crash in France in 2003 is 5732 / 59625919 = 0.000096. The probability on not dying in a car crash is 0.999904. The probability of not dying in a car crash for the next five years, assuming conditions stay the same as in 2003, would be (0.999904)5 = 0.99952. Therefore, the probability of dying in a car crash over the next five years is: 1 – P(not dying) = 1 – 0.99952 = 0.00048.

download full file at http://testbankcafe.com Probability of at least one success from many independent trials. Success Probs 1 2 3 4 5

2-50.

0.000096 0.000096 0.000096 0.000096 0.000096

Prob. of at least one success

0.0005

P(at least one drives home safely) = 1 – P(none drive home safely) = 1 – (0.50)(0.75)(0.80) = 1 – 0.30 = 0.70 Success Probs 1 2 3

2-51.

0.5 0.25 0.2

Prob. of at least one success

0.7000

Since 1/4 of the items are in any particular quartile, and assuming independent random samples

with replacement so that all four choices have the same probability of being in the top quartile, P(all four in top quartile) = (1/4)4 = 1/256 = 0.0039 P(at least one from bottom quartile) = 1 – P(all four from top three quartiles) = 1 – (3/4)4 = 175/256 = 0.684 2-52.

(55)(30)(21)(13) = 450,450 sets of representatives

2-53.

9! = (9)(8)(7)(6)(5)(4)(3)(2)(1) = 362,880 different orders.

2-54.

nPr = n! / (n–r)! = 15! / (15-8)! = (15)(14)(13)(12)(11)(10)(9)(8) = 259,459,200 Permutation n 15

2-55.

r 8

nPr 259459200

6P3 = 6! / (6–3)! = (6)(5)(4) = 120 ordered choices

7! / [(7–2)!2!] = 21 pairs. Combination n r 7 2

nCr 21

2.57.

Only one possible combination of 3 elements chosen from the 14 parts consists of the 3 faulty ones. So since any 3-element combination is equally likely to be picked, the probability = 1 / (14! / 3!11!) = 1 / 364 = 0.00275

2-58.

Only one of the combinations wins, so the probability of guessing it is 1 / (36! / 6!30!) = 1 / 1,947,792 = 0.000000513

2-59.

How many ways of guessing a set of 6 of the numbers from 1 to 36 will have 5 correct and 1 wrong? If W = {WI, W2, - - -, W6} is the winning combination, then there are 6 choices of which wi is not in the guessed combination, and 30 possible wrong guesses in place of wi (since the one wrong guess can be any of the numbers from 1 to 36 that are not in W). So (6) (30) = 180 possible combinations match exactly 5 of the winning numbers. Thus the probability of making such a guess is = 180/1,947,792 = 0.0000924

2-60.

Let T,R be the events: successful takeover, resignation of a board member. P(T | R) = 0.65 P(T | R) = 0.30

P(R) = 0.70

P(T) = P(T | R)P(R) + P(T | R)P(R) = (.65)(.7) + (.30)(.30) = 0.545 2-61.

Let A,S be the events: the drug is approved, the drug has side effects. P(A | S) = 0.5 P(A | S ) = 0.95

P(S) = 0.20 P( S ) = 0.8

P(A) = P(A | S)P(S) + P(A | S )P( S ) = (.5)(.2) + (.95)(.80) = 0.86 2-62.

Let D,B be the events: deal is concluded, competitor makes a bid. P(D | B) = 0.25

P ( D | B ) = 0.45

P(B) = 0.40 P(B ) = 0.6

P(D) = P(D | B)P(B) + P(D | B)P( B) = (.25)(.4) + (.45)(.6) = 0.37 2-63.

Let S,E be the events: property is sold, economy improves. P(S | E) = 0.9

P(S | E) = 0.50

P(E) = 0.70

P(S) = P(S | E)P(E) + P(S | E)P(E) = (.9)(.7) + (.50)(.30) = 0.78

Let F,A be the events: ships sail full this summer, dollar appreciates against European currencies. P(F | A) = 0.75 P ( F | A = 0.92 P(A) = 0.23 P(F) = P(F | A)P(A) + P(F | A)P(A) = (.75)(.23) + (.92)(.77) = 0.8809

2-65.

Let O,G be the events: door should open, green light appears. P(O) = 0.9 P(G | O) = 0.98

P(G | O ) = 0.05

P(G | O) P(O)

(.98)(.9)

P(O | G) =

= P(G | O) P(O) + P(G | O )P( O )

(.98)(.9) + (.05)(.1)

= 0.9944 2-66.

Let S,E be the events: the alarm sounds, there is an emergency situation. P(E) = 0.004

P(S | E) = 0.95 P(S | E) = 0.02 P(S | E) P(E)

P(E | S) =

(.95)(.004) =

P(S | E) P(E) + P(S | E)P(E)

(.95)(.004) + (.02)(.996)

= 0.1602 2-67.

Let I,H,M,L be the events: indicator rises, economic situation is high, medium, low. P(I | H) = 0.6 P(I | M) = 0.3 P(I | L) = 0.1 P(H) = 0.15 P(M) = 0.7 P(L) = 0.15 P(I | H) P(H)

(.6)(.15)

P(H | I) =

= P(I | H) P(H) + P(I |M) P(M) + P(I | L) P(L)

(.6)(.15)+ (.3)(.7)+ (.1)(.15)

= 0.2857 2-68.

Let I,O be the events: test indicates oil, oil really is present P(O) = 0.4

P(I | O) = 0.85

P(I | O) = 0.10

P(I | O) P(O) P(O | I) =

(.85)(.4) =

P(I | O) P(O) + P(I | O )P( O )

(.85)(.4) + (.10)(.6)

= 0.85

Let I,S be the events: test indicates success, product really is successful. P(S) = 0.6

P(I | S) = 0.75

P(I | S ) = 0.15

P(I | S) P(S)

(.75)(.6)

P(S | I) =

= P(I | S) P(S) + P(I | S )P( S )

(.75)(.6) + (.15)(.4)

= 0.8824 2.70.

Let MM, FF, MF denote the events of reaching the two men, the two women, the married couple (respectively). Let W be the event that a woman answers the door. Then: P(MF  W) = 1/6 P(MF) = 1/3 P(W | MF) = ½ P(MM) = 1/3

P(W | MM) = 0

P(MM  W) = 0

P(FF) = 1/3

P(W | FF) = 1

P(FF  W) = 1/3

P(W) = ½

P(MF | W) =  P(MF  W)/P(W) = [ (1/6) / (1/2) ] = 1/3 2-71.

P(  1 out of 5 misdirected) = 1 – P (none misdirected) = 1 – [P(a single call not misdirected)]5 = 1 – (199/200)5 = 0.0248

2-72.

(0.02)(1/200) = 1/10,000 = 0.0001

2-73.

P(A  B) = 0.4 + 0.3 – 0.1 = 0.6

2.74.

Assume independence in each manager’s decision. a) P(none leave) = (1 – 0.5)25 = 2.9 x 10-8 = approximately 0.000 b) P(all leave) = (0.5)25 = 2.9 x 10-8 = approximately 0.000 c) P(at least one leaves) = 1 – (2.9 x 10-8 )= 0.999 = approximately 1.0

2-75.

P(C  S) = 0.15 + 0.10 – 0.05 = 0.20

2-76.

Let C, J be the events: pass CPA exam, get job offer. P(C) = 0.6 P(C  J) = 0.4 P(J | C) = P(C  J) / P(C) = 0.4/0.6 = 0.667

2-77.

P(A) = 0.20

P(A  B) = 0.12

P(B | A) = P(A  B) / P(A) = 0.12/0.20 = 0.60

Let P, I be the events: production increases, interest rates decline more than half a point. P(P | I) = 0.72 P(I) = 0.25 P(P  I) = P(P | I ) P(I) = (.72)(.25) = 0.18

2-79.

Let EQ, SD be the events: engineering quality, sporty design rated among most important features. P ( EQ  SD ) = 0.25 P(EQ) = 0.35 P(SD) = 0.50 (.35)(.50) = .175  .25, therefore the two events are not independent

2-80.

The assumption of independence is justified by random sampling. P(all 3 people consider EQ) = (.35)3 = 0.0429 P(at least one person considers EQ) = 1 – P(none of the three consider EQ) = 1 – (.65)3 = 0.7254

2-81.

P(exposed to at least one mode of advertising) = 1 – P (not exposed to any of the 3) = 1 – (.90)(.85)(.80) = 0.388 Success Probs 1 2 3

0.1 0.15 0.2

Prob. of at least one success

0.3880

2-82.

Let S, R be the events: see the ad in the Wall Street Journal, remember it. P(S) = 0.6 P(R | S) = 0.85 P ( S  R ) = P(R | S) P(S) = (.85)(.6) = 0.51 51% of the people see and remember the advertisement

2-83.

P(defective) = 0.1 5 chips are chosen at random. P(none of the 5 are defective) = (0.90)5 = 0.59049 P(at least one is defective) = 1 – P(none are defective) = 0.40951

2-84.

P(at least one color) = 1 – P(none of the colors) By independence: = 1 – (1– 0.3)(1 – 0.2)(1 – 0.15) = 0.524

0.3 0.2 0.15

Prob. of at least one success

0.5240

2-85.

(2/3)5 = 0.132

2-86.

Let S, F, N, U be the events: subsidiary will be successful, political situation is favorable, neutral, unfavorable. P(S | F) = 0.55 P(S / N) = 0.3 P(S / U) = 0.1 P(F) = 0.6 P(N) = 0.2 P(U) = 0.2 P(S) = P(S | F) P(F) + P(S | N) P(N) + P(S | U)P(U) = (0.55)(0.6) + (0.3)(0.2) + (0.1)(0.2) = 0.41

2-87.

Let L, A be the events: legislation is passed, authorization is granted. P(L  A) = 0.5 P(L) = 0.75 P(A | L) = P(L  A) / P(L) = 0.5/0.75 = 0.6667

2-88.

Let D, H be the events: customer defaults, economy is high. P(D | H ) = 0.13 P(D | H) = 0.04 P(H) = 0.65

The choice is probably not random.

P(D) = P(D | H) P(H) + P(D | H)P(H) = (.04)(.65) + (.13)(.35) = 0.0715 2-89. a) P(assembly line worker) = P(skilled worker on assembly line) + P(unskilled worker on assembly line) = (0.30)(0.15) + (0.70)(0.30) = 0.045 + 0.21 = 0.255 b) P(unskilled worker | assembly line worker) = 0.21 / 0.255 = 0.8235

P(at least one booking) = 1 – P(none) = 1 – (0.92)20 = 0.8113 Success Probs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08

Prob. of at least one success

0.8113

2-91.

P(none are in error) = (0.95)10 = 0.5987

2-92.

Let A, F, So, J, Se be the events: student got an A, student is a freshman, sophomore, junior, senior P(Se | A) P(A | Se) P(Se) = P(A | Se) P(Se) + P(A |F) P(F) + P(A | So) P(So) + P(A | J)P(J) (.40)(.15) =

= 0.2034 (.40)(.15) + (.20)(.30) + (.30)(.35) + (.35)(.20)

2-93.

Let S, C be the events: product is successful, competitor produces similar product. P(S | C) = 0.67 P(S | C) = 0.42 P(C) = 0.35 P(S) = (.67)(.65) + (.42)(.35) = 0.5825

Assume independence. P(at least 1 orders drink) = 1 – P(none order drink) = 1 – (0.92)13 = 0.6617 Probability of at least one success from many independent trials. Success Probs 1 2 3 4 5 6 7 8 9 10 11 12 13

0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08

Prob. of at least one success

0.6617

2-95.

Conceptually, in the hole-card game there is a card missing from the space of cards available for play by the players-the hole card itself, which is removed from play at the beginning. So, for example, if a player wants to make a draw and needs anything lower than a 4 in order not to go over 21, she can't be sure of the probability that the next card is in fact smaller than a 4 (given the cards already drawn and known to everyone), since the hole card may or may not be one of the remaining cards smaller than 4. In other words, the probability space represented by the deck (in a randomized order) used for the main play isn't exactly specified in the hole-card game (from any player's point of view), while this space is always based on the full 52-card deck in the regular game. Practically speaking, however, the probabilities involved vary only slightly, and their role in the outcome of the game should be more or less unnoticeable when averaged over a span of games.

2-96.

The probability of dying in a car crash in the U.S. is 40676 / 280000000 = 0.000145. The probability of dying in a car crash in France is 0.000096 (from problem 2-49). The rate in the U.S. is 1.5 times larger than the rate in France. The probability of not dying in a car crash in the U.S. is 0.999855. The probability of dying in a car crash in the U.S. over the next 20 years is: 1- P(not dying) = 1 – (0.999855)20 = 0.002896.

2-97.

Assume a large population so that the sampling can be considered as being done with replacement (i.e., removing one item does not appreciably alter the remaining number of data points on either side of the median). Then the first item drawn is on a particular side of the median, after which the second item has a 1/2 probability of coming from the other side, since half of the points in the population are on one side and half are on the other by the definition of median. Thus the probability that the median will lie between the two points drawn is 1/2.

Making the same assumptions as in Problem 2-97, first count the number of ways that all n elements drawn could lie on the same side of the median: since the choices are independent (this being a random sample) and each of the n – 1 choices after the first has probability 1/2 of being on the same side of the median as the first choice, the number of such ways is (1/2) n-1. The median will lie somewhere between the smallest and largest values drawn exactly when the above situation does not occur, so the desired probability is 1 – (1/2).

2-99.

P (at least one paper accepted) = 1 – P (no papers accepted) = 1 – P(first paper rejected) P(second paper rejected) ... = 1 – [1 – .14][1 – (.14)(.9)][1 – (.14)(.9)2] [1 – (.14)(.9)3] ... = 1 – (.86)(.874)(.8866)(.89794)(.908146) ... Using 800 iterations on a computer: = 0.766660928

2-100. We need a method that is fair even if the caller knows which way the coin is biased (since otherwise a single random call would still have probability 1/2 of matching the actual outcome). Here it is: the caller pre-selects one of H-T or T-H as the guessed sequence of outcomes of two consecutive flips. The coin is then flipped twice: if it results in two different outcomes, the caller attends the meeting iff his or her choice of the order of outcomes was correct. But if the two flips have the same outcome, then the coin is flipped twice again, repeated as needed until a pair of different outcomes is obtained. Note that even if the coin is much more likely to come up heads (and even if the caller knows this), the sequence H-T is equally likely as T-H: they both have probability P(H) P(T) , since the two tosses are independent. 2-101. open close total

Kwik Save 893 107 1000

Somerfieldtotal 0 424 424

893 531 1424

a. 107 / 531 = .202 b. 531 / 1424 + 1000 / 1424 – 107 / 1424 = 1.00 c. 0 2-102. 0.089 = 1700 /19100 2-103. Since there are 38 members, a simple majority would be 20 votes. P(yes) = 0.25

P(pass) = 0.0002 P(Northwest compete successfully) = 0.00014 2-104. S = Species Survives G = Project goes ahead Given Probabilities: P(–S | G) = 0.6 P(G) = 0.7 P(S) = P(S | G) P(G) + P(S | –G) P(–G) = (0.4)(0.7) + (1.0)(0.3) = 0.58

P(S | –G) = 1.0

2.105. (Use template: Bayesian Revision.xls) Bayesian Revision based on Empirical Conditional Probabilities Machine A Machine B Machine C s1 s2 s3 s4s5s6 Prior Probability 0.26 0.38 0.36 Conditional Probabilities s1 defective P(I1 | .) 0.08 good P(I2 | .) 0.92 P(I3 | .) P(I4 | .) P(I5 | .) Total 1

s7

s8

s4s5s6

s7

s8

0 0 0

0

0

s2 0.0190 0.3610

s3 s4s5s6 0.0144 0.3456

s7

s8

s2 0.3506 0.3817

s3 s4s5s6 0.2657 0.3654

s7

s8

s2 0.05 0.95

s3

1

1

Total 1

0.04 0.96

Joint Probabilities s1 I1 0.0208 I2 0.2392 I3 I4 I5

Marginal 0.0542 0.9458

Posterior Probabilities s1 P(. | I1) 0.3838 P(. | I2) 0.2529 P(. | I3) P(. | I4) P(. | I5)

1) Given the part is defective, the probability it is from machine A is 0.3838. 2) Given the part is good, the probability that it is from machine B is 0.3817. Case 2: Job Applications 1) P(at least one offer) = 1 – P(no offers) = 1 – (0.62) (0.65) (0.72) (0.80) (0.82) (0.82) (0.83) (0.86) (0.86) (0.92) = 1 – 0.0881 = 0.9119 2) No. It is better to apply to company 6, since it has the same probability of success, but is cheaper. 3) \$1430. P(at least one offer) = 1 – P(no offers from 2,5,8,9) = 1 – (0.65)(0.82)(0.86)(0.86) = 1 – (0.3942) = 0.6058

download full file at http://testbankcafe.com 4) Through trial and error we arrive at a total cost of \$2190 with a 0.7644 probability of getting an offer from either company 2,3,6,7,8 or 9. 5) Through trial and error we arrive at a total cost of \$1500 with a 0.6329 probability of getting an offer from either company 2,7,8, 9 or 10.