a) PROBABILITY
2-12. P(exploring the products) = 1790 / 2385 = 0.7505 b) P(purchase) = 387 / 2385 = 0.1623 c) P(purchase | explored the products) = 387 / 1790 = 0.2162 d) (a) approximately 75 percent of the visitors go beyond the homepage. Based on murder statistics for a given time period: total murders per 100,000 population. More likely to occur than not to occur. P(first shopper detected) + P(second detected) - P(both detected) = 0.98 + 0.94 - 0.93 = 0.99 P(F) + P(>50) P(F & >50) = 12/20 + 2/20 2/20 = 0.6 P(< 30) = 2/20 = 0.1
CHAPTER 2
2-14. 2-16. 2-18. 2-20. 2-22. 2-24.
P ( S B ) = P ( S ) + P ( B ) - P ( S B ) = 0.85 + 0.33 - 0.28 = 0.90
a. 1268 / 2074, found by: 597 / 2074 + 962 / 2074 291 / 2074 = 1268 / 2074 b. 230 / 2074 c. 419 / 2074, found by: 230 / 2074 + 189 / 2074 = 419 / 2074 d. 306 / 1112 e. 189 / 427 f. 773 / 2074, found by: 310 / 2074 + 291 / 2074 + 172 / 2074 g. 773 / 1647, found by: (310 + 291 + 172) / (648 + 597 + 402) = 773 / 1647 30%: The three foreign banks make up approximately 10% of the market and SBC Warburg has approximately a 3% share. 3 / 10 = .30 P(M | R) = 0.32 Given that P(M R) = 0.80 and P(R) = 0.4, P(M | R) = P(M / R) P(R) = (0.80)(0.40) = 0.32 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 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 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
2-26.
2-28. 2-30.
2-32. 2-34.
2-36.
For any single issue, the probability that it is foreign underwritten is 10% and the probability that it is not foreign underwritten is 90%. The probability that none of the six issues is foreign underwritten would be (.90)6 = .5314. The probability that at least one is foreign underwritten is: P( 1 foreign underwritten) = 1 P(none are foreign underwritten) = 1 (.90)6 = .4686
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2-38.
2-40. 2-42. 2-44.
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 P(device works satisfactorily) = 1 P(both components fail) = 1 (0.02)(0.1) = 1 0.002 = 0.998 P(reads at least one paper) = 0.31 + 0.20 0.09 = 0.42 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) P(E)P( S ) = (3/10)(3/10) = 0.09
2-46.
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. 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 (55)(30)(21)(13) = 450,450 sets of representatives nPr = n! / (nr)! = 15! / (15-8)! = (15)(14)(13)(12)(11)(10)(9)(8) = 259,459,200 7! / [(72)!2!] = 21 pairs. Only one of the combinations wins, so the probability of guessing it is 1 / (36! / 6!30!) = 1 / 1,947,792 = 0.000000513 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 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 2-62. Let D,B be the events: deal is concluded, competitor makes a bid. P(D | B) = 0.25
2-48.
2-52. 2-54. 2-56. 2-58. 2-59.
2-60.
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 2-64. Let F,A be the events: ships sail full this summer, dollar appreciates against European currencies. P(F | A) = 0.75 2-66.
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 Let S,E be the events: the alarm sounds, there is an emergency situation.
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P(E) = 0.004
P(S | E) = 0.95 P(S | E) = 0.02 P(S | E) P(E) (.95)(.004) = P(S | E) P(E) + P(S | E)P(E) (.95)(.004) + (.02)(.996)
P(E | S) = = 0.1602 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 (.85)(.4) = P(I | O) P(O) + P(I | O)P(O ) (.85)(.4) + (.10)(.6) P(I | O) P(O) P(O | I) =
= 0.85 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) = P(MM) = 1/3 P(FF) = 1/3 P(W | MM) = 0 P(W | FF) = 1 P(MM W) = 0 P(FF W) = 1/3
P(MF | W) = [ P(MF W)/P(W)] = [ (1/6) / (1/2) ] = 1/3 2-74. 2-76. a. P(probability all work in retailing) = 0.12 x 0.12 x 0.09 x 0.09 = 0.000117 b. P(at least one works in retail) = 1 (.88)(.88)(.91)(.91) = 0.359 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 2-78. 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 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 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 Let D, H be the events: customer defaults, economy is high.
P(W) =
2-84. 2-86.
2-88.
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P(D | H) = 0.04
P(D | H) = 0.13
P(H) = 0.65
P(D) = P(D | H) P(H) + P(D | H)P(H) = (.04)(.65) + (.13)(.35) = 0.0715 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) = 2-94. = 0.2034 (.40)(.15) + (.20)(.30) + (.30)(.35) + (.35)(.20) P (at least one error) = 1 P(all 1000 entries correct) = 1 (0.9992)1000 = 0.5508
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. 2-104. 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
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