2-1. Objective and Subjective

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 is the event that the baby is either a girl, or is over 5 pounds (of either sex). The intersection 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
1 / 2
3
4
5
6
2 / 3
4
5
6
3 / 4
5
6
4 / 5
6
5 / 6
6 / none

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

2-8. is the event that a randomly chosen person is exposed to the ad on the radio or the ad on television, or both. is the event that a randomly chosen person is exposed to the ad on the radio and the ad on television.

2-9. : purchase stock or bonds, or both. : 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. The third sample space is correct. It lists all possible outcomes in tossing a coin twice. The first two sample spaces do not contain all the possible outcomes of the experiment.

2-12.

a) 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.

2-13.

a) P(Ace of Spades) = 1 / 52

b) P(Friend’s card is Ace of Spades| that you have drawn a card first) = 1 / 52

c) 1 / 51

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 = 0.02 + 0.015 - 0 = 0.035

c. . 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. = 0.25 + 0.34 – 0.10 = 0.49

2-22. = 0.85 + 0.33 - 0.28 = 0.90

2-23. = 380/550 + 412/550 - 357/550 = 0.7909

2-24. 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

2-25. = (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 Given that P(R | B) = 0.85 and P(B) = 0.40,

= P(R | B) P(B) = (.85)(.40) = 0.34

2-28. P(M | R) = 0.32 Given that = 0.80 and P(R) = 0.4,

P(M | R) = P(M / R) P(R) = (0.80)(0.40) = 0.32

2-29. P(D | L) = 0.60 Given that = 0.12 and P(L) = 0.20,

P(D | L) = / 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(D) = (.25)(.10) = 0.025

2-31. a. P(M) = 198 / 1976 = 0.1002

b. P(E) = 408/1976 = 0.2065

c. Mutually exclusive events: = P(M) + P(S) = (198 + 968) / 1976 = 0.59

d. P(H) = 284/1976 = 0.144

e. P(S | H) = / P(H) = (128 / 1976) / (284 / 1976) = 0.451

f. P(P | E) = / P(E) = (233 / 1976) / (408 / 1976) = 0.571

g. P(W | O) = / P(O) = (99 / 1976) / (590 / 1976) = 0.168

h. = P(E) + P(O) – = (408 + 590 – 100) / 1976 = 0.454

i. = P(H) + P(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(H) = (.94)(.65) = 0.611

2-33. a. P(I) = 119 / 246 = 0.484

b. P(D) = 112/ 246 = 0.455

c. = 34 / 246 = 0.138

d. = 49/ 246 = 0.199

e. P(D | I) = / P(I) = 0.138 / 0.484 = 0.285

f. = = (85 / 246) / (134 / 246) = 0.634

g. = P(D) + P(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. = 3 / 10 = 0.30

c. = = (2 / 10) / (3 / 10) = 2/3 = 0.667

d. P(S | E) = = (1/10)/ (3/10) = 1/3 = 0.333

2-35. a. P(Male | 2003) = P(Male ∩ 2003) / P(2003) = 0.59 / (0.59 + 0.65) = 0.4758

b. P(Male | 2002) = P(Male ∩ 2002) / P(2002) = 0.47 / (0.47 + 0.55) = 0.4608

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 / 0.65 / Prob. of at least one success / 0.9947
2 / 0.65
3 / 0.65
4 / 0.65
5 / 0.65

2-37. P(at least one job) = 1 – P(no jobs) = 1 – (by independence) = 0.8143

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)

For any single cellphone user, the probability that he/she is a Verizon customer is (36 / 148.4) = 24.26%. The probability that the user is not a Verizon customer is 75.74%. The probability that none of the six cellphone users are Verizon customers would be (0.7574)6 = 0.1888. Therefore, the probability that at least one of the cellphone users is a Verizon customer is:

P(≥ 1 Verizon customer) = 1 – P(none are Verizon customers) = 1- 0.1888 = 0.8112

Probability of at least one success from many independent trials.
Success Probs
1 / 0.2426 / Prob. of at least one success / 0.8112
2 / 0.2426
3 / 0.2426
4 / 0.2426
5 / 0.2426
6 / 0.2426

2-43. Problem 2-43:

E1: The chosen box is A, P(A) = 0.50

E2: The chosen coin is a dime, P(dime)=0.167

E3: The outcome of the coin toss is Heads, P(H)=0.50

1. Are events E1 and E2 independent? No

P(E1)P(E2) = (0.50)(0.167) = 0.833

while: P(E1∩E2) = P(E1|E2)P(E2) = (1.0)(0.167) = 0.167

They are not equal.

2. Are events E1 and E3 independent? Yes

P(E1)P(E3) = (0.50)(0.5) = 0.25

while: P(E1∩E3) = P(E1|E3)P(E3) = (0.5)(0.5) = 0.25

They are equal.

3. Are events E2 and E3 independent? Yes

P(E2)P(E3) = (0.167)(0.5) = 0.0833

while: P(E2∩E3) = P(E2|E3)P(E3) = (0.167)(0.5) = 0.0833

They are equal.

2-44. P(H)P(M) = (284/1976)(198/1976) = 0.01440

= 29/1976 = 0.01468 ¹ 0.01440

The two events are not independent (but they are close)

2-45. P(D)P(I) = (112/246)(119/246) = 0.2202

= 34/246 = 0.1382 ¹ 0.2202

The two events are not independent.

2-46. = (3/10)(3/10) = 0.09

= 2/10 = 0.20 ¹ 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.

Probability of at least one success from many independent trials.
Success Probs
1 / 0.000096 / Prob. of at least one success / 0.0005
2 / 0.000096
3 / 0.000096
4 / 0.000096
5 / 0.000096

2-50. 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 / 0.5 / Prob. of at least one success / 0.7000
2 / 0.25
3 / 0.2

2-51. 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 / r / nPr
15 / 8 / 259459200

2-55. 6P3 = 6! / (6–3)! = (6)(5)(4) = 120 ordered choices

2-56. 7! / [(7–2)!2!] = 21 pairs.

Combination
n / r / nCr
7 / 2 / 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 = 0.30 P(R) = 0.70