XS="100%" YS="100%"/> D <ART W="50.27pt" H="25.672pt" XS="100%" YS="100%"/> F <ART W="50.27pt" H="24.387pt" XS="100%" YS="100%"/> 1.0 Thus, the probability of a student receiving a C in the course is 0.30 = 30%. The probability of a student receiving a C may also be calculated using the following equation: <ART FILE="02_14eq06.eps" H="22.047pt" XS="100%" YS="100%"/> <ART FILE="02_14eq07.eps" W="44.51pt" H="20.802pt" XS="100%" YS="100%"/>= 0.30 2-15. a. P(H) = 1/2 = 0.5 b. P(T  H) = P(T) = 0.5 c. P(TT) = P(T) P(T) = (0.5)(0.5) = 0.25 d. P(TH) = P(T) P(H) = (0.5)(0.5) = 0.25 e. We first calculate P(TH) = 0.25, then calculate P(HT) = (0.5)(0.5) = 0.25. To find the probability of either one occurring, we simply add the two probabilities. The solution is 0.50. f. At least one head means that we have either HT, TH, or HH. Since each of these have a probability of 0.25, their total probability of occurring is 0.75. On the other hand, the complement of the outcome “at least one head” is “two tails.” Thus, we could have also computed the probability from 1 – P(TT) = 1 – 0.25 = 0.75. 2-16. The distribution of chips is as follows: Red 8 Green 10 White 22 = 20 a. The probability of drawing a white chip on the first draw is
1. [P(MB over K) and P(M over MB)], or 2. [P(MB over M) and P(K over MB)] P(1) = (0.6)(0.2) = 0.12 P(2) = (0.8)(0.4) = 0.32 Probability = P(1) + P(2) = 0.12 + 0.32 = 0.44 d. Probability = 1 – winning every game = 1 – answer to part (a) = 1 – 0.08 = 0.92 2-25. a. Probability = P(K over M) = 0.2. b. Probability = P(K over MB) = 0.4. c. Probability = [P(K over M) and P(MB over K)] or [P(K over MB) and P(M over K)] = (0.2)(0.6) + (0.4)(0.8) = 0.12 + 0.32 = 0.44 d. Probability = [P(K over MB) and P(K over M)] = (0.4)(0.2) = 0.08 e. Probability = P(MB over K) and P(M over K) = (0.6)(0.8) = 0.48 f. No. They do not appear to be a very good team. 2-26. The probability of Dick hitting the bull’s-eye: P(D) = 0.90 The probability of Sally hitting the bull’s-eye: P(S) = 0.95 a. The probability of either Dick or Sally hitting the bull’s-eye: P(D or S) = P(D) + P(S) – P(D)P(S) = 0.90 + 0.95 – (0.90)(0.95)
= 0.995 b. P(D and S) = P(D)P(S) = (0.9)(0.95) = 0.855 c. It was assumed that the events are independent. This assumption seems to be justified. Dick’s performance shouldn’t influence Sally’s performance. 2-27. In the sample of 1,000 people, 650 people were from Laketown and 350 from . Thirteen of those with cancer were from Laketown. Six of those with cancer were from . a. The probability of a person from Laketown having cancer: <ART FILE="02_27eq02.eps" W="19.157pt" H="20.38pt" XS="100%" YS="100%"/> = 0.020 The probability of a person having cancer: <ART FILE="02_27eq03.eps" W="71pt" H="18.953pt" XS="100%" YS="100%"/> = 0.019 Not independent. b. I would rather live in . 2-28. <ART FILE="02_28eq01.eps" W="149.625pt" H="22.984pt" XS="100%" YS="100%"/> where A = fair die = F = unfair die = L B = getting a 3 <ART FILE="02_28eq04.eps" W="47.664pt" H="19.522pt" XS="100%" YS="100%"/> = 0.22 Therefore, P(L) = 1 – 0.22 = 0.78 2-29. Parts (a) and (c) are probability distributions because the probability values for each event are between 0 and 1, and the sum of the probability values for the events is 1. 2-30. X P(X) X · P(X) 0 0.05 0.00 1 0.15 0.15
2-48. Let W = candidate wins the election L = candidate loses the election PW = poll predicts win PL = poll predicts loss P(W) = 0.50 P(L) = 0.50 P(PW | W) = 0.80 P(PL | W) = 0.20 P(PW | L) = 0.10 P(PL | L) = 0.90 2-49. Let S = successful restaurant U = unsuccessful restaurant PS = model predicts successful restaurant PU = predicts unsuccessful restaurant P(S) = 0.70 P(U) = 0.30 P(PS | S) = 0.90 P(PU | S) = 0.10 P(PS | U) = 0.20 P(PU | U) = 0.80 2-50. Let D = Default on loan; D' = No default; = Loan rejected; R = Loan approved Given: P(D) = 0.2 P(D') = 0.8 P(R | D) = 0.9 P(R' | D') = 0.7 (a) P(R | D') = 1 – 0.7 = 0.3 (b) <ART FILE="DM.02.001.eps" W="160.695pt" H="52.459pt" XS="100%" YS="100%"/> 2-51. (a) F0.05, 5, 10 = 3.33 (b) F0.05, 8.7 = 3.73 (c) F0.05, 3, 5 = 5.41 (d) F0.05, 10. 4 = 5.96 2-52. (a) F0.01, 15, 6 = 7.56
Using the formulas for the binomial: E(X) = np = (5)(.2) = 1.0 2 = np(1 – p) = 5(.2)(.8) = 0.80 The equation produced equivalent results. 2-65. a. n = 10; p = .25; q = .75; <ART FILE="02_57eq01.eps" W="51.378pt" H="19.375pt" XS="100%" YS="100%"/> P(X) X <ART FILE="02_57eq02.eps" W="83.227pt" H="19.375pt" XS="100%" YS="100%"/> .0563 0 <ART FILE="02_57eq03.eps" W="79.766pt" H="19.375pt" XS="100%" YS="100%"/> .1877 1 <ART FILE="02_57eq04.eps" W="83.604pt" H="19.375pt" XS="100%" YS="100%"/> .2816 2 <ART FILE="02_57eq05.eps" W="84.195pt" H="19.375pt" XS="100%" YS="100%"/> .2503 3 <ART FILE="02_57eq06.eps" W="84.195pt" H="19.375pt" .1460 4
XS="100%" YS="100%"/> <ART FILE="02_57eq07.eps" W="83.9pt" H="19.375pt" XS="100%" YS="100%"/> .0584 5 .0162 6 <ART FILE="02_57eq09.eps" W="84.49pt" H="19.375pt" XS="100%" YS="100%"/> .0031 7 <ART FILE="02_57eq10.eps" W="84.195pt" H="19.375pt" XS="100%" YS="100%"/> .0004 8 <ART FILE="02_57eq11.eps" H="19.375pt" XS="100%" YS="100%"/> .00003 9 <ART FILE="02_57eq12.eps" H="19.375pt" XS="100%" YS="100%"/> .0000 10 b. E(X) = (10).25 = 2.5 2 = npq = (10)(.25)(.75) = 1.875 c. Expected weekly profit: $125. 12.5