= 0.995 b. and = P(D 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. In the sample of 1,000 people, 650 people were from Laketown and 350 from . Thirteen of 2-27. 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 = fair die = A F = unfair die = L = getting a 3 B <ART FILE="02_28eq04.eps" W="47.664pt" H="19.522pt" XS="100%" YS="100%"/> = 0.22 Therefore, = 1 0.22 = 0.78 P(L) – 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
= 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
Solution-Manual-for-Quantitative-Analysis-for-Management-11E-11th-Edition.docx