1. [P(MB over and over or K) P(M MB)], 2. [P(MB over and over M) P(K MB)] = (0.6)(0.2) = 0.12 P(1) = (0.8)(0.4) = 0.32 P(2) 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 = over = 0.2. P(K M) b. Probability = over = 0.4. P(K MB) c. Probability = [P(K over and over or M) P(MB K)] [P(K over and over MB) P(M K)] = (0.2)(0.6) + (0.4)(0.8) = 0.12 + 0.32 = 0.44 d. Probability = [P(K over and over MB) P(K M)] = (0.4)(0.2) = 0.08 e. Probability = over and over P(MB K) P(M 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: = 0.90 P(D) The probability of Sally hitting the bull’s-eye: = 0.95 P(S) a. The probability of either Dick or Sally hitting the bull’s-eye: or = + P(D S) P(D) P(S) P(D)P(S) – = 0.90 + 0.95 (0.90)(0.95) –
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)
Solution-Manual-for-Quantitative-Analysis-for-Management-11E-11th-Edition.docx