Some_solutions_for_Chapter_2

Some_solutions_for_Chapter_2 - manager will leave. P(L1) =...

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Some solutions for Chapter 2 – Probability 2-71. One out of 200 calls is misdirected Probability that a random call misdirected is 1/200 = 0.005 Or probability that a single call NOT misdirected is 199/200 = 0.995. Five calls are taken independently , denote C1, C2, C3, C4, C5 are five call is not misdirected. Probability that at least one call misdirected is: ) = 1 – = 1 – = 0.0248 ~ 0.025 . 2-72. (0.02)(1/200) = 1/10,000 = 0.0001 2-73. B) P(A = P(A) + P(B) – P(A ∩ B) = 0.4 + 0.3 – 0.1 = 0.6 2-74. Assume independence in each manager’s decision. Denote L1, L2, . . ., L25 are events that each
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Unformatted text preview: manager will leave. P(L1) = P(L2) = . . . = P(L25) = 0.5 a) P(none leave) =P( (1 0.5) 25 = 2.9 x 10-8 = approximately 0.000 b) P(all leave) = (0.5) 25 = 2.9 x 10-8 = approximately 0.000 c) P(at least one leaves) = 1 = 1 - (2.9 x 10-8 )= 0.999 = approximately 1.0 2-75. Let C is the event that a consumer will buy computer. S is event the event that consumer will buy software. We have: P(C) = 0.15, P(S) =0.1 and P( C S) = 0.05. P(buy computer OR software OR both ) = P (C U S) = P(C) + P(S) - P( C S) = 0.2...
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