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Unformatted text preview: 18 Solutions Manual for Statistical Inference 1.37 a. Using the same events A, B, C and W as in Example 1.3.4, we have P (W) = P (WA)P (A) + P (WB)P (B) + P (WC)P (C) 1 1 1 +1 = +0 +1 = . 3 3 3 3 Thus, P (AW) =
P (AW) P (W) = /3 (+1)/3 = +1 where,
1 3 1 3 1 3 +1 = < +1 > +1 if = 1 2 if < 1 2 if > 1 . 2 b. By Exercise 1.35, P (W) is a probability function. A, B and C are a partition. So P (AW) + P (BW) + P (CW) = 1. But, P (BW) = 0. Thus, P (AW) + P (CW) = 1. Since P (AW) = 1/3, P (CW) = 2/3. (This could be calculated directly, as in Example 1.3.4.) So if A can swap fates with C, his chance of survival becomes 2/3. 1.38 a. P (A) = P (A B) + P (A B c ) from Theorem 1.2.11a. But (A B c ) B c and P (B c ) = 1  P (B) = 0. So P (A B c ) = 0, and P (A) = P (A B). Thus, P (AB) = . b. A B implies A B = A. Thus, P (BA) = And also, P (AB) = P (A B) P (A) = . P (B) P (B) P (A B) P (A) = = 1. P (A) P (A) P (A) P (A B) = = P (A) P (B) 1 c. If A and B are mutually exclusive, then P (A B) = P (A) + P (B) and A (A B) = A. Thus, P (A (A B)) P (A) P (AA B) = = . P (A B) P (A) + P (B) d. P (A B C) = P (A (B C)) = P (AB C)P (B C) = P (AB C)P (BC)P (C). 1.39 a. Suppose A and B are mutually exclusive. Then A B = and P (A B) = 0. If A and B are independent, then 0 = P (A B) = P (A)P (B). But this cannot be since P (A) > 0 and P (B) > 0. Thus A and B cannot be independent. b. If A and B are independent and both have positive probability, then 0 < P (A)P (B) = P (A B). This implies A B = , that is, A and B are not mutually exclusive. 1.40 a. P (Ac B) = P (Ac B)P (B) = [1  P (AB)]P (B) = [1  P (A)]P (B) = P (Ac )P (B) , where the third equality follows from the independence of A and B. b. P (Ac B c ) = P (Ac )  P (Ac B) = P (Ac )  P (Ac )P (B) = P (Ac )P (B c ). ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics

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