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Dr. Hackney STA Solutions pg 11

Dr. Hackney STA Solutions pg 11 - 1-8 Solutions Manual for...

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1-8 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 ( W| A ) P ( A ) + P ( W| B ) P ( B ) + P ( W| C ) P ( C ) = γ 1 3 + 0 1 3 + 1 1 3 = γ +1 3 . Thus, P ( A |W ) = P ( A ∩W ) P ( W ) = γ/ 3 ( γ +1) / 3 = γ γ +1 where, γ γ +1 = 1 3 if γ = 1 2 γ γ +1 < 1 3 if γ < 1 2 γ γ +1 > 1 3 if γ > 1 2 . b. By Exercise 1.35, P ( ·|W ) is a probability function. A , B and C are a partition. So P ( A |W ) + P ( B |W ) + P ( C |W ) = 1 . But, P ( B |W ) = 0. Thus, P ( A |W ) + P ( C |W ) = 1. Since P ( A |W ) = 1 / 3, P ( C |W ) = 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 ( A | B ) = P ( A B ) P ( B ) = P ( A ) 1 = P ( A ) . b. A B implies A B = A . Thus, P ( B | A ) = P ( A B ) P ( A ) = P ( A ) P ( A ) = 1 . And also, P ( A | B ) = P ( A B ) P ( B ) = P ( A ) P ( B ) . 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 ( A B )) P ( A B ) = P ( A ) P ( A ) + P ( B ) . d. P ( A B C ) = P ( A ( B C )) = P ( A | B C ) P ( B C ) = P ( A | B C ) P ( B | C ) P ( C ) . 1.39 a. Suppose A and B are mutually exclusive. Then A
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