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hw3sols09 - Stat 330 1 Solution to Homework 3 Independence...

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Stat 330 Solution to Homework 3 1 Independence, Baron p. 40 Events A and B are independent. Show, intuitively, and mathematically, that (a) Their complements are also independent. We know, that P ( A B ) = P ( A ) P ( B ) . We want to ﬁnd out about the probability of the intersection of the complements. Intuitively, the complements should be independent, since the occurrence of the event or the negative of the events should not matter in the case of independence. Mathematically, we get: P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = = P ( A ) + P ( B ) - (1 - P ( A B )) = = P ( A ) + P ( B ) - (1 - P ( A ) · P ( B )) = = 1 - P ( A ) + P ( B ) - 1 + P ( A ) · P ( B )) = = P ( A ) ( 1 - P ( B )) + P ( B ) = P ( A ) · P ( B ) (b) If they are disjoint, then P ( A ) = 0 or P ( B ) = 0. Since we know, that P ( A B ) = P ( A ) P ( B ) the statement follows directly, if we also assume, that the events are disjoint, i.e. P ( A B ) = 0 (c) If they are exhaustive (i.e. A B = Ω), then P ( A ) = 1 or P ( B ) = 1. From 1 = P ( A B ) we get, 1 = P ( A B ) = 1 - P ( A B ) = 1 - P ( A ) · P ( B ) Therefore P ( A ) · P ( B ) = 0 , which implies that either P ( A ) = 0 or P ( B ) = 0 . Then either P ( A ) = 1 or P ( B ) = 1 . (4 points) 2 Bayes’ Theorem (a) Suppose that a barrel contains many small plastic eggs. Some eggs are painted red and some are painted

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hw3sols09 - Stat 330 1 Solution to Homework 3 Independence...

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