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Unformatted text preview: exclusiveness) P p u i =1 A i P = s i =1 P ( A i ) = 0 . If A is the event of choosing a rational and B is the event of choosing an irrational number, then A c = B and so P ( B ) = P ( A c ) = 1P ( A ) = 1. pg. 36 problems: 5. The sample space consists of sequences of the form HTHTHTHT. .. or THTHTHTH until it ends with HH or TT , at which point it terminates. Thus we have HH , TT , HTT , THH , HTHH , THTT , . . . , S . To make this explicit, I would say: S = { a 1 a 2 a k : k { 2 , 3 , . . . } , a i { H, T } , a i1 n = a i for any i < k, a k = a k1 } . 9. Let A be the red wine group and B be the white wine group. Then, P ( B ) = . 4 and P ( A ) = . 5 and P ( A B ) = . 7. We want P ( A B ) = P ( A )+ P ( B )P ( A B ) = . 5+ . 4. 7 = . 2. 1 11. We want P ( A c B c ) in terms of P ( A ), P ( B ), and P ( AB ). We have by De Morgan P ( A c B c ) = P (( A B ) c ) = 1P ( A B ) = 1P ( A )P ( B ) + P ( AB ) . 2...
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 Spring '08
 Anderson
 Math, Probability

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