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Unformatted text preview: = 0.2873 3.116 Optional Exercises 3.18 3.26 a. P ( A c ) = P ( E 3 ) + P ( E 6 ) + P ( E 8 ) = 0.2 + 0.3 + 0.03 = 0.53 b. P ( B c ) = P ( E 1 ) + P ( E 7 ) + P ( E 8 ) = 0.10 + 0.06 + 0.03 = .19 c. P ( A c ∩ B ) = P ( E 3 ) + P ( E 6 ) = 0.2 + 0.3 = 0.5 d. P ( A ∪ B ) = P ( E 1 ) + P ( E 2 ) + P ( E 3 ) + P ( E 4 ) + P ( E 5 ) + P ( E 6 ) + P ( E 7 ) = .10 + .05 + .20 + .20 + .06 + .30 + .06 = .97 e. P ( A ∩ B ) = P ( E 2 ) + P ( E 4 ) + P ( E 5 ) = .05 + .20 + .06 = .31 f. P ( A c ∪ B c ) = P ( E 1 ) + P ( E 7 ) + P ( E 3 ) + P ( E 6 ) + P ( E 8 ) = .10 + .06 + .20 + .30 + 0.03 = .69 g. No. A and B are mutually exclusive if P ( A ∩ B ) = 0. Here, P ( A ∩ B ) = .31. 3.32 3.38 3.52 3.54 Thus P(B c  A) = 0.10 We require to determine P(A ∩ B c ) = P(A) * P(B c  A) = 0.9 * 0.10 = 0.09...
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 Spring '10
 woo
 Haplogroup O, CVSA, incorrect results, new lie detector, Quantitative Methods Solutions

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