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# homework_3 - St John’s University Department of...

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Unformatted text preview: St. John’s University 02/23/09 Department of Mathematics and Computer Science Course MTH 1210 Homework #3 PROBLEMS INVOLVING BAYES’ FORMULA P(B i )*P(A|B i ) P(B i )*P(A|B i ) P(B i )*P(A|B i ) P(B i |A) = --------------------- = ------------------------ = ------------------- (5.9) P(A) k k Σ P(B i )*P(A|B i ) Σ P(A∩B i ) i =1 i =1 In a special case, when there are only two mutually exclusive complementary hypotheses _ B 1 = B and B 2 = B associated with the occurrence of event A , the formula (5.9) takes the forms: _ _ P(B|A) = P(B)*P(A|B) / [P(B)*P(A|B) + P(B)*P(A|B)] (5.9a) or _ _ _ _ _ P(B|A) = P(B)*P(A|B) / [P(B)*P(A|B) + P(B)*P(A|B)] (5.9b) Using Tree Diagrams for Solving Bayes’ Theorem Problems P(A|B 1 ) P(B 1 ) /--------------- A P(A∩B 1 ) = P(B 1 )*P(A|B 1 )------------ B 1 * ……………. ← Branches for conditional probabilities of other events / \---------------- ← Ā, C, D… that may be associated with the set of / hypotheses { B i } / P(A|B 2 ) / P(B 2 ) /-------------- A P(A∩B 2 ) = P(B 2 )*P(A|B 2 ) / ------------ B 2 * …………… / \--------------- * …………………………... A set of branches that initiate from a single \ point (*) have a total probability of 1. \ …………………………… \ \ …………………………… \ \ P(A|B k ) \ P(B k ) /-------------- A P(A∩B k ) = P(B k )*P(A|B k )--------- B k * ………….....
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homework_3 - St John’s University Department of...

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