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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(AB i ) P(B i )*P(AB i ) P(B i )*P(AB i ) P(B i A) =  =  =  (5.9) P(A) k k Σ P(B i )*P(AB 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(BA) = P(B)*P(AB) / [P(B)*P(AB) + P(B)*P(AB)] (5.9a) or _ _ _ _ _ P(BA) = P(B)*P(AB) / [P(B)*P(AB) + P(B)*P(AB)] (5.9b) Using Tree Diagrams for Solving Bayes’ Theorem Problems P(AB 1 ) P(B 1 ) / A P(A∩B 1 ) = P(B 1 )*P(AB 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(AB 2 ) / P(B 2 ) / A P(A∩B 2 ) = P(B 2 )*P(AB 2 ) /  B 2 * …………… / \ * …………………………... A set of branches that initiate from a single \ point (*) have a total probability of 1. \ …………………………… \ \ …………………………… \ \ P(AB k ) \ P(B k ) / A P(A∩B k ) = P(B k )*P(AB k ) B k * ………….....
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 Spring '05
 Kzaer
 Math, Conditional Probability, Probability, Probability theory, Type I and type II errors, Medical test

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