homework_3_ans - St. Johns University Department of...

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St. John’s University Mar 15’09 Department of Mathematics and Computer Science Course MTH 1210 Homework #3 (answers) PROBLE MS 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 * …………. . \--------------- 1. Start a tree diagram with branches representing hypotheses B 1 , B 2 , …, B k associated with the occurrence of an event A . Label each branch with its corresponding probability P(B i ) . 2. From the end of each of these branches, draw a branch for the event A . Label this branch with the conditional probability of getting to it, P(A|B i ) . 1
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3. Now we have k different paths that result in event A . Next to each path put its probability P( A ∩B i ) = P(B i )*P(A|B i ) – the product of the probability P(B i ) that the first branch B i occurs, and the conditional probability P(A|B i ) that the second branch occurs. 4. In accordance with Bayes’ formula (5.9), the probability of interest P(B i | A ) is found by dividing the probability P(B i )*P(A|B i ) of the branch for B i by the sum of the probabilities P( A ∩B i ) of all the branches producing the given event under consideration ( A ) (which is written in the expression P(B i | A ) for the asked for Bayes’ probability to the right from the vertical separating bar). Multiply probabilities along any branch related to the event under consideration
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This note was uploaded on 11/24/2011 for the course MATH 3000 taught by Professor Kzaer during the Spring '05 term at St. Johns College MD.

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homework_3_ans - St. Johns University Department of...

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