Bayes' theorem in medicine_stud

Bayes' theorem in medicine_stud - Bayes theorem in...

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Bayes’ theorem in medicine_ stud .doc // This file prepares students for // homework #3 and quizzes #4c, d, e. // It continues the file // Conditional_probability_2_stud.doc Feb 22’10 BAYES’ THEOREM // See handout #5 Suppose an event A can occur only with one of k mutually exclusive events B 1 , B 2 , …, B k that represent a set of all possible outcomes (a sample space) of an experiment. The events B i are called hypotheses associated with the occurrence of an event A . M.V.: The events B i represent actually our hypotheses about possible reasons due to which an event A may occur . Since the events AB i = A∩B i and AB j = A∩B j are mutually exclusive at i ≠ j , the event A is a sum of events A 1 = A∩B 1 , A 2 = A∩B 2 , …, A k = A∩B k , and in accordance with the theorem (5.2) of addition of probabilities we can write k k P(A) = Σ P(A i ) = Σ P(A∩B i ) (5.6) i =1 i =1 In accordance with the product rule (5.4a) P(A∩B i ) = P(B i )*P(A|B i ) ; thus, we can represent (5.6) as k k P(A) = Σ P(A i ) = Σ P(B i )*P(A|B i ) (5.7) i =1 i =1 The relation (5.7) is called a formula for total probability . It allows us to determine the probability P(A) of the event A , if the probability P(B i ) of each hypothesis B i and conditional probability P(A|B i ) of an event A given event B i are known. When the probability P(B i ) and conditional probabilities P(A|B i ) are known, we can find the conditional probabilities P(B i |A) of hypotheses B i which represent for us main interest. By combining the relations (5.4a) and (5.4b) we obtain P( A B i ) = P( B i )*P(A| B i ) = P( A )*P( B i | A ) from which P( B i | A ) = P( B i )*P(A| B i )/P( A ) (5.8) Now by substituting the relation (5.7) in the denominator of (5.8) we obtain the 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 1
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In a special case, when there are only two mutually exclusive complementary hypotheses B 1 and B 2 the formula (5.9) leads to the following expressions. P(B 1 | A ) = P(B 1 )*P(A|B 1 ) / [P(B 1 )*P(A|B 1 ) + P(B 2 )*P(A|B 2 )] (5.9#) P(B 2 | A ) = P(B 2 )*P(A|B 2 ) / [P(B 1 )*P(A|B 1 ) + P(B 2 )*P(A|B 2 )] (5.9##) _ Sometimes it may be convenient to introduce new notations B 1 = B and B 2 = B in two latter formulas. In such a case these formulas take the forms: _ _ P(B| A ) = P(B)*P( A |B)/P( A ) = P(B)*P( A |B) / [P(B)*P( A |B) + P(B)*P( A |B)] (5.9a) _ _ _ _ _ _ _ P(B| A ) = P(B)*P( A |B)/P(A) = P(B)*P( A |B) / [P(B)*P( A |B) + P(B)*P( A |B)] (5.9b) //////////////////////////////// END OF HANDOUT #5 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ // See the file Example of using of BAYES’ formula.doc or beginning of the homework #3 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
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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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Bayes' theorem in medicine_stud - Bayes theorem in...

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