# In probability how do prove that P(A|B) / 1-P(A |B) = P(B|A)P(A) / P(B |Ā )P(Ā) ?

Thanks for any help.

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- stymLv 51 decade agoFavorite Answer
Law of conditional probability: P(A∩B) = P(B∩A) = P(A|B)P(B) = P(B|A)P(A)

Law of total probability: P(B) = P(B∩A) + P(B∩Ā)

Since P(A|B) = P(A∩B)/(P(B), you have:

P(A|B) / (1-P(A|B)) = P(A∩B)/(P(B)( 1- P(A∩B)/P(B))

P(A|B) / (1-P(A|B)) = P(A∩B)/(P(B)- P(A∩B))

But P(B)- P(A∩B) = P(B∩Ā), so:

P(A|B) / (1-P(A|B)) = P(A∩B)/P(B∩Ā) = P(B|A)P(A) / P(B|Ā)P(Ā)

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