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Given $P(A)$, $P(B)$, and $P(B\mid A^c)$, how do you find $P(B\mid A)$?

I need this to find $P(A\mid B)$ using Bayes' Theorem: $ P(A\mid B)=\frac{P(A)P(B\mid A)}{P(A)P(B\mid A)+P(A^c)P(B\mid A^c)} $ and $P(B\mid A)$ is the only one I can't seem to find the value for.

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    @newguy ... and what fixed value do _you_ think the _sum_ $S = P(B\mid A) + P(B \mid A^c)$ has so that it is possible to deduce the value of $P(B\mid A)$ as $S - P(B\mid A^c)$2012-11-18

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$P(A^C) = 1-P(A)$ $P(B\cap A^C)=P(B|A^C)P(A^C)$ $ P(A\cap B) = P(B) - P(B\cap A^C)$ $ P(B|A) = P(A\cap B) / P(A)$

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    You are right. But P(A \cap B)=0.8772 > P(A). This shouldn't happen. A question flaw perhaps?2012-11-18