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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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    Using the definition of conditional probability, can you find a relation between $P(B|A)$ and $P(B|A^c)$. Perhaps look at their sum...2012-11-18
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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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    If you want $P(A|B)$, replace last step denominator with $P(B)$.2012-11-18
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    so the given values were P(A)=.08, P(B)=.96, $P(B|A^c)=.09$ and $P(B|A)$ comes out to be > 1. im assuming this is not okay?2012-11-18
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    sorry, typo. fixed now. I can see how the equations make sense, but for some reasons the numbers do not.2012-11-18
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    The question was: Suppose that 8% of people have disease, that a person tests positive 96% of the time, and that a non-diseased person tests positive 9% of the time. What is the probability that a person who tests positive actually has the disease. Did I interpret values incorrectly?2012-11-18
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    I chose A to be has disease and B to be positive test. Isn't the question saying positive test given no disease?2012-11-18
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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