# Finding conditional probability

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... –  newguy Nov 18 '12 at 18:26
@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 $S - P(B\mid A^c)$ –  Dilip Sarwate Nov 18 '12 at 18:44

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

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If you want $P(A|B)$, replace last step denominator with $P(B)$. –  Gautam Shenoy Nov 18 '12 at 18:27
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? –  James Nov 18 '12 at 18:39
sorry, typo. fixed now. I can see how the equations make sense, but for some reasons the numbers do not. –  James Nov 18 '12 at 18:43
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? –  James Nov 18 '12 at 18:49
I chose A to be has disease and B to be positive test. Isn't the question saying positive test given no disease? –  James Nov 18 '12 at 18:57
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