Probability of A give B > A rare disease affects 1 in 10,000 individuals in the population. A blood test for the disease is positive for 95% of people with the disease. However, 0.4% of people without the disease will also test positive. Given that a person (selected at random from the general population) tests positive, what is the probability that she has the disease? Can someone help me figure this out? I know the probability of having the disease is 1/10,000. I know this rule: $\displaystyle P\left(A\mid B\right) = \frac{P\left(A \cap B\right)}{P(B)}$ but in this formula wouldn't $A$ be testing positive which is either $.95$ or $.004$?

Use the extended form of Bayes' theorem, since A is a binary virable (a person can either be sick or not sick): $\displaystyle P(A|B) = \frac{(P(B|A)*P(A)}{P(B|A)*P(A)+P(B|A^c)*P(A^c)}$ Note that the question is what the probablity that a person has the disease, given he tested positive. So A would be person having the disease, and B would be testing positive.