Q:

Disease test - probability for a person with a positive result to be healthy

Consider a disease that is on average caught by $1$ in $N$ people. Let's say that a disease test gives a false positive for $p\%$ of healthy people. For diseased people, it always gives a correct, positive answer.
What's the chance that a person who got a positive result is healthy?
I've tried using $P(A|B)=\frac{P(A \cap B)}{P(B)}$ but I don't know which is $A$ and $B$.

A:

Imagine 1000N people.  1000N/N= 1000 of those people have the disease, 1000N- 1000= 1000(N- 1) do not.  Of the 1000(N- 1) people who do not have the disease the test falsely reports that 1000p(N-1) do have the disease.  The test also reports all 1000 people who have the disease so reports a total of 1000+ 1000p(N-1)= 1000(1+ p(N-1)) people as having the disease.  Of the 1000(1+ p(N-1)) people reported to have the disease, 1000p(N-1) are actually healthy.  The probability that a person reported to have the disease is actually healthy is [1000p(N-1)]/[1000(1+ p(N-1))]= p(N-1)/(1+ p(N-1)).

