How to find probability of $P(B|A)$ using Bayes Rule?

Question

I have a problem that I'm stuck on for awhile that I'm hoping someone can help me out with.

I know $P(A|B)$ is $0.95$ and $P(\mathrm{not}\,A|\mathrm{not}\,B)$ is also $0.95$. I also know $P(B)$ is $0.0001$. How can I find $P(B|A)$? I know I am to use Bayes rule but I'm stuck on how to calculate $P(A)$ in order to use it. Can someone help me out?

Answer 1

$$ \begin{align} P(A) &= P(AB) + P(A\bar{B})\\ &= P(A|B)P(B) + P(A|\bar{B})P(\bar{B})\\ &= P(A|B)P(B) + [1-P(\bar{A}|\bar{B})][1-P(B)] \\ &= 0.95 * 0.00001 + [1-0.95][1-0.00001] \end{align} $$

Answer 2

P(AB)=P(A|B)×P(B) and P(A'B')= 1-P(A)P(B) hence P(A)={1-P(A'B')}/P(B) now P(B|A)={P(AB)}/P(A)