Probability of A given $P(A) + P(B), P(A|B),$ and $P(B|A)$.

Question

Given $P(A|B)= 0.5, P(B|A)=0.4,$ and $P(A) + P(B) = 0.9$ what is $P(A)=$ ?.

Answer 1

Let $a=\Pr(A)$ and $b=\Pr(B)$. We are told that $a+b=0.9$.

We have $\Pr(A|B)\Pr(B)=\Pr(A\cap B)$. Thus $0.5b=\Pr(A\cap B)$.

Similarly, $0.4a=\Pr(A\cap B)$.

It follows that $0.5b=0.4a$. Use this equation, together with $a+b=0.9$ to solve for $a$.

Answer 2

$$\frac{P[A]}{P[A]+P[B]}=\frac{\frac1{P[B]}}{\frac1{P[A]}+\frac1{P[B]}}=\frac{P[A\cap B]\cdot\frac1{P[B]}}{P[A\cap B]\cdot\frac1{P[A]}+P[A\cap B]\cdot\frac1{P[B]}} $$ $$ \text{hence}\qquad P[A]=\frac{P[A\mid B]}{P[A\mid B]+P[B\mid A]}\cdot(P[A]+P[B])$$