I have 2 independent events: $P(A) = 0.4\,;P(B) = 0.5$.
Probability that exactly one of them occurs: $0.4(1-0.5) + 0.5(1-0.4) = 0.5$.
My questions are: Given that exactly one event is successful, what is the probability that the event is A? How this would be calculated?
Given that exactly one is successful, the fact that $A$ is successful implies that $B$ is not. Let $E$ be the event that either $A\cap\lnot B$ or $\lnot A\cap B$ holds. Then we have $P(A|E)=\frac{P(A\cap E)}{P(E)}=\frac{P(A\cap\lnot B)}{P(A\cap\lnot B)+P(\lnot A\cap B)}=\frac{0.4\cdot 0.5}{0.4\cdot 0.5+0.6\cdot 0.5}$.