# Unconditional and independent probability

Two archers, independently, target a mark firing one arrow at a time. Probability that one archer will hit a mark is 0.8 and another is 0.4. After contest it is determined that one hit got in a mark. Find a probability that first archer scored it.

I have no idea from where to start on this. It would be logical that the result is 0.8, 'cause that is the probability to hit a mark of the first archer, but that's not the result wanted here. :(

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Per your statement, only one arrow hit the mark. That is possible when either archer A hit, and archer B missed, or the other way around. $$p_1 = \mathbb{P}\left(\text{A hit} \land \text{B miss}\right) = \underline{\quad\quad\quad}?$$ $$p_2 = \mathbb{P}\left(\text{B hit} \land \text{A miss}\right) = \underline{\quad\quad\quad}?$$ Since these events are exclusive (both can not happen at the same time), $\mathbb{P}\left(\text{only one hit}\right) = p_1 + p_2$. Now to answering the question $$\mathbb{P}\left( \text{A hit} | \text{only one hit}\right) = \frac{\mathbb{P}\left(\text{A hit} \land \text{B miss}\right)}{\mathbb{P}\left(\text{only one hit}\right)} = \frac{p_1}{p_1+p_2}$$