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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 $$ \mathbb{P}\left( \text{A hit} | \text{only one hit}\right) = \frac{\mathbb{P}\left(\text{A hit} \land \text{B miss}\right)}{\mathbb{P}\left(\text{only one hit}\right)} = \frac{p_1}{p_1+p_2} $$

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Thank you! You explained it very good! Now I understand this situation too! :D –  Takarakaka Jul 2 '12 at 20:09
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I think the question is to suppose that exactly one of the two archers hit the mark. Given that archer 1 has a probability of 0.8 to achieve it and archer 2 0.4 what is the probability of this outcome. It will happen if archer 1 hits and 2 misses or if two hits and 1 misses. So calculate the porbability for each fo those disjoint events and sum them. Each archer's result is independent of the other.

P{ 1 hits and 2 misses} =P(1 hits) P(2 misses)= P(1 hits)(1-P(2 hits)) =0.8x(1-0.4)=0.8x0.6=0.48. Now P{2 hits and 1 misses}=P(2 hits) P(1 misses)=P(2hits) (1-P(1 hits))=0.4(1-0.8)=0.4x(0.2)=0.08. So the probability that exactly one of the archers hits the target os 0.48 + 0.08 =0.56. Now given that there is a hit the probability that it is archer 1 is 0.48/0.56 =0.857.

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Thank you @Micheal . I have been just doing this task, solving it and great thing is that I came up with the same result! Yeah! –  Takarakaka Jul 2 '12 at 20:19
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Are you giving us a bunch of homework problems? –  Michael Chernick Jul 2 '12 at 20:21
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