Bayes' theorem help: "3 archers shot a target once...."

tonitoneff

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I got stuck, I couldn't apply Bayes' theorem after trying multiple times reading the theory for the theorem but i got the answer wrong every time. The question was: 3 archers shot a target once. Each archer has probability to hit from the first shot as follows: 0.8, 0.7, 0.6. The target was hit by 2 arrows, what is the probability that the arrows which hit belong to the 1st (0.8 hit probability) and the second archer (0.7 hit probability). I would love it if someone just writes down the solution step by step with a small logic explanation.:):confused:
 
I got stuck, I couldn't apply Bayes' theorem after trying multiple times reading the theory for the theorem but i got the answer wrong every time. The question was: 3 archers shot a target once. Each archer has probability to hit from the first shot as follows: 0.8, 0.7, 0.6. The target was hit by 2 arrows, what is the probability that the arrows which hit belong to the 1st (0.8 hit probability) and the second archer (0.7 hit probability). I would love it if someone just writes down the solution step by step with a small logic explanation.
NOTATION The three archers are \(\displaystyle A,~B,~\&~C\), so \(\displaystyle \mathcal{P}(AB\overline{C})=(0.8)(0.7)(0.4)\)
that is archers \(\displaystyle A~\&~B\) alone hit the target, i.e. \(\displaystyle C\) does not hit it.

Thus
you are looking for:\(\displaystyle \mathcal{P}[(AB\overline{C})|(AB\overline{C})\cup(A\overline{B}C) \cup(\overline{A}BC)] =\dfrac{\mathcal{P}[(AB\overline{C})}{\mathcal{P}(AB\overline{C})\cup \mathcal{P}(A\overline{B}C)\cup\mathcal{P}( \overline{A} BC)}\)
The probability that the first two hit given that only two do.
 
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