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3 hunters fire simultaneously at a boar. One bullet hits the boar. What probability is for each hunter to be the one who hit the boar, when hunter A hits with accuracy of 0.2, B 0.4, C 0.6?
I see 2 possible solutions:

The first one is simple. When I take 10 average bullets of hunter A, I've got 2 bullets that hit, 4 for B and 6 for C. If I take all 30 imaginary bullets, I put the bullets that missed away and have 12 left. This leads me to $$\frac{2}{12},\frac{4}{12},\frac{6}{12}=\frac1{6}, \frac1{3}, \frac1{2}$$The second one is that for each hunter, I take probability of the fact that his bullets hit and the other hunters' did not and divide it by probability that this occured or one of the others got lucky. This is basically dividing the 'good' possibilities by all possibilities. I get the numerator for the first hunter by multiplying his accuracy (0.2) by inverses of accuracies of the other guys (0.6 and 0.4). The denominator is the same for all of them and contains the sum of nominators for each hunter. So for A, it is $$\frac{0.2\cdot0.6\cdot0.4}{0.2\cdot0.6\cdot0.4+0.4\cdot0.8\cdot0.4+0.6\cdot0.8\cdot0.6}$$The results are $\frac{48}{464},\frac{128}{464},\frac{288}{464}=\frac{3}{29},\frac{8}{29},\frac{18}{29}$
Now, which one of these methods is correct? And maybe even more importantly, why the other one isn't?

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  • $\begingroup$ The procedure in the second calculation is correct. I would use a more formal conditional probability calculation, for security. $\endgroup$ – André Nicolas Nov 13 '13 at 22:29
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The second is correct. The error in the first is that you could have more than one bullet that hits, but the problem prohibits that. In the second you have considered the chance that two have missed as well as one hit. Your answers in the first will be the correct answer to the question: given that each hunter fires an equal number of bullets and we select one that hit at random, what is the chance it came from each one?

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