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I am having difficulty understanding how this problem is done:

Bill and George go target shooting together. Both shoot at a target at the same time. Suppose Bill hits the target with probability $0.7$, whereas George, independently, hits the target with probability $0.4$. Given that exactly one shot hit the target, what is the probability that it was George's shot?

I have the solution, it's $\frac{2}{9}$, but I have a test this week and am having difficulty understanding the methods used. Can anyone explain please?

Thanks.

Edit: After reading the solution, I was able to understand. But the second part of the question confuses me because it seems like it is the same as the first part, how can I differentiate their meanings? Here is the second part:

Given that the target is hit, what is the probability that George hit it? I know we're dealing with different sample spaces...

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    $\begingroup$ The second part is basically the same, except instead of knowing either Bill or George hit the target, we also have the case that both of them hit the target. So instead of dividing by P(Bill, not George) + P(not Bill, George), you also need to add P(Bill, George). $\endgroup$
    – Danica
    Commented Feb 12, 2013 at 6:22
  • $\begingroup$ @Dougal Oh okay, I see now. Thank you for your help. $\endgroup$
    – Alti
    Commented Feb 12, 2013 at 6:41

1 Answer 1

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You know that exactly one shot hit the target. There are two possible cases.
(1) Either Bill hit it, and George didn't.
(2) Or George hit it and Bill didn't.
$$ P(1) = 0.7 \cdot (1-0.4) =0.42\\ P(2) = (1-0.7) \cdot 0.4 = 0.12 $$ Now, we know that exactly one out of events 1 or 2 definitely happened.

So,
$$ P(1\mid 1 \text{ or } 2) = \frac{P(1)}{P(1)+P(2)} = \frac{0.42}{0.42 + 0.12} = \frac29. $$ Similarly,
$$ P(2\mid 1 \text{ or } 2) = \frac79. $$

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