Given that the target is hit, what is the probability that George hit it?

Question

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...

Answer 1

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,
$$ P(2\mid 1 \text{ or } 2) = \frac79. $$