Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
1  
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). –  Dougal Feb 12 '13 at 6:22
    
@Dougal Oh okay, I see now. Thank you for your help. –  Alti Feb 12 '13 at 6:41
add comment

1 Answer

up vote 2 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.