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Bill and Gates go target shooting together. Both shoot at a target at the same time. Suppose Bill hits the target with probability $0.7$ ,and gates independently hits the target with probability $0.4$. Given that target is hit, what is the probability Gates hit it?

I tried using: $$P(A\cup B)=P(A)+P(B)-P(A).P(B)$$ and the answer could be: $P(A\cup B)-P(B)$

Maybe Baye's theorem can be used?

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    $\begingroup$ What did you try? $\endgroup$
    – Did
    May 4, 2013 at 8:32
  • $\begingroup$ Using this for which A and which B? $\endgroup$
    – Did
    May 4, 2013 at 9:16
  • $\begingroup$ No, that the answer couldn't be that. Yes, Bayes Theorem could be used. You know that the target was hit. You could enumerate the four possibilities, eliminate the one which doesn't meet the criteria (nobody hit) and find the relative probability of P(B). That's all that Bays theorem is really doing, but its just as easy to do by hand as it were. $\endgroup$
    – Peter Webb
    May 4, 2013 at 9:18

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Indeed, use Bayes theorem.

If $B$ means 'Bill hit the target', $G$ means 'Gates hits the target' and $T$ means that 'the target is hit', you have $$P(G|T) = \frac{P(T|G)P(G)}{P(T)}$$

Note that $G \Rightarrow T$, hence, $P(T|G)=1$ and you know $P(G)$. You only have to find $P(T)$.

To find it, note that the probability to be hit is $1-P(\lnot T)$, where $P(\lnot T)$ is very easy to find. And it's done.

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  • $\begingroup$ is the ans. 0.7/0.82 $\endgroup$
    – jyoti
    May 4, 2013 at 10:48
  • $\begingroup$ Actually, the probability that Gates hits the target is 0.4 in your question. So it would be 0.4/0.82 $\endgroup$
    – mwoua
    May 4, 2013 at 11:43

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