Probability exercise (archers)

We´ve got 3 archers. Every archer shoots only once. The Probability that the first archer will hit the target is 0.7, second 0.8 and third 0.9.

• What is the probability that at least two of them will hit the target?

I've tried to complete it, I've calculated the Probability that at least one of them will hit the target and tried to delete from it options when one of them will hit the target.

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Firstly, what are the cases in which two or more will hit the target? for example two and three, but not one. Secondly, what is the chance that each of these combinations happen?In my example $(1-0.7)\times 0.8 \times 0.9$ Thirdly, what is the sum of these chances? That would be your answer. For further reading: Combinatorics and Binomial theorem

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Hint: Can you assess the probability that the first one is the only one who hits the target? The probability that none of them do?

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None of them is not a problem. Only the first one = 0.3*0.8*0.9, is it right? – user12392 Dec 5 '12 at 18:23
@user12392: No, that is the chance that the first misses and the second and third hit. – Ross Millikan Dec 5 '12 at 19:22

It would be something like $(0.3×0.8×0.9)+(0.7×0.2×0.9)+(0.7×0.8×0.1)+(0.7×0.8×0.9)$

a misses, b & c hit + b misses, a & c hit + c misses, a & b hit + all hit

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hmm right… i'll fix it thx. – Rob Dec 5 '12 at 19:44