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Answer :
Final answer:
Given that Bill hits the target with a probability of 0.7 and George hits the target with a probability of 0.4 and that exactly one shot hit the target, the probability that George hit the target is about 0.222.
Explanation:
The subject matter of this question deals with probability, specifically focusing on the concept of conditional probability. The question asks: Given that Bill hits the target with a probability of 0.7 and George hits the target with a probability of 0.4, what is the probability that George hit the target if we know that only one of them did?
The first step in solving this is recognizing that since only one shot hit the target, the possibilities are that either Bill hit and George missed, or George hit, and Bill missed. The probability of Bill hitting, and George missing is (0.7) * (1-0.4) = 0.42. The probability of George hitting and Bill missing is (0.4) * (1-0.7)=0.12. These are the only two scenarios we're interested in because the problem states that exactly one shot hit the target.
To determine the probability that it was George's shot that hit, given that only one hit, we divide the probability of George hitting and Bill missing by the total probability of one hit. So, the answer is the probability of George hitting and Bill missing divided by the sum of the probabilities of both events, and 0.12/ (0.42 + 0.12) = 0.222 approximately.
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