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.
(a) Given that exactly one shot hit the target, what is the probability that it was
George’s shot?
(b) Given that the target is hit, what is the probability that George hit it?
(a) The probability of exactly one shot hitting the target can be calculated by finding the sum of the probabilities of Bill missing the target and George hitting it, and vice versa. The probability of Bill missing the target is 1-0.7 = 0.3 and the probability of George hitting the target is 0.4. The combined probability of these two scenarios is 0.3 * 0.4 = 0.12. The same probability applies to Bill hitting the target and George missing it. The combined probability of exactly one shot hitting the target is 0.12 * 2 = 0.24. Hence, the probability that it was George’s shot given exactly one shot hit the target is 0.12/0.24 = 0.5.
(b) Given that the target is hit, the probability that George hit it can be calculated using Bayes’ theorem. Bayes’ theorem states that the probability of event A given event B is equal to the probability of event B given event A multiplied by the probability of event A, divided by the probability of event B. In this scenario, event A is that George hit the target and event B is that the target is hit. Hence, the probability of George hitting the target given that the target is hit is 0.4/(0.7 * 0.4 + 0.3 * 0.6) = 0.4/0.46 = 0.87.
The probability that George hit the target given exactly one shot hit the target is 0.5, and the probability that George hit the target given that the target is hit is 0.87. These probabilities give us a better understanding of the likelihood of events in a target shooting scenario with two shooters.
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