Thank you for visiting Bill and George go target shooting together Both shoot at a target at the same time Suppose Bill hits the target with a probability of. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Final answer:
For part (a), the probability that George hit the target, given that only one hit the target, is 0.4. For part (b), the probability that George hit the target, given that the target was hit, is approximately 0.6896.
Explanation:
For part (a) of the question: 'Given that exactly one shot hit the target, what is the probability that it was George's shot?', we need to analyze the two scenarios where only one shot can hit the target. That is either Bill hits and George misses or George hits and Bill misses. So the probability that George hit the target if only one hit target is: P(George and not Bill) divided by P(only one hit). The probability of the event 'George and not Bill' is P(George) * P(not Bill) = 0.4 * (1-0.7) = 0.12. The probability of the event 'only one hit' is P(Bill and not George) + P(George and not Bill) = 0.7 * (1-0.4) + 0.4 * (1-0.7) = 0.18 + 0.12 = 0.3. Therefore, P(George | only one hit) = P(George and not Bill) / P(only one hit) = 0.12 / 0.3 = 0.4.
For part (b) of the question: 'Given that the target is hit, what is the probability that George hit it?', we know that the target can be hit if both shots hit or if only one hits. The probability George hit given that the target is hit is therefore P(George and hit) / P(hit). But P(George and hit) = P(George) = 0.4 and P(hit) = P(both hit) + P(only one hit) = 0.7*0.4 + 0.3 = 0.58. Hence, P(George | hit) = 0.4 / 0.58 = approx. 0.6896
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