Bill and George go target shooting together. | vedclass

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(B) Let $B$ be the event that Bill hits the target and $G$ be the event that George hits the target.Given $P(B) = 0.7$ and $P(G) = 0.4$.The probabilit

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(B) Let $B$ be the event that Bill hits the target and $G$ be the event that George hits the target.Given $P(B) = 0.7$ and $P(G) = 0.4$.The probability that Bill misses is $P(B') = 1 - 0.7 = 0.3$.The probability that George misses is $P(G') = 1 - 0.4 = 0.6$.Since the events are independent,the probability that exactly one shot hits the target is the sum of the probabilities of two mutually exclusive cases: (Bill hits and George misses) or (Bill misses and George hits).$P(	ext{Exactly one hit}) = P(B \cap G') + P(B' \cap G) = P(B)P(G') + P(B')P(G)$$= (0.7 	imes 0.6) + (0.3 	imes 0.4) = 0.42 + 0.12 = 0.54$.We want to find the conditional probability that it was George's shot,given that exactly one hit occurred.$P(G 	ext{ hit} | 	ext{Exactly one hit}) = \frac{P(B' \cap G)}{P(	ext{Exactly one hit})} = \frac{0.12}{0.54} = \frac{12}{54} = \frac{2}{9}$.Thus,the correct option is $B$.

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 $0.4$. Given that exactly one shot hit the target,what is the probability that it was George's shot?

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