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(D) Let $A, B, C,$ and $D$ be the events that the four persons hit the target respectively. Given probabilities are $P(A) = \frac{1}{2}, P(B) = \frac{

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$\frac{4}{5}$

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(D) Let $A, B, C,$ and $D$ be the events that the four persons hit the target respectively. Given probabilities are $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}, P(D) = \frac{1}{5}$. The probability that the target is not hit by any of them is the probability that all four miss the target. $P(A') = 1 - \frac{1}{2} = \frac{1}{2}$ $P(B') = 1 - \frac{1}{3} = \frac{2}{3}$ $P(C') = 1 - \frac{1}{4} = \frac{3}{4}$ $P(D') = 1 - \frac{1}{5} = \frac{4}{5}$ Since the events are independent,the probability that none hit the target is: $P(	ext{None hit}) = P(A') 	imes P(B') 	imes P(C') 	imes P(D') = \frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4} 	imes \frac{4}{5} = \frac{1}{5}$. The probability that the target is hit at least once is: $P(	ext{Target is hit}) = 1 - P(	ext{None hit}) = 1 - \frac{1}{5} = \frac{4}{5}$.

Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively. If all hit at the target independently,then the probability that the target would be hit,is

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