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Question

(A) Let the probabilities of hitting the target by four persons be $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,$P(C) = \frac{1}{4}$,and $P(D) = \frac{1}

Vedclass · Accepted Answer

$\frac{25}{32}$

Vedclass · Answer

(A) Let the probabilities of hitting the target by four persons be $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,$P(C) = \frac{1}{4}$,and $P(D) = \frac{1}{8}$.The probability that the target is hit is $1 - P(	ext{none of them hits the target})$.Since the events are independent,$P(	ext{none hits}) = P(\overline{A}) \cdot P(\overline{B}) \cdot P(\overline{C}) \cdot P(\overline{D})$.$P(\overline{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\overline{B}) = 1 - \frac{1}{3} = \frac{2}{3}$,$P(\overline{C}) = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(\overline{D}) = 1 - \frac{1}{8} = \frac{7}{8}$.$P(	ext{none hits}) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{7}{8} = \frac{1 \cdot 2 \cdot 3 \cdot 7}{2 \cdot 3 \cdot 4 \cdot 8} = \frac{42}{192} = \frac{7}{32}$.Therefore,the probability that the target is hit is $1 - \frac{7}{32} = \frac{25}{32}$.

Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{8}$ respectively. If all hit at the target independently,then the probability that the target would be hit is:

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