The probability of hitting a target by three | vedclass

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Question

(A) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,and $P(C) = \frac{1}{4}$ be the probabilities of hitting the target by the three marksmen.Then,the p

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$\frac{11}{24}$

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(A) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,and $P(C) = \frac{1}{4}$ be the probabilities of hitting the target by the three marksmen.Then,the probabilities of missing the target are $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$,and $P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4}$.The event that one and only one of them hits the target can happen in three mutually exclusive ways: ($A$ hits,$B$ misses,$C$ misses) or ($A$ misses,$B$ hits,$C$ misses) or ($A$ misses,$B$ misses,$C$ hits).Required probability $= P(A)P(\bar{B})P(\bar{C}) + P(\bar{A})P(B)P(\bar{C}) + P(\bar{A})P(\bar{B})P(C)$$= (\frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4}) + (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{3}{4}) + (\frac{1}{2} 	imes \frac{2}{3} 	imes \frac{1}{4})$$= \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24}$.

The probability of hitting a target by three marksmen are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. The probability that one and only one of them will hit the target when they fire simultaneously,is

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