Probability of hitting a target by three marksmen are frac{1}{2},,frac{1}{3} and frac{1}{4} respecti

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14.Probability

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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

A

$\frac{{11}}{{24}}$

B

$\frac{1}{{12}}$

C

$\frac{1}{8}$

D

None of these

Solution

(a) Here $P(A) = \frac{1}{2},$ $P(B) = \frac{1}{3},$ $P(C) = \frac{1}{4}$

Hence 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).$

Standard 11

Mathematics

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