The probability of hitting a target by three | vedclass

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Question

(A) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.Then $P(\overline{A}) = \frac{1}{2}, P(\overline{B}) = \frac{2}{3}, P(\overline{C

Vedclass · Accepted Answer

$\frac{13}{24}$

Vedclass · Answer

(A) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.Then $P(\overline{A}) = \frac{1}{2}, P(\overline{B}) = \frac{2}{3}, P(\overline{C}) = \frac{3}{4}$.$\lambda$ is the probability that exactly two hit the target:$\lambda = P(A)P(B)P(\overline{C}) + P(A)P(\overline{B})P(C) + P(\overline{A})P(B)P(C)$$\lambda = (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{3}{4}) + (\frac{1}{2} 	imes \frac{2}{3} 	imes \frac{1}{4}) + (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{4})$$\lambda = \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$.$\mu$ is the probability that at least two hit the target:$\mu = \lambda + P(A)P(B)P(C)$$\mu = \frac{6}{24} + (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{4}) = \frac{6}{24} + \frac{1}{24} = \frac{7}{24}$.$\lambda + \mu = \frac{6}{24} + \frac{7}{24} = \frac{13}{24}$.

The probability of hitting a target by three marksmen is $\frac{1}{2}, \frac{1}{3},$ and $\frac{1}{4}$ respectively. If the probability that exactly two of them will hit the target is $\lambda$ and that at least two of them hit the target is $\mu,$ then $\lambda + \mu$ is equal to :-

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