The probability of solving a problem by three | vedclass

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(B) Given,$P(A) = \frac{1}{2}, P(B) = \frac{1}{4}, P(C) = \frac{1}{3}$.Therefore,the probabilities of not solving the problem are $P(\bar{A}) = 1 - \f

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$\frac{1}{4}$

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(B) Given,$P(A) = \frac{1}{2}, P(B) = \frac{1}{4}, P(C) = \frac{1}{3}$.Therefore,the probabilities of not solving the problem are $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\bar{B}) = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(\bar{C}) = 1 - \frac{1}{3} = \frac{2}{3}$.The problem is solved by exactly two persons if ($A$ and $B$ solve,$C$ fails) or ($A$ and $C$ solve,$B$ fails) or ($B$ and $C$ solve,$A$ fails).Required Probability $= P(A)P(B)P(\bar{C}) + P(A)P(\bar{B})P(C) + P(\bar{A})P(B)P(C)$.$= (\frac{1}{2} 	imes \frac{1}{4} 	imes \frac{2}{3}) + (\frac{1}{2} 	imes \frac{3}{4} 	imes \frac{1}{3}) + (\frac{1}{2} 	imes \frac{1}{4} 	imes \frac{1}{3})$.$= \frac{2}{24} + \frac{3}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$.

The probability of solving a problem by three persons $A, B$ and $C$ independently is $\frac{1}{2}, \frac{1}{4}$ and $\frac{1}{3}$ respectively. Then the probability that the problem is solved by exactly two of them is

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