The probabilities of A, B, and C solving a pr | vedclass

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(B) Let $P(A) = \frac{1}{3}, P(B) = \frac{2}{7}, P(C) = \frac{3}{8}$.Then the probabilities of not solving the problem are $P(A') = 1 - \frac{1}{3} =

Vedclass · Accepted Answer

$\frac{25}{56}$

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(B) Let $P(A) = \frac{1}{3}, P(B) = \frac{2}{7}, P(C) = \frac{3}{8}$.Then the probabilities of not solving the problem are $P(A') = 1 - \frac{1}{3} = \frac{2}{3}, P(B') = 1 - \frac{2}{7} = \frac{5}{7},$ and $P(C') = 1 - \frac{3}{8} = \frac{5}{8}$.Exactly one of them solves the problem if ($A$ solves and $B, C$ do not) $OR$ ($B$ solves and $A, C$ do not) $OR$ ($C$ solves and $A, B$ do not).Required probability $= P(A)P(B')P(C') + P(A')P(B)P(C') + P(A')P(B')P(C)$$= (\frac{1}{3} 	imes \frac{5}{7} 	imes \frac{5}{8}) + (\frac{2}{3} 	imes \frac{2}{7} 	imes \frac{5}{8}) + (\frac{2}{3} 	imes \frac{5}{7} 	imes \frac{3}{8})$$= \frac{25}{168} + \frac{20}{168} + \frac{30}{168} = \frac{75}{168} = \frac{25}{56}$.

The probabilities of $A, B,$ and $C$ solving a problem are $\frac{1}{3}, \frac{2}{7},$ and $\frac{3}{8}$ respectively. If all three try to solve the problem simultaneously,the probability that exactly one of them will solve it is:

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