The probability of A, B, C solving a problem | vedclass

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(B) Let $P(A) = p_1 = \frac{1}{3}$,$P(B) = p_2 = \frac{2}{7}$,and $P(C) = p_3 = \frac{3}{8}$.Then the probabilities of not solving the problem are $q_

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$\frac{25}{56}$

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(B) Let $P(A) = p_1 = \frac{1}{3}$,$P(B) = p_2 = \frac{2}{7}$,and $P(C) = p_3 = \frac{3}{8}$.Then the probabilities of not solving the problem are $q_1 = 1 - p_1 = \frac{2}{3}$,$q_2 = 1 - p_2 = \frac{5}{7}$,and $q_3 = 1 - p_3 = \frac{5}{8}$.Exactly one person 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 $= p_1 q_2 q_3 + q_1 p_2 q_3 + q_1 q_2 p_3$.$= (\frac{1}{3} 	imes \frac{5}{7} 	imes \frac{5}{8}) + (\frac{2}{3} 	imes \frac{2}{7} 	imes \frac{5}{8}) + (\frac{2}{3} 	imes \frac{5}{7} 	imes \frac{3}{8})$.$= \frac{25}{168} + \frac{20}{168} + \frac{30}{168} = \frac{75}{168}$.Simplifying by dividing by $3$,we get $\frac{25}{56}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

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

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