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(B) Let $P(A), P(B),$ and $P(C)$ be the probabilities of students $A, B,$ and $C$ solving the problem respectively.$P(A) = \frac{1}{2}, P(B) = \frac{1

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$\frac{3}{4}$

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(B) Let $P(A), P(B),$ and $P(C)$ be the probabilities of students $A, B,$ and $C$ solving the problem respectively.$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$The probability that the problem is not solved by any of them is the probability that all three fail to solve it.$P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$$P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$$P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4}$Since the students try independently,the probability that none of them solve the problem is:$P(	ext{none solve}) = P(A') 	imes P(B') 	imes P(C') = \frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4} = \frac{1}{4}$The probability that the problem is solved is the complement of the probability that it is not solved:$P(	ext{solved}) = 1 - P(	ext{none solve}) = 1 - \frac{1}{4} = \frac{3}{4}$

$A$ problem in statistics is given to three students $A, B$ and $C$. Their probabilities of solving the problem are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. If all of them try independently,then the probability that the problem is solved is

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