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(C) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4},$ and $P(D) = \frac{1}{5}$ be the probabilities of the $4$ students solving the pr

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$\frac{4}{5}$

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(C) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4},$ and $P(D) = \frac{1}{5}$ be the probabilities of the $4$ students solving the problem individually. The probability that the problem is solved by at least one student is given by $1 - P(	ext{none of the students solve the problem}).$ The probability that a student fails to solve the problem is $P(\bar{X}) = 1 - P(X).$ Thus, $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}, P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}, P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4},$ and $P(\bar{D}) = 1 - \frac{1}{5} = \frac{4}{5}.$ Since the events are independent, the probability that none of them solve the problem is $P(\bar{A} \cap \bar{B} \cap \bar{C} \cap \bar{D}) = P(\bar{A}) 	imes P(\bar{B}) 	imes P(\bar{C}) 	imes P(\bar{D}).$ $P(	ext{none}) = \frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4} 	imes \frac{4}{5} = \frac{1}{5}.$ Therefore, the probability that the problem is solved by at least one student is $1 - \frac{1}{5} = \frac{4}{5}.$

$A$ problem in mathematics is given to $4$ students whose chances of solving it individually are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4},$ and $\frac{1}{5}.$ The probability that the problem will be solved by at least one student is

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