Let the respective events of solving the problem be denoted by A, B, C.

Then P(A) = \(\frac{1}{2}\) , P(B) = \(\frac{1}{3} \) , P(C) = \(\frac{1}{4}\)

Clearly, A, B, C are independent events and the problem will be considered to have been solved if at least one student solves it.

∴ Required probability = P(A or B or C) = \(P(A\,\cup\,B\,\cup\,C)\) = \(1- P(\overline{A})P(\overline{B})P(\overline{C})\)

\(\overline{A}, \overline{B},\overline{C}\) are the respective events of not solving the problem.

Also,

\(P(\overline{A}) =1-P(A) =1- \frac{1}{2} =\frac{1}{2}\) ,

\(P(\overline{B}) =1-P(B) =1- \frac{1}{3} =\frac{2}{3}\)

\(P(\overline{C}) =1-P(C) =1- \frac{1}{4} =\frac{3}{4}\)

∴** Required probability **= \(1- P(\overline{A})P(\overline{B})P(\overline{C}) =1- \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4} =1-\frac{1}{4} \)

**\(=\;\frac{3}{4}\)**