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 = \(1- P(\overline{A})P(\overline{B})P(\overline{C}) =1- \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4} =1-\frac{1}{4} \)
\(=\;\frac{3}{4}\)
