A problem in statistics is given to three students \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). 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 problem is solved, is

\(\mathrm{P}(\mathrm{A})=\frac{1}{2}\)
\(\therefore \mathrm{P}\left(\mathrm{A}^{\prime}\right) =1-\frac{1}{2}=\frac{1}{2}\)
\(\mathrm{P}(\mathrm{B}) =\frac{1}{3}\)
\(\therefore \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3} \)
\( \mathrm{P}(\mathrm{C})=\frac{1}{4} \)
\( \therefore \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{1}{4}=\frac{3}{4} \)
\( \therefore \mathrm{P}(\mathrm{Problem} \text { is not solve }) \)
\( =\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right) \)
\( =\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right) \)
\( \ldots\left[\because \mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{C}^{\prime} \text { are independent }\right]\)
\(=\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\)
\(=\frac{1}{4}\)
\(\therefore \text { P(the problem will be solve) } \)
\( =1-\mathrm{P} \text { (Problem is not solved) } \)
\( =1-\frac{1}{4} \)
\( =\frac{3}{4}\)