**Given :** let A , B and C be three students whose chances of solving a problem is given i.e ,

P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{4}\) and P(C) = \(\frac{1}{6}.\)

\(\Rightarrow P(\overline A)=\frac{2}{3},\) P(\(\overline B\)) = \(\frac{3}{4}\) and P(\(\overline C\)) = \(\frac{5}{6}\)

**To Find: **The probability that excatly one of them will solve it .

Now, P(excatly one of them will solve it)

= P(A and not B and not c) +P (B and not A and not C) +P (C and not A and not B)

= \(\frac{31}{72}\)

**Therefore ,** The probability that excatly one of them will solve the problem is\(\frac{31}{72}\)