Let \(E_{1}, E_{2}, E_{3}\) be the eventsthat the problem issolved by \(A, B, C\) respectively and let \(\mathrm{p}_{1}, \mathrm{P}_{2}, \mathrm{p}_{3}\) be corresponding probabilities. Then,

\(\mathrm{P}_{1}=P\left(\mathrm{E}_{1}\right)=\frac{1}{3}, \mathrm{P}_{2}=P\left(\mathrm{E}_{2}\right)=\frac{2}{7}, \mathrm{P}_{3}=\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{3}{8}, \mathrm{q}_{1}=\mathrm{P}\left(\overline{\boldsymbol{E}}_{1}\right)=1-\frac{1}{3}=\frac{2}{3}\),

\(\mathrm{q}_{2}=\mathrm{P}\left(\bar{E}_{2}\right)=1-\frac{2}{7}=\frac{5}{7}, \mathrm{q}_{3}=\mathrm{P}\left(\overline{\boldsymbol{E}}_{3}\right)=1-\frac{3}{8}=\frac{5}{8} .\)

The problem will be solved by exactly one of them if it happens in the following mutually exclusive ways:

(1) A solves and \(B\), and \(C\) do not solve;

(2) B solves and \(A\), and \(C\) do not solve;

(3) C solves and \(A\), and \(B\) do not solve;

Required probability \(=p_{1} q_{2} q_{3}+q_{1} p_{2} q_{3}+q_{1} q_{2} p_{3}\)

\(=\frac{1}{3} \times \frac{5}{7} \times \frac{5}{8}+\frac{2}{3} \times \frac{2}{7} \times \frac{5}{8}+\frac{2}{3} \times \frac{5}{7} \times \frac{3}{8}=\frac{25}{168}+\frac{5}{42}+\frac{5}{28}=\)

\(\frac{25}{56}\).