Three students \(S_1, S_2\) and \(S_1\) are given a problem to solve. Consider the following events :
\(U:\) At least one of \(S_1, S_2\) and \(S_3\) can solve the problem,
\(V: S_1\) can solve the problem, given that neither \(\mathrm{S}_2\) nor \(\mathrm{S}_3\) can solve the problem,
\(W: S_2\) can solve the problem and \(S_3\) cannot solve the problem,
\(T: S_3\) can solve the problem.
For any event \(E\), let \(P(E)\) denote the probability of \(E\).
If \(P(U)=\frac{1}{2}, P(V)=\frac{1}{10}\) and \(P(W)=\frac{1}{12}\), then \(P(T)\) is equal to

\(\begin{aligned}
& \mathrm{P}(\mathrm{U})=1-\mathrm{P}\left(\mathrm{~S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} \\
& \Rightarrow \mathrm{P}\left(\mathrm{~S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} ; \mathrm{P}\left(\mathrm{~S}_1^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{~S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{~S}_3^{\prime}\right)=\frac{1}{2} \\
& \Rightarrow\left(1-\mathrm{P}\left(\mathrm{~S}_1\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{2} \quad \ldots(1)
\end{aligned}\)
\(\begin{aligned} & \mathrm{P}(\mathrm{V})=\frac{\mathrm{P}\left(\mathrm{S}_1 \cap \mathrm{~S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}{\mathrm{P}\left(\mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}=\frac{1}{10} \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right) \cdot \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \mathrm{P}\left(\mathrm{S}_3^{\prime}\right)=\frac{1}{10} \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{S}_3^{\prime}\right) \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right)=\frac{1}{10}\end{aligned}\)
\(\begin{aligned}
& \mathrm{P}(\mathrm{~W})=\mathrm{P}\left(\mathrm{~S}_2 \cap \mathrm{~S}_3^{\prime}\right)=\frac{1}{12} \\
& \mathrm{P}\left(\mathrm{~S}_2\right) \cdot \mathrm{P}\left(\mathrm{~S}_3^{\prime}\right)=\frac{1}{12} \\
& \mathrm{P}\left(\mathrm{~S}_2\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{12}...(2)
\end{aligned}\)
Eq. (1)
\(\left(1-\frac{1}{10}\right)\left(1-\mathrm{P}\left(\mathrm{~S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{2}\)
\(\left(1-\mathrm{P}\left(\mathrm{~S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{5}{9}....(3)\)
\(\begin{aligned}
& \frac{\text { Eq. }(2)}{\text { Eq. }(3)} \Rightarrow \frac{\mathrm{P}\left(\mathrm{~S}_2\right)}{1-\mathrm{P}\left(\mathrm{~S}_2\right)}=\frac{1}{12} \times \frac{9}{5} \\
& \mathrm{P}\left(\mathrm{~S}_2\right)=\frac{3}{23}
\end{aligned}\)
Put in Eq. (2)
\(\begin{aligned}
& \frac{3}{23}\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{12} \\
& 1-\mathrm{P}\left(\mathrm{~S}_3\right)=\frac{23}{36} \\
& \mathrm{P}\left(\mathrm{~S}_3\right)=\frac{13}{36} \\
& \mathrm{P}(\mathrm{~T})=\frac{13}{36}
\end{aligned}\)