Q18 / 48 · JEE Advanced 2025 (Paper 1) · Mathematics
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
- Correct answerA$\frac{13}{36}$
- B$\frac{1}{3}$
- C$\frac{19}{60}$
- D$\frac{1}{4}$
Solution
$\begin{aligned}
& \mathrm{P}(\mathrm{U})=1-\mathrm{P}\left(\mathrm{~S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} \\
& \Rightarrow \mathrm{P}\left(\mathrm{~S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} ; \mathrm{P}\left(\mathrm{~S}_1^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{~S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{~S}_3^{\prime}\right)=\frac{1}{2} \\
& \Rightarrow\left(1-\mathrm{P}\left(\mathrm{~S}_1\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{2} \quad \ldots(1)
\end{aligned}$
$\begin{aligned} & \mathrm{P}(\mathrm{V})=\frac{\mathrm{P}\left(\mathrm{S}_1 \cap \mathrm{~S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}{\mathrm{P}\left(\mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}=\frac{1}{10} \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right) \cdot \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \mathrm{P}\left(\mathrm{S}_3^{\prime}\right)=\frac{1}{10} \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{S}_3^{\prime}\right) \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right)=\frac{1}{10}\end{aligned}$
$\begin{aligned}
& \mathrm{P}(\mathrm{~W})=\mathrm{P}\left(\mathrm{~S}_2 \cap \mathrm{~S}_3^{\prime}\right)=\frac{1}{12} \\
& \mathrm{P}\left(\mathrm{~S}_2\right) \cdot \mathrm{P}\left(\mathrm{~S}_3^{\prime}\right)=\frac{1}{12} \\
& \mathrm{P}\left(\mathrm{~S}_2\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{12}...(2)
\end{aligned}$
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)
$\begin{aligned}
& \frac{3}{23}\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{12} \\
& 1-\mathrm{P}\left(\mathrm{~S}_3\right)=\frac{23}{36} \\
& \mathrm{P}\left(\mathrm{~S}_3\right)=\frac{13}{36} \\
& \mathrm{P}(\mathrm{~T})=\frac{13}{36}
\end{aligned}$
Hence, the correct answer is option: 'A'.