Three students $S_{1}, S_{2}$ and $S_{3}$ 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

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Probability XII

Three students

S_{1}, S_{2}

and

S_{3}

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

S_{2}

nor

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

Answer: (A)

From

P (U) = \frac{1}{2}

, we have

P (S_{1}^{'} \cap S_{2}^{'} \cap S_{3}^{'}) = \frac{1}{2}

, so

(1 - P (S_{1})) (1 - P (S_{2})) (1 - P (S_{3})) = \frac{1}{2}

From

P (V) = \frac{1}{10}

, we get

P (S_{1}) = \frac{1}{10}

, so

(1 - \frac{1}{10}) (1 - P (S_{2})) (1 - P (S_{3})) = \frac{1}{2}

(1 - P (S_{2})) (1 - P (S_{3})) = \frac{5}{9}

From

P (W) = \frac{1}{12}

, we have

P (S_{2}) (1 - P (S_{3})) = \frac{1}{12}

.

Dividing these equations:

\frac{P ( S _{2} )}{1 - P ( S _{2} )} = \frac{1}{12} \times \frac{9}{5} = \frac{3}{20}

, giving

P (S_{2}) = \frac{3}{23}

.

Substituting back:

\frac{3}{23} (1 - P (S_{3})) = \frac{1}{12}

, so

1 - P (S_{3}) = \frac{23}{36}

.

Therefore,

P (S_{3}) = P (T) = \frac{13}{36}

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option