JEE Advanced 2025 · Paper 1 · Mathematics Q2

MathematicsSingle correct+3 / -1Question 2 of 48

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

$P(U) = \frac12,\qquad P(V) = \frac{1}{10},\qquad \text{and}\qquad P(W) = \frac{1}{12},$

then $P(T)$ is equal to

\dfrac{13}{36}

\dfrac13

\dfrac{19}{60}

\dfrac14

Answer and worked solution

Official answer: A

Let $p_1,p_2,p_3$ be the probabilities that $S_1,S_2,S_3$ solve the problem, independently.

Translate the events.

$P(V)=P(S_1\mid \bar S_2\cap\bar S_3)=p_1=\frac1{10}$ .
$P(U)=1-(1-p_1)(1-p_2)(1-p_3)=\frac12$ , so $(1-p_2)(1-p_3)=\frac{1/2}{9/10}=\frac59$ .
$P(W)=p_2(1-p_3)=\frac1{12}$ .

Solve for $p_3$ . Since $p_2(1-p_3)=(1-p_3)-(1-p_2)(1-p_3)$ :
$1-p_3=\frac1{12}+\frac59=\frac{23}{36}\implies P(T)=p_3=\frac{13}{36}.$

JEE Advanced 2025 · Paper 1, Mathematics question 2