JEE Advanced 2025 — Paper 1 — MATH-I solution

JEE Advanced2025Paper 1MATH-I

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

A
$\frac{13}{36}$
B
$\frac{1}{3}$
C
$\frac{19}{60}$
D
$\frac{1}{4}$

Solution

Write $p_{i} = P (S_{i})$ and $q_{i} = 1 - p_{i}$ for $i = 1, 2, 3$ , assuming independence.

From $P (U) = 1 - q_{1} q_{2} q_{3} = \frac{1}{2}$ , we get $q_{1} q_{2} q_{3} = \frac{1}{2}$ . $(1)$

For the conditional event $V$ : $P (V) = \frac{p _{1} q _{2} q _{3}}{q _{2} q _{3}} = p_{1} = \frac{1}{10}$ , so $q_{1} = \frac{9}{10}$ .

From $P (W) = p_{2} q_{3} = \frac{1}{12}$ . $(2)$

Substituting $q_{1} = \frac{9}{10}$ in $(1)$ : $q_{2} q_{3} = \frac{1}{2} \cdot \frac{10}{9} = \frac{5}{9}$ . $(3)$

Dividing $(2)$ by $(3)$ : $\frac{p _{2} q _{3}}{q _{2} q _{3}} = \frac{p _{2}}{q _{2}} = \frac{1}{12} \cdot \frac{9}{5} = \frac{3}{20}$ .

Hence $p_{2} = \frac{3}{23}$ and $q_{2} = \frac{20}{23}$ . Plugging back into $(2)$ : $\frac{3}{23} \cdot q_{3} = \frac{1}{12} \Rightarrow q_{3} = \frac{23}{36}$ .

Therefore $P (T) = p_{3} = 1 - \frac{23}{36} = \frac{13}{36}$ .

$∴$ The correct option is (A).

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