A Problem From JEE 2025

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A Problem From JEE 2025

image containing math problem statement

Here is a problem on conditional probability from the toughest exam in India, the JEE. This problem appeared in the JEE 2025.

Problem contributed by debosmita1729

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Let $p_1, p_2, p_3$ be the probabilities that students $S_1, S_2, S_3$ solve the problem, respectively. Assuming independence:

$P(V) = P(S_1 \mid S_2^c \cap S_3^c) = p_1 = \dfrac{1}{10}$

$P(W) = P(S_2 \cap S_3^c) = p_2(1 - p_3) = \dfrac{1}{12}$

$P(U) = 1 - (1 - p_1)(1 - p_2)(1 - p_3) = \dfrac{1}{2}$

Substitute $p_1 = \dfrac{1}{10}$ into the equation for $P(U)$:

$1 - \dfrac{9}{10}(1 - p_2)(1 - p_3) = \dfrac{1}{2}$

$\Rightarrow (1 - p_2)(1 - p_3) = \dfrac{5}{9}$

Expanding $(1 - p_2)(1 - p_3)$ gives $(1 - p_3) - p_2(1 - p_3)$.

Substituting $P(W) = \dfrac{1}{12}$, we get:

$(1 - p_3) - \dfrac{1}{12} = \dfrac{5}{9}$

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$ \Rightarrow 1 - p_3 = \dfrac{20}{36} + \dfrac{3}{36} = \dfrac{23}{36}$

$\Rightarrow p_3 = 1 - \dfrac{23}{36} = \dfrac{13}{36}$

Thus, $P(T) = p_3 = \dfrac{13}{36}$.