Probability
PYP_JEE_ADV_2025_P1
Grade None
Question:
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
$$P(U) = \dfrac{1}{2}, \quad P(V) = \dfrac{1}{10}, \quad \text{and} \quad P(W) = \dfrac{1}{12},$$
then $P(T)$ is equal to
$\dfrac{13}{36}$
$\dfrac{1}{3}$
$\dfrac{19}{60}$
$\dfrac{1}{4}$
Step-by-Step Solution
Key Concept: Independent events, conditional probability reduces to marginal when events are independent
Let $p_1, p_2, p_3$ be probabilities that $S_1, S_2, S_3$ solve the problem (independently).
$P(U) = 1 - (1-p_1)(1-p_2)(1-p_3) = \dfrac{1}{2}$, so $(1-p_1)(1-p_2)(1-p_3) = \dfrac{1}{2}$.
$P(V) = P(S_1 \text{ solves} | S_2, S_3 \text{ don't}) = p_1 = \dfrac{1}{10}$ (since events are independent).
$P(W) = p_2(1-p_3) = \dfrac{1}{12}$.
From $p_1 = \dfrac{1}{10}$: $(1-p_1) = \dfrac{9}{10}$.
So $(1-p_2)(1-p_3) = \dfrac{1}{2} \cdot \dfrac{10}{9} = \dfrac{5}{9}$.
Let $q_2 = 1-p_2$, $q_3 = 1-p_3$. Then $q_2 q_3 = \dfrac{5}{9}$ and $(1-q_2)(q_3) = p_2 q_3 = \dfrac{1}{12}$, i.e., $q_3 - q_2 q_3 = \dfrac{1}{12}$, so $q_3 - \dfrac{5}{9} = \dfrac{1}{12}$, giving $q_3 = \dfrac{5}{9} + \dfrac{1}{12} = \dfrac{20+3}{36} = \dfrac{23}{36}$.
Thus $p_3 = P(T) = 1 - \dfrac{23}{36} = \dfrac{13}{36}$.
Correct Answer: A