JEE Advanced 2025 Paper 1 Q02 Mathematics P&C and Probability Probability Medium

JEE Advanced 2025 Paper 1 · Q02 · Probability

Three students $S_1, S_2, S_3$ are given a problem to solve. Consider the events: $U$: At least one of $S_1, S_2, S_3$ 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. Given $P(U) = \frac{1}{2}$, $P(V) = \frac{1}{10}$, $P(W) = \frac{1}{12}$, find $P(T)$.

  1. A. $\frac{13}{36}$
  2. B. $\frac{1}{3}$
  3. C. $\frac{19}{60}$
  4. D. $\frac{1}{4}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

Let $p_i = P(S_i \text{ solves})$, and assume independence. $P(V) = P(S_1 \text{ solves} \mid S_2, S_3 \text{ fail}) = p_1 = \frac{1}{10}$ (independence). $P(U) = 1 - (1-p_1)(1-p_2)(1-p_3) = \frac{1}{2}$ $\Rightarrow$ $(9/10)(1-p_2)(1-p_3) = 1/2$ $\Rightarrow$ $(1-p_2)(1-p_3) = 5/9$. $P(W) = p_2 (1-p_3) = \frac{1}{12}$ $\Rightarrow$ $p_2 = 1 / (12(1-p_3))$. Substitute $(1-p_2) = 1 - 1/(12(1-p_3))$ into $(1-p_2)(1-p_3) = 5/9$: $(12(1-p_3) - 1)/(12(1-p_3)) \cdot (1-p_3) = 5/9$, giving $12(1-p_3) - 1 = 20/3$, so $1-p_3 = 23/36$, and $p_3 = \frac{13}{36}$. Answer: $\frac{13}{36}$.

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