Probability · JEE Advanced previous year questions

2025 2 questions

1. Q22025 · 18 May 2025 · Paper 1

   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 $$P(U)=\frac{1}{2},P(V)=\frac{1}{10},\text{and}P(W)=\frac{1}{12},$$ then $P(T)$ is equal to

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

2. Q112025 · 18 May 2025 · Paper 2· Numerical answer

   A factory has a total of three manufacturing units, $M_1,M_2$, and $M_3$, which produce bulbs independent of each other. The units $M_1,M_2$, and $M_3$ produce bulbs in the proportions of $2:2:1$, respectively. It is known that 20% of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_1$, 15% are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_2$ is $\frac{2}{5}$. If a bulb is chosen randomly from the bulbs produced by $M_3$, then the probability that it is defective is ____.

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   Answer: 0.27 to 0.33