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JEE Advanced Mathematics Probability 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Mathematics Question (2025) — Solution

Question

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 2 5 . If a bulb is chosen randomly from the bulbs produced by M_3, then the probability that it is defective is ____.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Now given probability aligned & P ( Produced by M _2 defective )= 40 100 14- x 40 20 100 = 2 5 \\ & 14- x 40 = 1 5 \\ & 14- x =8 \\ & x =6 aligned So, the required probability = 6 20 =0.3

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