JEE Advanced 2025 Mathematics question

MathematicsProbabilityJEE Advanced 2025

A factory has a total of three manufacturing units, , , and , which produce bulbs independent of each other. The units , , and 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 , 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 is . If a bulb is chosen randomly from the bulbs produced by , then the probability that it is defective is .

Answer: 0.3

Solution

P(M₁) = P(M₂) = 0.4, P(M₃) = 0.2 and P(D) = 0.2. P(M₂ ∩ D) = P(M₂|D)P(D) = 0.4 × 0.2 = 0.08. P(M₁ ∩ D)

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Question ID VK-035476