JEE Advanced 2025 — Paper 2 — MATH-III solution

JEE Advanced2025Paper 2MATH-III

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 ________.

Your answer

Solution

Step 1: Scale to $100$ bulbs.

By the production ratio $2 : 2 : 1$ , $M_{1}$ makes $40$ , $M_{2}$ makes $40$ , $M_{3}$ makes $20$ . Total defectives $= 20$ .

Step 2: Defectives from $M_{1}$ .

$15%$ of $40 = 6$ defectives from $M_{1}$ .

Step 3: Defectives from $M_{2}$ via Bayes.

$P (M_{2} ∣ def) = \frac{defectives from M _{2}}{20} = \frac{2}{5}$ , so defectives from $M_{2} = 8$ .

Step 4: Defectives from $M_{3}$ .

$20 - 6 - 8 = 6$ defectives out of $20$ from $M_{3}$ , giving $P (def ∣ M_{3}) = \frac{6}{20} = 0.30$ .

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