JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Probability XII

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}

\frac{2}{5}

.
If a bulb is chosen randomly from the bulbs produced by

M_{3}

, then the probability that it is defective is ____.

Answer: 0.3

Solution

Let the proportions be:

P (M_{1}) = \frac{2}{5}

P (M_{2}) = \frac{2}{5}

P (M_{3}) = \frac{1}{5}

.

Let

D

denote defective. Given:

$P (D) = 0.20$
$P (D ∣ M_{1}) = 0.15$
$P (M_{2} ∣ D) = \frac{2}{5}$

Let

P (D ∣ M_{2}) = p_{2}

and

P (D ∣ M_{3}) = p_{3}

.

By total probability:

P (D) = P (M_{1}) P (D ∣ M_{1}) + P (M_{2}) P (D ∣ M_{2}) + P (M_{3}) P (D ∣ M_{3})

0.20 = \frac{2}{5} \cdot 0.15 + \frac{2}{5} \cdot p_{2} + \frac{1}{5} \cdot p_{3}

0.20 = 0.06 + 0.4 p_{2} + 0.2 p_{3}

0.14 = 0.4 p_{2} + 0.2 p_{3} ...(i)

By Bayes' theorem:

P (M_{2} ∣ D) = \frac{P ( M _{2} ) P ( D ∣ M _{2} )}{P ( D )} = \frac{\frac{2}{5} \cdot p _{2}}{0.20} = \frac{2 p _{2}}{1} = 2 p_{2}

Given

P (M_{2} ∣ D) = \frac{2}{5}

2 p_{2} = \frac{2}{5} \Rightarrow p_{2} = \frac{1}{5} = 0.20

Substituting into (i):

0.14 = 0.4 (0.20) + 0.2 p_{3} = 0.08 + 0.2 p_{3}

0.06 = 0.2 p_{3} \Rightarrow p_{3} = 0.30

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer

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JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Probability XII

Solution

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