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← JEE Advanced 2025 · Paper 2
MathematicsNumerical+4Question 11 of 48

JEE Advanced 2025 · Paper 2, Mathematics question 11

A factory has a total of three manufacturing units, M1M_1, M2M_2, and M3M_3, which produce bulbs independent of each other. The units M1M_1, M2M_2, and M3M_3 produce bulbs in the proportions of 2:2:12 : 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 M1M_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 M2M_2 is 25\dfrac25.

If a bulb is chosen randomly from the bulbs produced by M3M_3, then the probability that it is defective is _____.

Answer and worked solution

Official answer: 0.27

Prior proportions. P(M1)=P(M2)=0.4P(M_1)=P(M_2)=0.4 and P(M3)=0.2P(M_3)=0.2. Overall P(D)=0.2P(D)=0.2.

Defectives from each unit.

  • M1M_1: 0.4×0.15=0.060.4\times0.15=0.06.
  • M2M_2: P(M2∩D)=25×0.2=0.08P(M_2\cap D)=\frac25\times0.2=0.08.
  • M3M_3: 0.2−0.06−0.08=0.060.2-0.06-0.08=0.06.

Answer.
P(D∣M3)=0.060.2=0.3.P(D\mid M_3)=\frac{0.06}{0.2}=0.3.