Question 11

NUMERICALMEDIUM

A factory has a total of three manufacturing units, M1,M2M_1, M_2, and M3M_3, which produce bulbs independent of each other. The units M1,M2M_1, M_2, and M3M_3 produce bulbs in the proportions of 2:2:12:2:1, respectively. It is known that 20%20\% of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by M1,15%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 M2M_2 is 25\frac{2}{5}. If a bulb is chosen randomly from the bulbs produced by M3M_3, then the probability that it is defective is ___________.

Correct Answer: 0.3

Detailed Solution

Let P(M1)=2/5P(M_1) = 2/5, P(M2)=2/5P(M_2) = 2/5, P(M3)=1/5P(M_3) = 1/5. Let DD be the event that a bulb is defective. Given P(D)=0.20P(D) = 0.20 and P(D∣M1)=0.15P(D|M_1) = 0.15. Also given P(M2∣D)=2/5P(M_2|D) = 2/5. By Bayes' Theorem: P(M2∣D)=P(M2)P(D∣M2)P(D)  ⟹  25=(2/5)P(D∣M2)0.20  ⟹  P(D∣M2)=0.20P(M_2|D) = \frac{P(M_2)P(D|M_2)}{P(D)} \implies \frac{2}{5} = \frac{(2/5)P(D|M_2)}{0.20} \implies P(D|M_2) = 0.20. By the law of total probability: P(D)=P(M1)P(D∣M1)+P(M2)P(D∣M2)+P(M3)P(D∣M3)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=(2/5)(0.15)+(2/5)(0.20)+(1/5)P(D∣M3)0.20 = (2/5)(0.15) + (2/5)(0.20) + (1/5)P(D|M_3). 0.20=0.06+0.08+0.2P(D∣M3)  ⟹  0.06=0.2P(D∣M3)  ⟹  P(D∣M3)=0.30.20 = 0.06 + 0.08 + 0.2 P(D|M_3) \implies 0.06 = 0.2 P(D|M_3) \implies P(D|M_3) = 0.3.

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