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Worked Examples · Example 19

Q.In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

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Appeared in past exams:AP EAPCET 2025· Set eng-2025-05-21-AN· 1mreworded
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This is a classic Bayes’ theorem problem. We are given the prior probabilities of each machine producing a bolt and the conditional probabilities of a bolt being defective given the machine. The probability that a defective bolt came from machine B is approximately 0.406 (or 40.6%).


Why Bayes’ theorem?

We are asked: Given that a bolt is defective, what is the chance it came from machine B? This is a reverse probability — we know the chance of a defect given the machine, but we want the chance of the machine given a defect. That’s exactly what Bayes’ theorem handles.

The key insight: the total probability of a defective bolt is a weighted average of the defect rates of the three machines, weighted by their production shares. Then, the share of that total that comes from machine B is the answer.


Step-by-step solution

1. Define events clearly

Let:

  • AA, BB, CC = event that a randomly chosen bolt is made by machine A, B, or C respectively.
  • DD = event that the bolt is defective.

We are given:

  • P(A)=0.25P(A) = 0.25, P(B)=0.35P(B) = 0.35, P(C)=0.40P(C) = 0.40
  • P(D∣A)=0.05P(D \mid A) = 0.05, P(D∣B)=0.04P(D \mid B) = 0.04, P(D∣C)=0.02P(D \mid C) = 0.02

We want P(B∣D)P(B \mid D).

2. Find the total probability of a defective bolt

By the law of total probability:

P(D)=P(A)P(D∣A)+P(B)P(D∣B)+P(C)P(D∣C)P(D) = P(A)P(D \mid A) + P(B)P(D \mid B) + P(C)P(D \mid C)

Substitute:

P(D)=(0.25×0.05)+(0.35×0.04)+(0.40×0.02)P(D) = (0.25 \times 0.05) + (0.35 \times 0.04) + (0.40 \times 0.02)

Compute each term:

  • 0.25×0.05=0.01250.25 \times 0.05 = 0.0125
  • 0.35×0.04=0.01400.35 \times 0.04 = 0.0140
  • 0.40×0.02=0.00800.40 \times 0.02 = 0.0080

Sum:

P(D)=0.0125+0.0140+0.0080=0.0345P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345

So 3.45% of all bolts are defective.

3. Apply Bayes’ theorem

Bayes’ theorem for P(B∣D)P(B \mid D):

P(B∣D)=P(B) P(D∣B)P(D)P(B \mid D) = \frac{P(B) \, P(D \mid B)}{P(D)}

Plug in:

P(B∣D)=0.35×0.040.0345=0.01400.0345P(B \mid D) = \frac{0.35 \times 0.04}{0.0345} = \frac{0.0140}{0.0345}

4. Simplify the fraction …

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