Example-2: In a bolt factory machines A, B and C manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4, 2 percent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What are the probabilities that it was manufactured by machines A, B and C?See answer

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Example 2 In a bolt factory machines A B and C manufacture respectively 25 35 and 40 of the total Of their output 5 4 2 percent are defective bolts A bolt is drawn at random from the product and is…

Question

Example-2: In a bolt factory machines A, B and C manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4, 2 percent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What are the probabilities that it was manufactured by machines A, B and C?

Basic Answer

Step 1: Define the probabilities of each machine producing the bolt

• Machine A produces 25% of the bolts: $P(A)=0.25$
• Machine B produces 35% of the bolts: $P(B)=0.35$
• Machine C produces 40% of the bolts: $P(C)=0.40$

Step 2: Define the probabilities of a bolt being defective given it was produced by each machine

• Probability of a defective bolt from Machine A: $P(D|A)=0.05$
• Probability of a defective bolt from Machine B: $P(D|B)=0.04$
• Probability of a defective bolt from Machine C: $P(D|C)=0.02$

Step 3: Calculate the total probability of drawing a defective bolt

Using the law of total probability:
$$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$$
$$P(D)=(0.05\times 0.25)+(0.04\times 0.35)+(0.02\times 0.40)$$
$$P(D)=0.0125+0.014+0.008$$
$$P(D)=0.0345$$

Step 4: Use Bayes’ Theorem to find the probability that the defective bolt was produced by each machine

• Probability that the defective bolt was from Machine A:

$$P(A|D)=\frac{P(D|A)P(A)}{P(D)}=\frac{0.05\times 0.25}{0.0345}\approx 0.362$$

• Probability that the defective bolt was from Machine B:

$$P(B|D)=\frac{P(D|B)P(B)}{P(D)}=\frac{0.04\times 0.35}{0.0345}\approx 0.406$$

• Probability that the defective bolt was from Machine C:

$$P(C|D)=\frac{P(D|C)P(C)}{P(D)}=\frac{0.02\times 0.40}{0.0345}\approx 0.232$$

Final Answer

The probabilities that the defective bolt was manufactured by machines A, B, and C are approximately:

• Machine A: 0.362
• Machine B: 0.406
• Machine C: 0.232