Question

In a toy factory, the machines A, B, and C are used to manufacture 30%, 40%, and 30% of the output respectively. The probability that a toy produced by machines A, B, and C is defective is 2%, 3%, and 1% respectively. A toy is found to be defective. The probability that it was manufactured by machine B is:

• A. \( \frac{4}{5} \)
• B. \( \frac{2}{9} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{4}{7} \)

Hint:

Bayes' Theorem}
Use when reverse probability is asked
Total probability helps in denominator
Multiply prior and conditional probabilities

Answer 1

The Correct Option is D

Let: \[ P(A) = 0.3, \quad P(B) = 0.4, \quad P(C) = 0.3 \] \[ P(D|A) = 0.02, \quad P(D|B) = 0.03, \quad P(D|C) = 0.01 \] Total probability of defect: \[ P(D) = 0.3 \cdot 0.02 + 0.4 \cdot 0.03 + 0.3 \cdot 0.01 = 0.006 + 0.012 + 0.003 = 0.021 \] Bayes’ theorem: \[ P(B|D) = \frac{P(B) \cdot P(D|B)}{P(D)} = \frac{0.4 \cdot 0.03}{0.021} = \frac{0.012}{0.021} = \frac{4}{7} \]