Question

A manufacturing company has 3 units A, B, and C which produce 25%, 35%, 40% of bulbs respectively. 5%, 4%, and 2% of their production is defective. If a bulb is found defective, the probability it came from B is

• A. \(\dfrac{28}{69}\)
• B. \(\dfrac{28}{71}\)
• C. \(\dfrac{29}{67}\)

Hint:

Apply Bayes’ theorem with weighted probabilities to find conditional outcomes.

Answer 1

The Correct Option is A

Bayes’ Theorem:
Let D = defective, then: \[ P(B|D) = \dfrac{P(B)P(D|B)}{\sum P(i)P(D|i)} = \dfrac{0.35 \cdot 0.04}{0.25 \cdot 0.05 + 0.35 \cdot 0.04 + 0.4 \cdot 0.02} = \dfrac{0.014}{0.0125 + 0.014 + 0.008} = \dfrac{0.014}{0.0345} = \dfrac{28}{69} \]