CA Foundation · Quantitative Aptitude · Probability
Factory A makes 60% of the bulbs supplied to a Pune dealer and Factory B makes the other 40%. Of Factory A's bulbs 2% are defective, and of Factory B's bulbs 5% are defective. A bulb picked at random is found defective. What is the probability that it came from Factory A?
The probability is 3/8. Factory A contributes 0.6 times 0.02, which is 0.012, to defective bulbs, and Factory B contributes 0.4 times 0.05, which is 0.020. Applying Bayes' theorem, 0.012 divided by the total 0.032 equals 3/8.
- A3/8Correct
- B3/5
- C5/8
- D0.012
Explanation
P(defective and A) = 0.6 x 0.02 = 0.012 and P(defective and B) = 0.4 x 0.05 = 0.020. Total P(defective) = 0.032. By Bayes' theorem, P(A | defective) = 0.012/0.032 = 3/8. The value 3/5 is only the prior share of Factory A and ignores the defect information.
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