CA Foundation · Quantitative Aptitude · Probability
In a Pune factory, machines A, B and C produce 50%, 30% and 20% of the total output, and their defect rates are 2%, 3% and 5% respectively. An item picked at random is found defective. What is the probability that it was produced by machine C?
The probability is 10/29. Total defect probability is 0.01 + 0.009 + 0.01 = 0.029, and machine C contributes 0.010 of it. Bayes' theorem gives 0.010 divided by 0.029, which is 10/29, higher than C's 1/5 output share because C has the highest defect rate.
- A1/5
- B10/29Correct
- C9/29
- D1/3
Explanation
P(defective) = 0.5(0.02) + 0.3(0.03) + 0.2(0.05) = 0.01 + 0.009 + 0.010 = 0.029. By Bayes' theorem, P(C | defective) = 0.010/0.029 = 10/29. The prior share 1/5 is wrong because it ignores the higher defect rate of machine C.
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