Two-Machine Defect Testing: Bayes and Independence
A bulb assembly unit in Guwahati has two machines. Machine M1 makes 60% of daily output and M2 makes 40%. Historical records show 2% of M1 bulbs and 5% of M2 bulbs are defective. An inspector picks one bulb at random from the combined output and finds it defective. The floor supervisor says the bulb probably came from M2 because M2 has the higher defect rate. A quick test flags 90% of defective bulbs correctly but also flags 8% of good bulbs. A retailer rejects a lot if the probability that a randomly chosen bulb is defective exceeds 3%. The day's output is 5,000 bulbs. The supervisor wants the probability the defective bulb came from M2, the overall defect rate, and whether the lot passes the 3% rule. He also claims the test result and the machine are independent because 'defectiveness is random'.
1. Find the overall defective probability and P(M2|D). Is the supervisor's conclusion supported? Does the lot pass the 3% rule?
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Let D be defective. By the law of total probability, P(D)=P(M1)P(D|M1)+P(M2)P(D|M2)=0.6(0.02)+0.4(0.05)=0.012+0.020=0.032. Thus 3.2% of the day's 5,000 bulbs, about 160 bulbs, are expected to be defective. By Bayes' theorem, P(M2|D)=P(M2)P(D|M2)/P(D)=0.020/0.032=0.625. So conditional on being defective, the bulb is more likely to have come from M2 (62.5%) than M1 (37.5%). The supervisor's conclusion is numerically supported, but the reasoning must use the defect rates as well as the production shares. Since 3.2% exceeds 3%, the lot fails the retailer's rule.
2. Using the test, compute P(D|T+). Interpret the result and explain why it is not simply 90%.
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We have P(T+|D)=0.90 and P(T+|G)=0.08, where G means good. First P(T+)=P(T+|D)P(D)+P(T+|G)P(G)=0.90(0.032)+0.08(0.968)=0.0288+0.07744=0.10624. Then P(D|T+)=P(T+|D)P(D)/P(T+)=0.0288/0.10624≈0.271. A positive test therefore means about a 27% chance that the bulb is defective, not 90%. The 90% figure is the sensitivity P(T+|D), the probability of a positive test given defect. Because most bulbs are good, the 8% false-positive rate applied to the large good group produces many false positives, so the posterior probability of defect remains modest.
3. Test the independence claim by comparing P(D|M1) with P(D|M2), and by comparing P(T+|M1) with P(T+|M2).
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Defect status and machine are not independent because P(D|M1)=0.02 while P(D|M2)=0.05; the probability of defect changes with the machine. For the test result and machine, compute P(T+|M1)=0.90(0.02)+0.08(0.98)=0.018+0.0784=0.0964, and P(T+|M2)=0.90(0.05)+0.08(0.95)=0.045+0.076=0.121. Since 0.0964≠0.121, P(T+|M)=P(T+) does not hold for both machines, so the test result and machine are not independent. The supervisor's statement that 'defectiveness is random' does not justify independence; randomness does not remove dependence between machine and defect or test outcome.