Flagged Bulbs and the Base-Rate Trap
An LED bulb plant in Noida runs three machines. Machine A makes 50% of the output with a 1.5% defect rate, machine B makes 30% with a 3% defect rate, and machine C makes 20% with a 6% defect rate. Every bulb passes an automatic tester before packing. The tester flags 90% of genuinely defective bulbs, but it also flags 4% of good bulbs because dust on the contact pins mimics a fault. The quality head wants three numbers before the monthly review: the true defect rate of the output, the chance that a bulb flagged by the tester is genuinely defective, and - once a bulb has been flagged - which machine most probably made it, so that maintenance money can be aimed at the right line. The production manager has already concluded that 'C must be the culprit, it has the worst defect rate'.
1. Compute the true defect rate of the plant's output.
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Use the theorem of total probability with D = defective and M = machine.
P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) = 0.50(0.015) + 0.30(0.03) + 0.20(0.06) = 0.0075 + 0.0090 + 0.0120 = 0.0285.
So 2.85% of bulbs are defective, about 1 in 35. Note the contributions: A and B together account for 0.0165 of the 0.0285, i.e. 58% of all defects, even though their defect rates are the lowest. Volume, not rate, drives the defect count.
2. Given that a bulb is flagged by the tester, what is the probability that it is genuinely defective? Interpret the number.
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Let + be 'flagged'. P(+|D) = 0.90 and P(+|D′) = 0.04, with P(D) = 0.0285, P(D′) = 0.9715.
P(+) = 0.0285(0.90) + 0.9715(0.04) = 0.02565 + 0.03886 = 0.06451.
P(D|+) = P(+|D)P(D)/P(+) = 0.02565/0.06451 = 0.3976.
So a flagged bulb is only about 39.8% likely to be defective - roughly six in ten flagged bulbs are actually good. The reason is the base rate: only 2.85% of bulbs are defective, so the 4% false-positive rate applied to the huge good population generates 0.0389 of flag mass against 0.0257 from real defects. The tester flags 6.45% of output but fewer than half of those flags are true. Any decision to scrap flagged bulbs destroys good stock, so a second, more specific test before scrapping is the sensible judgement.
3. Given that a bulb is flagged, which machine most probably made it? Check the production manager's claim.
Read the worked answer
Condition on the machine and then on the flag: P(+|A) = 0.90(0.015) + 0.04(0.985) = 0.0135 + 0.0394 = 0.0529 ⇒ P(A ∩ +) = 0.0529(0.50) = 0.02645 P(+|B) = 0.90(0.03) + 0.04(0.97) = 0.0270 + 0.0388 = 0.0658 ⇒ P(B ∩ +) = 0.0658(0.30) = 0.01974 P(+|C) = 0.90(0.06) + 0.04(0.94) = 0.0540 + 0.0376 = 0.0916 ⇒ P(C ∩ +) = 0.0916(0.20) = 0.01832
These sum to 0.06451, agreeing with P(+). Dividing: P(A|+) = 0.02645/0.06451 = 0.410; P(B|+) = 0.306; P(C|+) = 0.284.
A flagged bulb most probably came from A (41%), not C (28%). The manager reasoned from defect rates alone; the correct comparison includes how many bulbs each machine contributes and the false-positive stream that every good bulb can enter. C still deserves attention on a per-bulb basis - its 6% rate is four times A's - but if the aim is to cut the number of flags, A's line is the biggest single source.