Defective LED Bulbs: Defect Rate Is Not the Same as Blame Share
A factory uses three machines A, B and C to make LED bulbs. Machine A makes 50% of the daily output, B makes 30% and C makes 20%. Historical quality records show defect rates of 2% for A, 3% for B and 5% for C. A quality inspector picks one bulb at random from the day's combined output and finds it defective. The production manager says, 'Machine C has the highest defect rate, so it is most likely that this defective bulb came from C.' The maintenance team wants to know the overall defective rate to set a daily rejection budget, and the manager wants the probability that the defective bulb came from each machine. A second bulb is then picked at random with replacement from the same day's output. You are asked to use conditional probability, check the manager's claim, and decide what the posterior probabilities imply for where to focus repairs.
1. Find the overall probability that a randomly chosen bulb is defective.
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Use the law of total probability. Let D be the event that a bulb is defective.
P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) = 0.50(0.02) + 0.30(0.03) + 0.20(0.05) = 0.010 + 0.009 + 0.010 = 0.029.
The overall defective rate is 2.9%, i.e. 29 defective bulbs per 1,000 produced, before any sorting costs.
2. Given that the chosen bulb is defective, find the probability it came from each machine. Does the manager's claim hold?
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By Bayes' theorem:
P(A|D) = 0.010/0.029 = 10/29 ≈ 0.345 P(B|D) = 0.009/0.029 = 9/29 ≈ 0.310 P(C|D) = 0.010/0.029 = 10/29 ≈ 0.345.
The manager's claim is not correct as stated. C has the highest defect rate, but it makes only 20% of output, so its posterior probability is about 34.5% and ties with A; it is not the most likely source in the sense of a majority. A and C each contribute 1.0% of total output as defective, while B contributes 0.9%.
For repairs, C is attractive because a small output share produces a large defective share, so reducing C's 5% rate could cut defects efficiently. A also matters because its large volume generates the same number of defectives. Posterior probabilities, not defect rates alone, should guide priorities.
3. A second bulb is picked with replacement. Find the probability it is also defective, and the probability that both defective bulbs came from C. Why does replacement matter?
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With replacement, the second draw has the same unconditional defective probability:
P(D₂) = 0.029.
Because the draws are independent, P(D₁ and D₂) = 0.029² = 0.000841, about 0.0841%.
Given that both are defective,
P(both from C | both D) = [P(C)² P(D|C)²] / [P(D)²] = (0.2² × 0.05²) / (0.029²) = 0.0001 / 0.000841 ≈ 0.1189.
Equivalently, conditional on a defective bulb the posterior for C is 10/29, and independence across draws gives (10/29)² = 100/841 ≈ 0.119, about 11.9%.
Replacement matters because it restores independence and keeps the machine proportions fixed. Without replacement from a finite day's production, the probabilities would change slightly after the first bulb is removed.