Warranty versus inspection for LED bulbs from two machines
An LED bulb assembly unit buys bulbs from two machines. Machine A makes 60% of daily output and has a 2% defect rate; machine B makes 40% and has a 5% defect rate. A finished bulb is selected at random. Selling a defective bulb leads to a warranty replacement costing ₹120. The quality manager proposes 100% inspection, which would cost ₹3.50 per bulb and catch every defective; each defective found would be reworked for ₹40 instead of incurring the ₹120 warranty cost. The production head wants to know the overall defect rate, the chance that a defective bulb came from machine B, whether inspection is financially justified, and how to interpret a sample of 10 bulbs from machine B in which the expected number of defectives is only 0.5.
1. Find the overall defect rate and the probability that a defective bulb came from machine B. Interpret the result.
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Let D be defective, A be from machine A and B be from machine B. [ P(D) = P(A)P(D|A) + P(B)P(D|B) = 0.60(0.02) + 0.40(0.05) = 0.012 + 0.020 = 0.032. ] So the overall defect rate is 3.2%.
By Bayes' theorem: [ P(B|D) = \frac{P(B)P(D|B)}{P(D)} = \frac{0.40(0.05)}{0.032} = \frac{0.020}{0.032} = 0.625. ] Given that a bulb is defective, there is a 62.5% chance it came from machine B. Although B produces fewer bulbs, its defect rate is 2.5 times A's, so it accounts for most defective output.
2. Compare the expected warranty cost per bulb with the expected cost of 100% inspection. At ₹3.50 per inspection, should the unit inspect? Find the break-even inspection cost.
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Without inspection, the expected warranty cost per bulb is [ 0.032 \times 120 = ₹3.84. ]
With inspection costing ₹c per bulb, the expected cost per bulb is [ c + 0.032 \times 40 = c + 1.28, ] because each defective bulb found is reworked for ₹40 instead of triggering a ₹120 warranty claim.
At c = ₹3.50, the inspection cost is 3.50 + 1.28 = ₹4.78 per bulb, which is ₹0.94 higher than the no-inspection cost of ₹3.84. Inspection is therefore not justified on these figures.
Break-even: c + 1.28 = 3.84 gives c = ₹2.56. Inspection would be worthwhile only if the testing cost is below ₹2.56 per bulb, assuming 100% detection and no other benefits.
3. In a sample of 10 bulbs from machine B, find the probability of at most one defective. Comment on the statement that the expected number of defectives is 0.5.
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For machine B, p = 0.05 and n = 10, so X ~ Binomial(10, 0.05). [ P(X \le 1) = P(X=0) + P(X=1) = 0.95^{10} + \binom{10}{1}(0.05)(0.95)^9. ] Using 0.95^{10} ≈ 0.5987 and 0.95^9 ≈ 0.6302: [ P(X \le 1) \approx 0.5987 + 10(0.05)(0.6302) = 0.5987 + 0.3151 = 0.9138. ] So there is about a 91.4% chance of at most one defective in 10 bulbs.
The expected number is np = 0.5, but this is a long-run average, not a prediction for a particular sample. In fact, P(X=0) ≈ 59.9% and P(X=1) ≈ 31.5%, so zero defectives is more likely than one. Saying 'we expect exactly 0.5 defective' misunderstands the meaning of expectation.