Defective Bulbs from Two Suppliers
A school buys LED bulbs from two suppliers. Supplier A provides 60% of the bulbs and has a 3% defect rate; Supplier B provides 40% and has an 8% defect rate. A bulb is chosen at random from the combined stock and found to be defective. The purchase officer wants to know which supplier is more likely to be responsible. Separately, a carton of 20 bulbs from Supplier A is inspected. The school's policy is to reject a carton if it contains 3 or more defective bulbs; the supplier argues that with a 3% defect rate such rejections should be rare. The officer must compute the probability that a carton from A is rejected, the expected number of defectives in 20 bulbs, and judge whether finding 3 defectives is strong evidence that A's defect rate has risen above 3%. The two suppliers' rates are assumed stable and independent.
1. Given that a randomly chosen bulb is defective, find the probability that it came from Supplier B.
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Let D be the event that a bulb is defective. By the law of total probability, P(D) = P(A)P(D|A) + P(B)P(D|B) = 0.60×0.03 + 0.40×0.08 = 0.018 + 0.032 = 0.050.
By Bayes' theorem, P(B|D) = P(B)P(D|B)/P(D) = 0.032/0.050 = 0.64. So 64% of defective bulbs come from Supplier B, even though B supplies only 40% of the stock. Supplier B is more likely to be responsible for a defective bulb because its defect rate is higher.
2. For a carton of 20 bulbs from Supplier A, find the expected number of defectives and the probability that the carton is rejected under the 3-or-more rule.
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Let X be the number of defectives in 20 bulbs from A. Then X ~ Binomial(n = 20, p = 0.03). The expected number is np = 20×0.03 = 0.6 defective bulbs.
The rejection probability is P(X ≥ 3) = 1 − P(X = 0) − P(X = 1) − P(X = 2). P(X = 0) = 0.97^20 ≈ 0.5438. P(X = 1) = C(20,1)(0.03)(0.97^19) ≈ 20×0.03×0.5606 ≈ 0.3364. P(X = 2) = C(20,2)(0.03²)(0.97^18) ≈ 190×0.0009×0.5780 ≈ 0.0988. Thus P(X ≥ 3) ≈ 1 − (0.5438 + 0.3364 + 0.0988) = 1 − 0.9790 = 0.0210, about 2.1%.
3. The inspector finds 3 defective bulbs in one carton. Is this strong evidence that Supplier A's defect rate has risen above 3%? Use a probability argument.
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Under the stated 3% rate, P(X = 3) = C(20,3)(0.03³)(0.97^17) ≈ 1140×0.000027×0.5958 ≈ 0.0183, and P(X ≥ 3) ≈ 0.0210. So seeing 3 or more defectives is a rare event, about 2.1%, if the rate is still 3%.
That is some evidence against the 3% claim, but it is not decisive from a single carton: rare events do happen, and the inspector selected only one carton of 20. A better judgement is to reject the carton under school policy but not yet conclude the rate has risen. The school should inspect several more cartons or a larger sample; if 3-or-more defectives recur at anywhere near the observed frequency, the evidence for an increased rate becomes strong.