Sampling plan or blame: two vendors and one rejected resistor lot
An electronics assembler buys 200-resistor lots from two vendors. Vendor A's process averages 2.0% defective; Vendor B's averages 5.0%. The incoming-inspection contract reads: draw 10 resistors at random without replacement from the lot, and reject the lot if two or more are defective. Because the stores area has no space for segregated bins, a rejected lot that arrives unlabelled cannot be traced to its vendor, and the purchase records for that day show that both vendors delivered. The quality manager wants three things: how often each vendor's lots will be rejected, whether a rejection is strong enough evidence to blame Vendor B, and whether raising the sample to 20 while keeping the two-or-more rule would justify the extra ₹18 per lot inspection charge. Scrap and rework of a lot that slips through costs the plant about ₹600.
1. Estimate the probability that a lot from each vendor is rejected by the 10-piece sampling plan, stating any approximation you use.
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Let X be the number of defectives in the sample and p the lot's defective fraction. The lot is rejected when X ≥ 2:
P(reject) = 1 − P(X = 0) − P(X = 1) = 1 − (1−p)¹⁰ − 10p(1−p)⁹.
Treating draws as independent is safe here because n/N = 10/200 = 0.05; the exact hypergeometric value differs by well under half a percentage point.
Vendor A, p = 0.02: 0.98¹⁰ = 0.8171; 10(0.02)(0.98)⁹ = 0.1668 → P(reject) = 0.0161 ≈ 1.6%.
Vendor B, p = 0.05: 0.95¹⁰ = 0.5987; 10(0.05)(0.95)⁹ = 0.3151 → P(reject) = 0.0862 ≈ 8.6%.
So the plan rejects roughly 1 lot in 62 from A and 1 lot in 12 from B. Note the plan's poor power: most bad lots are accepted, since a defective rate of 5% still passes 91 times in 100.
2. A lot has just been rejected. With equal prior odds, how likely is it that the lot came from Vendor B, and what should the manager actually do?
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Bayes with equal priors:
P(B | reject) = 0.0862 / (0.0862 + 0.0161) = 0.0862 / 0.1023 = 0.84.
Equivalently, the likelihood ratio is 0.0862 / 0.0161 ≈ 5.4 in favour of B.
Judgement. A rejection makes B about five times more likely than A, but 84% is not blame-grade evidence: roughly one rejection in six under this rule comes from the good vendor. The evidence is also asymmetric - the test catches only 8.6% of B's bad lots, so a passing lot says almost nothing.
The defensible actions are: quarantine and re-test the consignment, request batch certificates and process capability data from both vendors, and if the contract allows, sample every lot before mixing stock. Debiting Vendor B on a single rejection would be an over-read of a weak test. Note also that if the prior odds are not 1:1 (say B supplies 70% of volume), the posterior would rise further - the prior matters and should be stated.
3. Evaluate the proposal to raise the sample to 20 with the same two-or-more rule. Is ₹18 per lot worth it?
Read the worked answer
With n = 20:
A (p = 0.02): 1 − 0.98²⁰ − 20(0.02)(0.98)¹⁹ = 1 − 0.6676 − 0.2725 = 0.060. B (p = 0.05): 1 − 0.95²⁰ − 20(0.05)(0.95)¹⁹ = 1 − 0.3585 − 0.3774 = 0.264.
What improves. The absolute separation grows from about 7 to about 20 percentage points, and the chance of catching a B lot triples, so the plan discriminates far better in practice.
What worsens. Good lots from A are now rejected 6% of the time - roughly one in seventeen - so returns, paperwork and schedule disruption rise.
Cost check. Loss per accepted bad lot: with n = 10, 0.914 × ₹600 = ₹548; with n = 20, 0.736 × ₹600 = ₹442. Saving ≈ ₹106 per B lot against ₹18 of extra inspection, and even after adding the cost of false rejections of A lots the change looks favourable.
Judgement. Recommend the switch, but plug in the plant's own false-rejection cost and the true mix of A and B lots before amending the contract; the conclusion flips if a wrongly rejected lot is expensive to handle.