Resistor Lot Acceptance Under a Binomial Sampling Rule
A Pune electronics supplier receives resistors in large lots. The contract says a lot is accepted if a random sample of 12 resistors contains at most one defective. The supplier's current process has a defective rate of 8%, and defects are assumed independent. The quality manager wants the probability that a lot is accepted, the expected number of defectives in the sample, and the probability of exactly two defectives. The buyer is considering tightening the acceptance rule to zero defectives and claims this will reduce acceptance probability only slightly. The supplier says that would be unfair because the expected number of defectives is below 1. The manager must judge both claims using the binomial model, and also check whether the acceptance probability under the current rule falls below 0.80 if the defective rate drifts to 10%.
1. With p=0.08 and n=12, compute the probability that a lot is accepted under the at-most-one rule.
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Let X be the number of defectives in the sample, X~Binomial(n=12,p=0.08). Acceptance is X≤1. P(X=0)=C(12,0)(0.08)^0(0.92)^12≈0.3677. P(X=1)=C(12,1)(0.08)(0.92)^11=12×0.08×0.3996≈0.3836. P(X≤1)≈0.3677+0.3836=0.7513. So about 75.1% of lots are accepted. The calculation uses independence and a constant p, which are modelling assumptions.
2. Find the expected number of defectives and the probability of exactly two defectives. Assess the supplier's expected-below-1 argument.
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E(X)=np=12×0.08=0.96, so the supplier is correct that the expected count is below 1. P(X=2)=C(12,2)(0.08)^2(0.92)^10=66×0.0064×0.4344≈0.1835. An expected value below 1 does not mean a zero-defective rule is harmless. Under the current rule, P(X=2)≈18.4%, and under a zero-defective rule those lots would be rejected. The expected count is a long-run average, not a guarantee for each sample; the supplier's argument confuses the mean with the acceptance criterion.
3. Evaluate the buyer's claim that a zero-defective rule reduces acceptance only slightly, and check the p=0.10 case for the current rule.
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Under p=0.08, zero-defective acceptance is P(X=0)≈0.3677, compared with 0.7513 under X≤1. The reduction is about 0.3836, i.e. acceptance falls by roughly 51% relative to the current figure. That is not a slight reduction; the buyer's claim is false. At p=0.10, P(X=0)=0.9^12≈0.2824 and P(X=1)=12(0.1)(0.9)^11≈0.3766, so P(X≤1)≈0.6590. This is below 0.80, so the manager should flag that the current rule no longer meets the 0.80 acceptance benchmark if the defect rate drifts to 10%.