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Worked solution

A two-stage batch acceptance rule

In each batch of manufactured articles the proportion of defective articles is pp. From each batch a random sample of nine is taken and each of the nine articles is examined. If two or more of the nine articles are found to be defective that batch is rejected; otherwise it is accepted. Prove that the probability that a batch is accepted is (1−p)8(1+8p)(1-p)^8(1+8p).

It is decided to modify the sampling scheme so that, when one defective is found in the sample, a second sample of nine is taken and the batch rejected if this contains any defectives. With this exception, the original scheme is continued. Find an expression in terms of pp for the probability that a batch is accepted.

For this modified scheme evaluate the average number sampled per manufactured batch over a large number of batches when pp has the value 0.1.

[N81/II/12]

CJC · Discussion · Question 6

Answer

Original: (1−p)8(1+8p)(1-p)^8(1+8p). Modified: (1−p)9[1+9p(1−p)8](1-p)^9[1+9p(1-p)^8]. Average at p=0.1p=0.1: 12.512.5.

Original scheme

Let X be the number defective in a sample of nine, with X∼B(9,p)X\sim B(9,p). Acceptance means X<2. Thus P(accept)=P(X=0)+P(X=1)=(1−p)9+9p(1−p)8=(1−p)8[1−p+9p]=(1−p)8(1+8p)P(\mathrm{accept})=P(X=0)+P(X=1)=(1-p)^9+9p(1-p)^8=(1-p)^8[1-p+9p]=(1-p)^8(1+8p), as required.

How did this part go?

Modified scheme

Acceptance occurs with no defectives in the first sample, or exactly one in the first and none in the second. Under the same independent binomial-sampling model, P(accept)=(1−p)9+9p(1−p)8(1−p)9=(1−p)9[1+9p(1−p)8]P(\mathrm{accept})=(1-p)^9+9p(1-p)^8(1-p)^9=(1-p)^9[1+9p(1-p)^8].

How did this part go?

Average number sampled

Exactly one defective in the first sample leads to 18 articles examined. None or at least two leads to only nine; the latter is an immediate rejection, not an instruction to take another sample. The expected number is 9p(1−p)8(18)+[1−9p(1−p)8](9)=9+81p(1−p)89p(1-p)^8(18)+[1-9p(1-p)^8](9)=9+81p(1-p)^8. At p=0.1 this is 12.486784401, giving 12.5 articles per batch to 3 s.f.

How did this part go?

Common mistakes

A second sample is taken only after exactly one defective in the first.

Solution notes

The printed calculation assumes binomial samples with common defect probability and independent trials/stages. For finite sampling without replacement this is a model or large-batch approximation, not an exact conclusion from the stated proportion alone.

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