Question 2 (b)

Acceptance sampling of chipsets: accept, second sample or reject (binomial)

9 marks

Time to write: About 18 minutes
Topic: Binomial and Poisson distributions

Question

Question 2 (b) · 9 marks

On average, a machine produces 10% defective chipsets. When the quality manager visits to inspect, a batch is accepted if a sample of 7 chipsets drawn from it contains no defective chipsets, and rejected if the sample contains 3 or more defectives. Otherwise, a second sample is taken for a quality check by the manager. What is the probability that:

The batch will be accepted,
A second sample will be drawn,
The batch will be rejected.

Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.

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18 minfor 9 marks

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Read the data; note exactly what is asked 1 min
1 Binomial model, parameters and the decision rule 2 marks 4 min
2 (i) Probability that the batch is accepted 2 marks 4 min
3 (ii) Probability that a second sample is drawn 2½ marks 4 min
4 (iii) Probability that the batch is rejected 2½ marks 4 min
Check: arithmetic, units (₹, %), final answer underlined 1 min

1234

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Solution

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The distribution. Each of the 7 chipsets in the sample is either defective (a success for this count) or not, with the same chance p = 10% = 0.1 of being defective, and one chipset's condition does not affect another's. The number X of defectives in the sample is therefore binomial, X ~ B(7, 0.1), with q = 1 − p = 0.9 and P(X = x) = ⁷Cₓ pˣ q⁷⁻ˣ, x = 0, 1, …, 7. (The batch is taken to be large compared with the sample, so the chance of a defective stays the same from one chipset to the next.)

The quality manager's rule splits the values of X into three decisions:

X = 0: accept the batch;
X = 1 or 2: draw a second sample;
X ≥ 3: reject the batch.

Working table (probabilities to 4 decimal places)

x defectives	⁷Cₓ	pˣ q⁷⁻ˣ	P(X = x)	Decision
0	1	(0.9)⁷ = 0.4783	0.4783	Accept
1	7	(0.1)(0.9)⁶ = 0.0531	0.372	Second sample
2	21	(0.1)²(0.9)⁵ = 0.0059	0.124	Second sample
3 or more		1 − (sum of the rows above)	0.0257	Reject
Total			1

(i) The batch is accepted

P(X = 0) = (0.9)⁷ = 0.4783.

(ii) A second sample is drawn

P(X = 1 or 2) = P(X = 1) + P(X = 2) = 7 × 0.1 × 0.531441 + 21 × 0.01 × 0.59049 = 0.372 + 0.124 = 0.496.

(iii) The batch is rejected

P(X ≥ 3) = 1 − [P(X = 0) + P(X = 1) + P(X = 2)] = 1 − (0.4783 + 0.372 + 0.124) = 1 − 0.9743 = 0.0257.

The three decisions are mutually exclusive and cover every possible sample, so their probabilities add up to 1 (0.4783 + 0.496 + 0.0257 = 1).

Comment. With a 10% defective rate, fewer than half of the batches (about 48%) pass on the first sample, about half (49.6%) need a second sample, and only about 2.6% are rejected outright. The mean number of defectives in a sample of 7 is np = 7 × 0.1 = 0.7, which is why one or two defectives are so common and the second sample is drawn so often.

Where the 9 marks sit

#	Component	Marks
1	Binomial model, parameters and the decision rule	2
2	(i) Probability that the batch is accepted	2
3	(ii) Probability that a second sample is drawn	2.5
4	(iii) Probability that the batch is rejected	2.5
	Total	9

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