Chip Defects, Sampling and a Rapid Tester
A consumer-electronics plant runs two lines. Line L1 makes 60% of the chips and historically has a 2% defect rate; line L2 makes 40% with a 5% defect rate. Chips are mixed in the warehouse with no line marking. A batch of 500 chips from L1 is about to be packed; a quality engineer plans to inspect 10 chips from that batch without replacement and stop the line if any is defective. Separately, a new rapid tester is being evaluated: it flags 90% of truly defective chips and 5% of good chips. Management wants to know whether a positive rapid-test result is strong evidence of a defect, and whether screening every chip at ₹8 each is cheaper than paying ₹300 for each defective chip that reaches a customer. Assume a chip drawn at random from the warehouse.
1. Find the overall defect rate and the probability that a defective chip came from L2. Interpret the result for the line manager.
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Let D be defective, L1 and L2 the lines. P(L1) = 0.60, P(L2) = 0.40, P(D|L1) = 0.02, P(D|L2) = 0.05.
By the law of total probability, P(D) = 0.60(0.02) + 0.40(0.05) = 0.012 + 0.020 = 0.032, so 3.2% of warehouse chips are defective.
By Bayes' theorem, P(L2|D) = P(D|L2)P(L2)/P(D) = 0.020/0.032 = 0.625. Although L2 makes only 40% of chips, it accounts for 62.5% of all defects. The line manager should prioritise L2 for process improvement: its defect rate is 2.5 times L1's, and most defects originate there.
2. For the L1 batch of 500 chips containing 2% defectives, compute the probability that the 10-chip inspection finds at least one defective. Is the stop rule likely to trigger?
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The batch has 0.02(500) = 10 defective chips and 490 good chips. Sampling is without replacement, so the exact probability of no defective in 10 draws is P(no defective) = C(490,10)/C(500,10) = (490/500)(489/499)···(481/491).
The ratios are all close to 0.98, so the product is about 0.98^10 ≈ 0.8171. Hence P(at least one defective) ≈ 1 − 0.8171 = 0.1829, about 18.3%.
The binomial approximation with n = 10 and p = 0.02 gives the same value to two decimals: 1 − 0.98^10 ≈ 0.183. Because the batch is large relative to the sample, the without-replacement correction is negligible here.
The stop rule triggers only about once in five inspections. It is a weak detector of a 2% defect rate; a sample of 10 is too small for reliable screening, even though the exact and approximate models agree.
3. For a random warehouse chip, compute P(defective | rapid test positive). Use the overall defect rate from task 1. Explain why the answer may surprise a manager.
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Let + denote a positive rapid test. P(D) = 0.032, P(+|D) = 0.90, P(+|not D) = 0.05.
P(+) = P(+|D)P(D) + P(+|not D)P(not D) = 0.90(0.032) + 0.05(0.968) = 0.0288 + 0.0484 = 0.0772.
P(D|+) = 0.0288/0.0772 ≈ 0.373, about 37.3%.
A positive result is not strong evidence of a defect: nearly two-thirds of positives are false alarms. The reason is the base rate. Good chips greatly outnumber defective chips, so a 5% false-positive rate on 96.8% of the population produces 4.84 percentage points of false positives, compared with only 2.88 points of true positives.
For screening economics, expected warranty cost without screening is 0.032(300) = ₹9.60 per chip, which exceeds the ₹8 test cost. Full screening looks cheaper in expectation, but the rapid tester's false positives and any false negatives add rework and escape costs, so the manager should not treat the ₹8 figure as the whole cost.