Defective Items from Two Production Lines
A factory runs two production lines, A and B. Line A makes 60% of the daily output and has a defect rate of 2%. Line B makes the remaining 40% and has a defect rate of 5%. At the end of the day an inspector selects one item at random from the combined output and finds it defective. The line manager says: ‘The defective item is probably from A, because A makes most of the items.’ A quality engineer replies that the posterior probability depends on both the output share and the defect rate, and that B may actually be the more likely source. The two rates are stable for the day, and the combined output is large enough to treat sampling as independent when a batch of five items is checked. The engineer must produce a short note giving the overall defect rate, the probability that a defective item came from B, and the probability that a five-item sample contains at most one defective item.
1. Calculate the overall defect rate and the probability that a randomly chosen defective item came from line B. State whether the line manager’s intuition is correct.
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Let D be defective and B be line B. P(B) = 0.40, P(D|A) = 0.02, P(D|B) = 0.05. P(D) = P(A)P(D|A) + P(B)P(D|B) = 0.60 × 0.02 + 0.40 × 0.05 = 0.012 + 0.020 = 0.032, or 3.2%. Then P(B|D) = P(B)P(D|B)/P(D) = 0.40 × 0.05/0.032 = 0.020/0.032 = 0.625. So 62.5% of defective items come from B. The manager’s intuition is wrong: although A makes more items, B’s defect rate is high enough to make B the more likely source.
2. A trainee says: ‘Line B has a 5% defect rate, so if the item is defective there is a 5% chance it came from B.’ Identify the error and give the correct interpretation of the 5% figure.
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The 5% is P(defective | line B), not P(line B | defective). It is the defect rate within B’s output. The posterior P(B|D) also includes the prior share P(B) = 0.40 and the overall defect rate. Mixing up the conditional direction is the error; the correct posterior is 62.5%, not 5%. Also, P(B|D) is not P(B) = 40% either.
3. For a sample of five items from the combined output, find the probability that at most one is defective, stating any assumption needed.
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With p = 0.032 and a large lot, treat the number X of defectives in 5 items as Binomial(5, 0.032), assuming independent selections and a stable defect rate. P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.968^5 + 5(0.032)(0.968)^4 ≈ 0.8499 + 0.1405 = 0.9904. So the probability is about 0.990, or 99.0%. If the five items were drawn without replacement from a small batch, the hypergeometric distribution would be needed and the binomial value would be only an approximation.