Two presses, one misprint: Bayes, independence and a small-lot inspection
A school's examination cell gets its question papers printed by two presses. Press P prints 60% of the packets and press Q the remaining 40%. Past audits show that 2% of P's packets and 5% of Q's packets contain a misprint serious enough to force a reprint. On Monday a packet is picked at random from the day's mixed output and found to have such a misprint. The office assistant, told only that “P does most of our work”, concludes that the faulty packet most likely came from P and telephones P's manager to complain. Separately, the cell receives a sealed consignment of 20 packets from a single printer; the accompanying note states that exactly 3 of the 20 contain a misprint. The district inspector will draw 2 packets at random from the consignment, without replacement. The assistant wants to quote a single probability that the inspector finds at least one misprinted packet, and proposes to use “0.15 chance per packet, so 1 − 0.85²”.
1. Find the probability that the misprinted Monday packet came from press P, and assess the assistant's conclusion.
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Step 1 - events and data. Let P = printed by press P, Q = by press Q, D = defective (misprinted).
P(P) = 0.60, P(Q) = 0.40, P(D|P) = 0.02, P(D|Q) = 0.05.
Step 2 - total probability of a defect.
P(D) = 0.60(0.02) + 0.40(0.05) = 0.012 + 0.020 = 0.032.
Step 3 - Bayes' rule.
P(P|D) = P(D|P)P(P)/P(D) = 0.012/0.032 = 0.375. P(Q|D) = 0.020/0.032 = 0.625.
Step 4 - verdict. Given that the packet is faulty, it is more than twice as likely (0.625 vs 0.375) to have come from Q. The assistant has confused “P produces most packets” with “P produces most faulty packets”: P's 60% share buys it only 37.5% of the defects, because its defect rate is less than half Q's. Phoning P's manager is the wrong escalation - Q's process is the one to audit. A good sanity check is that the answer must lie between the two conditional defect rates' ordering, and it does.
2. Are the events “printed by press P” and “defective” independent? Show the test and interpret the result.
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Step 1 - definition. Two events A and B are independent exactly when P(A ∩ B) = P(A)P(B); any inequality proves dependence.
Step 2 - compute both sides.
P(P ∩ D) = P(P)P(D|P) = 0.60 × 0.02 = 0.012. P(P)P(D) = 0.60 × 0.032 = 0.0192.
Step 3 - compare. 0.012 ≠ 0.0192, so the events are not independent; equivalently P(D|P) = 0.02 while P(D) = 0.032, i.e. knowing the press changes the probability of a defect.
Step 4 - interpretation. This is the structural reason Bayes' rule is needed here at all. If printing press and defect were independent, both presses would share one defect rate, P(D|P) would equal P(D|Q), and the posterior would simply return the prior 0.60. The whole “which press is to blame” question exists only because the defect rates differ - and because they differ, the naive multiplication P(P)P(D) is not the probability of a P-printed defective packet.
3. Compute the correct probability that the inspector's 2 packets include at least one misprint, compare it with the assistant's 1 − 0.85², and say which the cell should quote.
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Step 1 - the correct model. The consignment is small (20 packets, exactly 3 defective) and the draw is without replacement, so this is sampling without replacement.
P(at least one defective) = 1 − P(both good) = 1 − [C(17, 2)/C(20, 2)].
Step 2 - evaluate. C(17, 2) = 136, C(20, 2) = 190.
1 − 136/190 = 54/190 = 27/95 ≈ 0.284.
The equivalent product form is (17/20)(16/19) = 272/380 = 68/95, so 1 − 68/95 = 27/95.
Step 3 - the assistant's number. Treating each draw as having probability 3/20 = 0.15 of being defective and as independent gives 1 − 0.85² = 0.2775.
Step 4 - judgement. Quote 27/95 ≈ 0.284, not 0.2775. The binomial-style calculation assumes the draws are independent, which is false without replacement: after a good packet is drawn, 3 of the remaining 19 are defective, so P(second defective | first good) = 3/19 ≈ 0.158, while P(second defective | first defective) = 2/19 ≈ 0.105. The draws are negatively associated, which is why the true figure is slightly higher. Here the gap is under 0.01, but the assistant's method becomes seriously wrong for a small lot (for 10 packets with 3 defective, the correct value is 1 − C(7,2)/C(10,2) = 0.533 against a binomial 0.51) - so the method, not just the number, should be corrected.