Grant-file audit: sampling without replacement
An audit team must review 20 grant files. Unknown to the team, 6 of the 20 contain a material error. The team will draw a random sample of 4 files without replacement and review them first. If the sample contains no error, the remaining 16 files will be checked only by an automated screen with a 40% chance of flagging each error file; if the sample contains at least one error, all remaining files will be reviewed manually. The audit manager estimates the chance of at least one error in the sample as 'roughly 1 in 5' and plans staffing on that basis. A team member says the chance is much higher and that the first three files already checked are all clean, so the fourth file is now less likely to be faulty. Evaluate both statements and compute the probability that the remaining files will need full manual review.
1. Compute the probability that the sample of 4 contains at least one error, and evaluate the manager's 'roughly 1 in 5' estimate.
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Use the complement. Total samples: C(20,4) = 4,845. Samples with no error: choose all 4 from the 14 clean files: C(14,4) = 1,001. P(no error) = 1,001 ÷ 4,845 = 0.2066, about 20.7%. Therefore P(at least one error) = 1 − 0.2066 = 0.7934, about 79.3%.
The manager has confused the two events. 'Roughly 1 in 5' is the probability of no error, not of at least one error. Staffing on a 20% chance of manual review would badly under-resource the team. The probability that the remaining files need full manual review is exactly the probability that the sample contains at least one error: 79.3%.
2. Calculate the probability that the sample contains exactly two error files.
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Number of ways to choose 2 error files from 6: C(6,2) = 15. Number of ways to choose 2 clean files from 14: C(14,2) = 91. Favourable samples = 15 × 91 = 1,365. Total samples = C(20,4) = 4,845. P(exactly 2 errors) = 1,365 ÷ 4,845 = 0.2817, about 28.2%.
This is a hypergeometric probability because sampling is without replacement: each draw changes the composition of the remaining files. The result is plausible: exactly two errors is the single most likely exact count, but it is still less likely than the combined event of at least one error.
3. Evaluate the team member's conditional statement: given that the first three files checked are clean, is the fourth file less likely to be faulty?
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Initially, the probability that any one file is faulty is 6/20 = 30%. After three clean files are removed, 17 files remain and all 6 error files are still among them. The probability that the fourth file is faulty is therefore 6/17 = 35.3%, which is higher than 30%, not lower.
The team member has the direction wrong. Removing clean files concentrates the remaining errors, so the fourth draw becomes more likely to be faulty, not less. Conditionally on the first three being clean, the probability that the sample of four contains at least one error is simply the probability that the fourth is faulty: 35.3%. The unconditional probability of manual review remains 79.3%; the conditional figure rises because the first three draws have already removed clean files.