In a QA lot of 50 units with 5 defectives, why is the count of defectives in 10 draws without replacement not binomial?

## Why the binomial does not apply A binomial count requires three things: a fixed number of trials, **independent** trials, and a **constant** success probability. Sampling without replacement satisfies the first and breaks the other two. In a lot of N = 50 units containing K = 5 defectives, the first draw is defective with probability 5/50 = 0.1. But the second draw's probability depends on the first: 4/49 ≈ 0.082 if the first was defective, 5/49 ≈ 0.102 if it was not. The draws are negatively dependent — pulling a defective makes the next one less likely — and no single p describes them all. ## The correct model The **hypergeometric** distribution counts successes in n draws taken without replacement from a finite population of N items containing K successes: `P(X = k) = C(K, k) · C(N-K, n-k) / C(N, n)`. The logic is pure counting: choose k of the K defectives and n-k of the N-K good units, divided by all ways to choose n of N. With N = 50, K = 5, n = 10: - `P(X = 0) = C(45, 10)/C(50, 10) ≈ 0.3106` - `P(X = 1) = 5·C(45, 9)/C(50, 10) ≈ 0.4313` - `P(X = 2) ≈ 0.2098` The corresponding Binomial(10, 0.1) values are 0.3487, 0.3874 and 0.1937 — visibly different, not a rounding-level discrepancy. ## Mean and variance The means agree exactly. The hypergeometric mean is `n·K/N = 10 × 0.1 = 1`, the same as the binomial's np = 1. This is a common surprise: dependence does not bias the expected count, because expectation is linear and each individual draw is marginally defective with probability K/N = 0.1. The variances differ: - Binomial: `np(1-p) = 10 × 0.1 × 0.9 = 0.9` - Hypergeometric: `np(1-p) · (N-n)/(N-1) = 0.9 × 40/49 ≈ 0.735` The extra factor `(N-n)/(N-1)` is the **finite population correction**. It is always ≤ 1, so sampling without replacement is always *less* variable. The intuition: the lot has exactly 5 defectives, and every one you draw is one fewer available to draw again, so extreme outcomes are damped. In the limit n = N you have sampled the whole lot and the count is exactly 5 with zero variance — the correction becomes (N-N)/(N-1) = 0, exactly as it should. ## Which direction the error runs With a smaller variance and the same mean, the hypergeometric puts less mass in both tails, including at zero: 0.311 against the binomial's 0.349. So P(catch at least one defective) is 0.689 exactly versus 0.651 under the binomial. Using the binomial here makes your QA sample look *worse* at detection than it is. That direction is not universal for every k, but the underdispersion is, and for acceptance-sampling decisions the difference between a 65 and a 69 percent detection rate is decision-relevant. ## When the binomial is good enough As N grows with K/N held fixed, `(N-n)/(N-1) → 1` and the hypergeometric converges to Binomial(n, K/N). The working rule is the **sampling fraction**: if n/N ≤ 0.1, the binomial approximation is normally fine. Here n/N = 10/50 = 0.2, twice the threshold, and the correction factor 40/49 ≈ 0.816 is an 18 percent variance reduction — too large to ignore. Sampling 10 units from a lot of 10,000 would be a completely different story: the correction is 9990/9999 ≈ 0.9991 and the binomial is effectively exact. ## The senior judgment The practical question is rarely "which formula is prettier" but "does the difference change the decision?" Three things to weigh: 1. **Lot size.** Small-lot acceptance sampling — pharma batches, low-volume hardware, an audit of 50 contracts — is exactly where the correction bites. High-volume inspection rarely needs it. 2. **What the number feeds.** A rough exploratory estimate tolerates the approximation; a documented acceptance plan with a stated detection guarantee does not. 3. **What the sampling actually was.** If units were drawn with replacement, or drawn from a continuously produced stream rather than a fixed lot, the binomial is genuinely correct and no correction is needed. Confirm the physical procedure before choosing the model. ## Common errors Assuming the means differ (they do not); believing sampling without replacement *increases* variance; applying the binomial to small-lot inspection without ever checking n/N; and forgetting that the correction depends on the sampling fraction, not on n alone — 100 drawn from 1,000,000 needs no correction, while 100 drawn from 200 badly does.