the hypergeometric-to-binomial relationship

The binomial and the hypergeometric are answering the same physical question — how many of the special items did I get in a sample of n? — but under two different rules. The binomial samples with replacement (or from a pool so big it makes no difference), so the success probability p stays fixed and trials are independent. The hypergeometric samples without replacement from a finite pool of N, so the probability drifts and trials are dependent. The relationship is simple to state: when the pool is huge compared with the sample, the two distributions become nearly identical.

Here is why, concretely. In the hypergeometric, the fraction of red items is p = K/N. On your first draw the chance of red is exactly p. On your second draw it is K/(N-1) or (K-1)/(N-1) depending on the first draw — close to p but not equal. If N is gigantic and n is small, removing a few items barely changes the proportion, so each draw is approximately an independent trial at probability p, which is precisely the binomial picture. You can also see it in the variances: the hypergeometric variance is the binomial variance n p (1 - p) multiplied by the finite population correction (N - n)/(N - 1). As N grows toward infinity with n fixed, that correction approaches 1 and the two variances coincide.

The practical rule of thumb is that the binomial is a fine approximation to the hypergeometric when the sample is at most about 5% to 10% of the pool. This matters because the hypergeometric pmf, full of large factorials, is awkward to compute, while the binomial is easy. So pollsters sampling 1000 people from a country of millions cheerfully use the binomial: the without-replacement reality is technically hypergeometric, but the difference is negligible.