Hypergeometric distribution

An urn with $N$ balls, $m$ of them red. You draw $K$ without replacement. How many reds do you expect to pull?

P (X = k) = \frac{( k m ) ( K - k N - m )}{( K N )}

Hypergeometric PMF Binomial approximation

Binomial (K, m / N)

N (population) 50 m (successes in pop) 20 K (items drawn) 10 n (samples) 1000

PMF

P (X = k) = \frac{( k m ) ( K - k N - m )}{( K N )}

. Dashed red marks the Binomial(K, m/N) approximation — the two converge as N grows with m/N fixed.

What to notice

Binomial convergence. The dashed red caps show Binomial(K, m/N). When the population is large relative to the sample size, the with-replacement and without-replacement distributions are nearly indistinguishable. Crank N up while keeping m/N fixed to watch them merge.
The finite-population correction. The hypergeometric variance carries an extra factor of $(N - K) / (N - 1)$ that the Binomial lacks. At K = N (you drew everything) the variance is zero — there’s no randomness left.
Support is narrower than it looks. You can only draw between $max (0, K - (N - m))$ and $min (K, m)$ successes. Extreme slider values make the support shrink to a single point.

Quality inspection. Pulling K items from a batch of N and counting defects.
Card games. “What’s the chance of drawing 3 aces in a 5-card hand?” is Hypergeometric(52, 4, 5).
Capture-recapture. Tag m fish, release them, then re-sample K; the count of tagged fish in the second sample is hypergeometric. Inverting gives the Lincoln-Petersen estimate of the population size.

E [X] = K m / N

Var (X) = K \frac{m}{N} \frac{N - m}{N} \frac{N - K}{N - 1}