Hypergeometric Probability: Drawing Without Replacement
You have a bag of 20 marbles: 12 red and 8 blue. You pick 5 without putting any back. What's the chance exactly 3 are red?
This is what the hypergeometric distribution handles. Unlike the binomial distribution, the trials are not independent — each draw changes the composition of the bag.
The Intuition
When you draw without replacement, the probability of a red on the second draw depends on the first: take a red first and fewer reds remain, so the next red is less likely. Hypergeometric probability captures exactly this dependency.
The setup: "I have a finite population split into two groups. I take a sample without replacement. What's the probability my sample has exactly k items from the first group?"
The Precise Statement
P(X=k)=(nN)(kK)(n−kN−K)
Where:
- N = total items in the population (20 marbles)
- K = number of "success" items (12 red)
- n = number drawn (sample size, 5)
- k = successes wanted in the sample (3 reds)
Why This Formula Makes Sense
The denominator (nN) counts all ways to choose n items from N — the equally likely outcomes. The numerator counts favourable ones:
- (kK): choose k reds from the K reds
- (n−kN−K): choose the remaining n−k from the N−K blues
Multiplying pairs each way of picking reds with each way of picking blues.
Worked Example
N=20, K=12, n=5, k=3:
P(exactly 3 reds)=(520)(312)(28)=15504220×28=155046160≈0.397
About 39.7%.
A common mistake is using the binomial formula here. Binomial assumes independent trials (drawing with replacement). With p=12/20=0.6 it gives (35)(0.6)3(0.4)2≈0.346 — close but wrong. The gap grows as the sample becomes a larger fraction of the population.
When to Use Hypergeometric …