Hypergeometric Distribution: Sampling Without Replacement

When do we need the hypergeometric distribution?

When you sample without replacement from a finite population split into two categories ("successes" and "failures"), the probability of each draw changes as items are removed. The binomial distribution assumes with-replacement (or equivalently, each trial has the same $p$), so it fails here.

The hypergeometric distribution handles this correctly.

The formula

Given:

• $N$ = total population size
• $K$ = number of successes in the population
• $n$ = sample size
• $k$ = observed successes

$P(X = k) = \dfrac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}}$

The numerator counts: ways to choose $k$ successes from the $K$ available, times ways to choose $n - k$ failures from the $N - K$ available. The denominator is the total number of samples.

Step-by-step solution

Setup: $N = 52$ cards, $K = 13$ hearts, $n = 5$ cards drawn, $k = 2$ hearts wanted.

Step 1 — Count favorable outcomes (2 hearts, 3 non-hearts):

• Choose 2 hearts from 13: $\binom{13}{2} = \dfrac{13 \cdot 12}{2} = 78$
• Choose 3 non-hearts from 39: $\binom{39}{3} = \dfrac{39 \cdot 38 \cdot 37}{6} = 9139$
• Multiply: $78 \cdot 9139 = 712{,}842$

Step 2 — Count total outcomes (any 5 cards from 52): $\binom{52}{5} = \dfrac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}{5!} = 2{,}598{,}960$

Step 3 — Divide: $P(X = 2) = \dfrac{712{,}842}{2{,}598{,}960} \approx \boxed{0.2743}$

So there's about a 27.4% chance of drawing exactly 2 hearts.

Sanity check

The mean of a hypergeometric is $E(X) = n \cdot \dfrac{K}{N} = 5 \cdot \dfrac{13}{52} = 1.25$. Since 2 is close to 1.25, a 27% probability is plausible — around the mode of the distribution.

Hypergeometric vs. binomial

• Hypergeometric (without replacement): the deck shrinks, probabilities shift after each draw.
• Binomial (with replacement, or infinite population): fixed $p$ each draw.

When $n$ is tiny compared to $N$, the two agree very closely. For our problem, the binomial approximation with $p = 13/52 = 0.25$ gives $\binom{5}{2}(0.25)^2(0.75)^3 \approx 0.2637,$ close to but not equal to the exact hypergeometric answer.

Common mistakes

• Using the binomial when drawing without replacement. It's almost right for large populations, but it's wrong for small ones like a deck.
• Mismatching the categories. Keep careful track: $K$ and $k$ refer to successes; $N - K$ and $n - k$ refer to failures.
• Forgetting $\binom{N}{n}$ in the denominator. Only normalizing by $N^n$ or leaving it out entirely are common slips.

Try it in the visualization

Sliders for $N$, $K$, $n$, and $k$ reshape the whole bar chart live. A deck of cards animates the draw, coloring hearts red as they are taken, showing why each subsequent probability changes.