Statistics Reference
Binomial vs Poisson vs Hypergeometric
All three distributions count events, and all three are routinely used where one of the others belongs. The choice comes down to three questions about how the data is generated — and getting it wrong changes real answers, as the worked example below shows by running one inspection scenario through all three models.
The Three-Question Decision Flow
- Is there a fixed number of trials, decided in advance? No fixed n — events just accumulate over time or space at some average rate — points to the Poisson distribution. A fixed n moves you to question 2.
- Do the trials stay independent with the same success probability? Yes — coin flips, or draws from an effectively infinite population — points to the binomial distribution. If each draw depletes a finite pool (no replacement), go to question 3.
- Is the sample a meaningful fraction of the population? If n/N is more than about 5%, use the exact hypergeometric distribution; below that, the binomial with p = K/N is a fine approximation.
A fourth relative answers a different question entirely: if you are waiting for the first success rather than counting successes, that is the geometric distribution.
Side by Side
| Binomial | Poisson | Hypergeometric | |
|---|---|---|---|
| Counts | Successes in n fixed trials | Events in a fixed window | Successes in n draws, no replacement |
| Parameters | n, p | λ | N, K, n |
| Key assumption | Independent trials, constant p | Steady rate, events one at a time | Finite pool, each draw depletes it |
| Mean | np | λ | nK/N |
| Variance | np(1−p) | λ | np(1−p)·(N−n)/(N−1) |
| Signature example | Conversions among n visitors | Calls per hour | Card hands, lottery matches |
Reading the variance row top to bottom (with p = K/N and the same mean): hypergeometric < binomial < Poisson. Removing replacement shrinks variability; removing the ceiling n adds it. That ordering is the entire modeling story in one line.
One Scenario, Three Answers
An inspector wants the probability of finding zero defects. Each model expects one defect on average — yet the three answers differ by up to 18%, because each encodes a different data-generating story.
- Hypergeometric — sampling a known finite lot. A batch of N = 50 contains K = 5 defective units; n = 10 are drawn without replacement. P(X = 0) = C(45,10)/C(50,10) = 0.3106. Mean = 10 × 5/50 = 1, variance ≈ 0.735.
- Binomial — a stable line, independent units. Each of n = 10 units independently has p = 0.10 of being defective. P(X = 0) = 0.9¹⁰ = 0.3487. Mean = 1, variance = 0.9.
- Poisson — defects arriving at a rate. Defects occur at λ = 1 per sample-sized window of production. P(X = 0) = e⁻¹ = 0.3679. Mean = 1, variance = 1.
Same expected count, three different clean-sample probabilities: 0.3106 vs 0.3487 vs 0.3679. The hypergeometric is lowest because drawing without replacement from a lot known to contain exactly 5 defects makes avoiding all of them genuinely harder; the Poisson is highest because an open-ended rate spreads its probability across more possible counts. If the inspector's acceptance rule is “pass on zero defects,” the modeling choice moves the pass rate by nearly 6 percentage points — enough to matter in any real quality system.
The Approximation Ladder
The three models are connected by limits, which is why the simpler ones so often stand in for the exact ones:
- Hypergeometric → binomial as the population grows relative to the sample (rule of thumb n/N ≤ 5%). The finite population correction (N−n)/(N−1) drifts to 1 and replacement stops mattering.
- Binomial → Poisson as n grows and p shrinks with np held moderate (rule of thumb n ≥ 100, np ≤ 10). The ceiling becomes irrelevant and only the rate λ = np survives.
- Both → normal for large counts — the basis of proportion z-tests (binomial: np ≥ 10 and n(1−p) ≥ 10; Poisson: λ ≥ 10 or so).
The example above deliberately violates the first rule (10/50 = 20%), which is why the binomial approximation misses the exact hypergeometric answer by 12% relative. Rules of thumb are summaries of when the error is small — checking against the exact calculator when in doubt costs nothing.
Quick Classification Practice
- Aces in a 5-card poker hand: finite deck, no replacement → hypergeometric (N = 52, K = 4, n = 5).
- Heads in 20 coin flips: fixed n, constant p, independent → binomial (n = 20, p = 0.5).
- Support tickets between 9 and 10 a.m.: no fixed n, steady rate → Poisson (λ = historical average).
- Winners in a 500-person raffle where 20 tickets were bought: finite pool, no replacement → hypergeometric.
- Typos per page of a manuscript: rate over space → Poisson.
- Calls until the first sale: not a count of successes at all — waiting time → geometric.
Try the Hypergeometric Distribution Calculator
The exact no-replacement probabilities from the worked example — population, successes, draws, and cumulative results with full working.
Frequently Asked Questions
How do I know whether to use the binomial or the Poisson distribution?
Ask whether a fixed number of trials exists. Checking 200 emails for bounces has n = 200 - binomial. Counting calls arriving per hour has no n at all, just an average rate - Poisson. The giveaway phrase for Poisson is 'per hour / per page / per square meter'; the giveaway for binomial is 'out of n attempts'.
What is the difference between the binomial and hypergeometric distributions?
Replacement. The binomial assumes every trial has the same success probability, which holds when you sample with replacement or from an effectively infinite population. The hypergeometric handles sampling without replacement from a finite population, where each draw changes the remaining pool. Drawing cards from a deck or inspecting units from a small batch is hypergeometric; flipping a coin n times is binomial.
When can the binomial approximate the hypergeometric?
When the sample is a small bite of the population - a common working rule is n/N <= 5%. Sampling 10 units from 10,000 barely changes the pool, so binomial with p = K/N works well. Sampling 10 from 50 removes 20% of the population and the approximation visibly fails: exact hypergeometric P(no defects) is 0.3106 versus the binomial's 0.3487.
When can the Poisson approximate the binomial?
When n is large and p is small with np moderate - a common rule is n >= 100 and np <= 10. In that regime the fixed ceiling n stops mattering and the count behaves like events at rate lambda = np. This is why rare defects, typos, and accidents are modeled with the Poisson even though a finite n technically exists.
What does it mean that the Poisson variance equals its mean?
It is the distribution's signature and a practical diagnostic. Binomial counts have variance np(1 - p), strictly below the mean np; hypergeometric variance is smaller still. If your real count data has variance far above its mean (overdispersion), none of the three fit - clustering or rate variation is present, and models like the negative binomial handle it.
Which distribution should I use for defects in manufacturing?
It depends on the sampling frame. Inspecting a sample drawn from a specific finite lot without replacement: hypergeometric. Testing n units where each independently fails with probability p (a continuous line with stable defect rate): binomial. Counting defects arising per hour or per square meter of material with no natural n: Poisson. The worked example in this article runs the same inspection through all three.