Binomial & Poisson Calculator
Exact probabilities for a fixed number of trials or an average event rate — P(X = k) and the cumulative P(X ≤ k).
Cumulative P(X ≤ 3) = 0.1719
Use the binomial distribution for a fixed number of independent yes/no trials, and the Poisson for counts arriving at an average rate. For 10 fair coin flips, the chance of exactly 3 heads is C(10, 3) · 0.5³ · 0.5⁷ = 0.1172 — about a 12% chance.
Two ways to model a count
Both distributions describe a count of events, but they answer different questions. The binomial applies when you run a fixed number of independent trials (n), each succeeding with the same probability (p), and you want the chance of exactly k successes. The Poisson applies when events happen at a known average rate (λ) over an interval and you want the chance of exactly k of them — with no fixed number of trials. Pick binomial when you can count the trials; pick Poisson when you only know the average.
binomial: C(n, k) = n! ÷ (k! · (n − k)!); Poisson uses P(X = k) = λᵏ · e⁻λ ÷ k!
Worked example
Flip a fair coin 10 times — what is the chance of exactly 3 heads? Here n = 10, k = 3, p = 0.5.
- 1 Count the combinations. C(10, 3) = 10! ÷ (3! · 7!) = 120 — the number of ways to place 3 heads among 10 flips.
- 2 Apply the probabilities. Each specific sequence of 3 heads and 7 tails has probability 0.5³ · 0.5⁷ = 0.5¹⁰ ≈ 0.000977.
- 3 Multiply. 120 × 0.000977 = 0.1172, so P(X = 3) ≈ 0.1172 (about 12%). The cumulative P(X ≤ 3) ≈ 0.1719.
Binomial vs. Poisson
When each applies, plus its mean and variance.
| Binomial | Poisson | |
|---|---|---|
| Use when | A fixed number of independent trials, each succeeding with probability p | Events occur at an average rate over an interval, no fixed trial count |
| Inputs | n (trials), k, p | λ (mean rate), k |
| Mean | n · p | λ |
| Variance | n · p · (1 − p) | λ |
| Example | Heads in 10 coin flips | Calls to a help desk per hour |
How they connect, and what they assume
Poisson approximates the binomial. When n is large and p is small, the binomial converges to a Poisson with λ = n · p. That makes Poisson a handy stand-in for rare events across many trials — modelling typos per page or defects per batch — where counting individual trials is impractical.
Both assume independence. Each binomial trial must be independent and share the same p; each Poisson event must occur independently at a constant average rate. If one outcome influences the next — drawing cards without replacement, or events that cluster — neither model fits, and you would reach for a hypergeometric or other distribution instead.