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Poisson Distribution Calculator

Enter the average rate λ and a number of events k to get the exact probability P(X=k) and every cumulative form (≤, <, ≥, >, between), plus the mean, variance, standard deviation and mode — with a PMF chart and the formula worked out on your own numbers. No signup.

By Induwara AshinsanaUpdated Jul 9, 2026
Poisson probabilityP(X = k), cumulative & distribution stats
NIST formula · cross-checked

Events per interval. Any value above 0, e.g. 2 or 1.5.

Whole number of events, 0 or more.

Probability type
Examples
Selected probability
P(X = 3) = 0.180447
18.04%≈ 1 in 6
P(X = 3)
18.04%
0.180447
P(X < 3)
67.67%
0.676676
P(X ≤ 3)
85.71%
0.857123
P(X ≥ 3)
32.33%
0.323324
P(X > 3)
14.29%
0.142877

Distribution summary

Mean μ = λ
2
Variance σ² = λ
2
Std dev σ = √λ
1.4142
Modes ⌊λ⌋
1 & 2
012345678
PMF over k = 0–8 — shaded bars are the selected region.

Show working

recurrence check Δ = 2.2e-16
  1. P(X = k) = e^(−λ) · λ^k / k!
  2. = e^(−2) · 2^3 / 3!
  3. e^(−2) = 0.135335 2^3 = 8 3! = 6
  4. P(X = 3) = 0.135335 · 8 / 6 = 0.180447
Sources cited

How it works

The Poisson distribution models the number of independent events that happen in a fixed interval when they occur at a known constant average rate λ. The random variable X counts the events. The chance of getting exactly k of them is given by the Poisson probability mass function (NIST §1.3.6.6.19):

P(X = k) = e^(−λ) · λᵏ / k!

Here λᵏ is the rate raised to the number of events, k! = k·(k−1)···1 is k factorial, and e^(−λ) normalises the whole distribution so the probabilities sum to 1. The cumulative questions are all sums of this one formula:

  • At most k: P(X ≤ k) = Σ P(X = i), i = 0…k
  • Less than k: P(X < k) = P(X ≤ k − 1)
  • At least k: P(X ≥ k) = 1 − P(X ≤ k − 1)
  • More than k: P(X > k) = 1 − P(X ≤ k)
  • Between: P(k ≤ X ≤ k₂) = P(X ≤ k₂) − P(X ≤ k − 1)

To stay exact for a large rate, the calculator evaluates every term in log-space using a Lanczos log-gamma (DLMF §5), since k! = Γ(k+1): ln P(X=k) = −λ + k·ln λ − lnΓ(k + 1), then P = exp(ln P). This avoids the overflow that a direct λᵏ or k! would hit past a couple of hundred events. Cumulative sums run from k = 0 and stop once they are past the peak and the remaining terms are negligible, and P(X ≥ k) is taken as the complement 1 − P(X ≤ k − 1). As an independent check, the tool also builds the CDF from the recurrence P(i) = P(i−1)·λ/i starting at P(0) = e^(−λ) and shows the two methods agree.

The distribution summary uses the defining Poisson identity that the mean and variance are equal: μ = λ, σ² = λ and so σ = √λ, with the mode at ⌊λ⌋ (two adjacent modes, λ−1 and λ, when λ is a whole number), all per Wolfram MathWorld. When λ ≥ 10 the shape is close enough to a bell curve that the normal approximation N(λ, λ) is reasonable, and the tool flags this with a continuity-correction reminder. For a large number of rare trials, Poisson(np) also approximates the binomial B(n, p) — see the Binomial Distribution Calculator for the discrete sibling.

Worked examples

Exactly 3 calls in an hour when the average is 2

λ = 2, k = 3, type =

  1. e^(−2) = 0.135335 2^3 = 8 3! = 6
  2. P(X = 3) = 0.135335 × 8 / 6 = 0.180447 ≈ 18.04%
  3. P(X ≤ 3) = e^(−2)·(1 + 2 + 2 + 1.333333) = 0.857123
  4. P(X ≥ 4) = 1 − P(X ≤ 3) = 0.142877; μ = σ² = 2, σ = 1.4142, modes 1 & 2

At least one defect in a batch averaging 5

λ = 5, k = 1, type ≥

  1. The complement of “at least one” is “none”: P(X ≥ 1) = 1 − P(X = 0)
  2. P(X = 0) = e^(−5) = 0.006738 ≈ 0.67%
  3. P(X ≥ 1) = 1 − 0.006738 = 0.993262 ≈ 99.33%
  4. μ = σ² = 5, σ = √5 = 2.2361, modes 4 & 5

Fractional rate — exactly 2 events when the average is 2.5

λ = 2.5, k = 2, type =

  1. e^(−2.5) = 0.082085 2.5^2 = 6.25 2! = 2
  2. P(X = 2) = 0.082085 × 6.25 / 2 = 0.256516 ≈ 25.65%
  3. λ is not a whole number, so there is a single mode ⌊2.5⌋ = 2
  4. μ = σ² = 2.5, σ = √2.5 = 1.5811

Frequently asked questions

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