Generated on: June 13, 2026 at 06:01 PM

Poisson Distribution Calculator

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Rate3
Events2

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PMF value 0

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Mathematics

Poisson Distribution Calculator

Calculate P(X=k), P(X≤k), and P(X≥k) for a Poisson distribution. Enter the rate parameter λ and the number of events k.

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\begin{aligned} P(X=k) &= \dfrac{e^{-\lambda} \lambda^k}{k!} \\ &= \dfrac{e^{-3} \cdot 3^{ 2 }}{ 2!} \\ &= ? \end{aligned}
\begin{aligned} \mu &= \lambda \\ &= ? \end{aligned}
\begin{aligned} \sigma^2 &= \lambda \\ &= ? \end{aligned}

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PMF value 0

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The Poisson distribution is a discrete probability distribution that counts the number of independent events occurring in a fixed interval when events happen at a known average rate. It is one of the most widely applied distributions in statistics, underpinning models for phone calls arriving at a call centre, customer arrivals at a queue, radioactive decay events, and web-server request rates.

Formula

Given a rate parameter λ > 0 and a non-negative integer k, the probability mass function (PMF) is:

P(X=k)=e−λ λkk!,k=0,1,2,…P(X = k) = \frac{e^{-\lambda}\,\lambda^k}{k!}, \quad k = 0, 1, 2, \ldots

The cumulative distribution function (CDF) is the sum of the PMF from 0 to k:

P(X≤k)=∑i=0ke−λ λii!P(X \leq k) = \sum_{i=0}^{k} \frac{e^{-\lambda}\,\lambda^i}{i!}

The complementary CDF (at-least probability) uses the identity:

P(X≥k)=1−P(X≤k−1)P(X \geq k) = 1 - P(X \leq k - 1)

The mean and variance of the Poisson distribution both equal λ:

μ=λ,σ2=λ\mu = \lambda, \qquad \sigma^2 = \lambda

Worked example

A help-desk team receives an average of 3 support tickets per hour. What is the probability that exactly 2 tickets arrive in the next hour?

Using λ = 3 and k = 2:

P(X=2)=e−3⋅322!=e−3⋅92≈0.2240P(X = 2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{e^{-3} \cdot 9}{2} \approx 0.2240

There is a 22.4 % chance that exactly 2 tickets arrive. The probability of receiving at most 2 tickets is:

P(X≤2)=e−3(1+3+4.5)=8.5 e−3≈0.4232P(X \leq 2) = e^{-3}(1 + 3 + 4.5) = 8.5\,e^{-3} \approx 0.4232

The probability of receiving 2 or more tickets (at least 2) is:

P(X≥2)=1−P(X≤1)=1−e−3(1+3)=1−4e−3≈0.8009P(X \geq 2) = 1 - P(X \leq 1) = 1 - e^{-3}(1 + 3) = 1 - 4e^{-3} \approx 0.8009

Formula derivation

The Poisson PMF is derived by taking the limit of the binomial distribution as the number of trials n → ∞ and the success probability p → 0, while keeping the product np = λ fixed:

(nk)pk(1−p)n−k  →n→∞,  np=λ  λkk!e−λ\binom{n}{k} p^k (1-p)^{n-k} \;\xrightarrow{n\to\infty,\;np=\lambda}\; \frac{\lambda^k}{k!} e^{-\lambda}

The key steps are: (1) the binomial coefficient n!/(k!(n−k)!) approaches nᵏ/k!; (2) pᵏ = (λ/n)ᵏ contributes λᵏ/nᵏ, cancelling the nᵏ; (3) (1 − λ/n)^n → e^−λ as n → ∞.

Why mean and variance both equal λ

For the binomial distribution, mean = np = λ and variance = np(1 − p). As p → 0, the factor (1 − p) → 1, so variance → np = λ as well. This equality — mean = variance = λ — is the signature property of the Poisson distribution. It implies that the standard deviation is √λ, and the coefficient of variation (σ/μ) is 1/√λ, decreasing as λ increases.

Numerical computation

For large k, computing λ^k and k! separately causes overflow. This calculator uses the log-PMF form:

log⁡P(X=k)=−λ+klog⁡λ−log⁡(k!)\log P(X = k) = -\lambda + k \log \lambda - \log(k!)

with log k! accumulated incrementally (log 2 + log 3 + … + log k), then exponentiating at the end. This remains accurate for k up to several hundred.

Relationship to other distributions

  • Binomial: The Poisson is the n → ∞ limit of Binomial(n, λ/n). The approximation is reliable when n ≥ 20 and p ≤ 0.05.
  • Exponential: The waiting time between successive Poisson events follows an Exponential(λ) distribution. If events occur at rate λ per unit time, the inter-arrival time has mean 1/λ.
  • Normal: For large λ (roughly λ ≥ 30), the Poisson distribution is well approximated by Normal(λ, λ), meaning (X − λ)/√λ is approximately standard normal.
  • Geometric: The geometric distribution models the trial number of the first success in repeated Bernoulli trials. Its continuous-time analogue is the exponential distribution, the same waiting-time distribution that connects to the Poisson counting process.

Practical applications

  • Call centres: estimating the probability that more than k calls arrive in a given minute, used to size staffing levels.
  • Epidemiology: modelling the number of disease cases in a region per week under a constant incidence rate.
  • Reliability engineering: predicting the number of component failures in an interval given a constant failure rate (the basis of the Poisson process model in reliability theory).
  • Network traffic: analysing packet arrival rates in routers and queuing systems (M/M/1 and related models assume Poisson arrivals).
  • Astronomy and physics: counting photons detected by a sensor, or radioactive decay events, where each particle decays independently with a fixed probability per unit time.

Frequently Asked Questions (FAQ)

When is the Poisson distribution appropriate?

Use the Poisson distribution when counting the number of events in a fixed interval of time or space, provided: events occur independently, the average rate λ is constant, and two events cannot occur at exactly the same instant. Classic examples include phone calls arriving at a call centre per hour, radioactive decay events per second, and typographical errors per page.

What does λ (lambda) represent?

λ is the average number of events expected in the chosen interval. If a server receives on average 5 requests per minute, λ = 5. The interval itself can be any fixed unit — seconds, metres, pages — as long as λ is adjusted proportionally. For half a minute, λ = 2.5; for two minutes, λ = 10.

How is the Poisson distribution related to the binomial distribution?

The Poisson distribution is the limiting case of the binomial distribution when the number of trials n → ∞ and the success probability p → 0, while their product np = λ remains constant. In practice, the Poisson approximation is accurate when n ≥ 20 and p ≤ 0.05 (or n ≥ 100 and np ≤ 10). It replaces the computationally heavy binomial formula with the simpler e^−λ λ^k / k!.

Why are the mean and variance both equal to λ?

This equality is a defining characteristic of the Poisson distribution and follows from its derivation as a limit of the binomial: mean = np = λ, and variance = np(1−p) → np = λ as p → 0. In practice it means the standard deviation is √λ and the coefficient of variation is 1/√λ. If observed data show variance much larger than the mean (overdispersion), a negative binomial model may fit better.

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