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Instrument MI-07-133 · Statistics

Poisson Distribution Calculator

How many calls, defects, or typos should you expect in a given stretch of time or space — and how likely is any specific count? The Poisson distribution answers exactly that.

Instrument MI-07-133
Sheet 1 OF 1
Rev A
Verified
Type 07 — Probability Distributions SER. 2026-07133

P(X = k) — probability of exactly k events

0.19536681

P(X=k) = lambda^k x e^(-lambda) / k!

The working Every figure verified twice
  1. pmf = pow(4, 4)·exp(0 − 4) ⁄ fact(4) = 0.19536681
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

The Poisson distribution models the count of events that happen in a fixed interval of time or space, when those events occur independently of one another at a known average rate λ (lambda). A call center that receives an average of 4 calls per hour, a stretch of highway with an average of 2 potholes per mile, or a page of text with an average of 0.5 typos — all of these are natural Poisson settings, because there's no fixed 'number of trials' the way there is for a binomial distribution, only a steady average rate over a continuous span.

The PMF, P(X = k) = λ^k × e^(−λ) / k!, gives the probability of observing exactly k events for a given λ. The λ^k term grows with k, but it's counterbalanced by the k! in the denominator, which grows much faster for large k — the combined effect is a distribution that rises from k = 0, peaks somewhere near λ, and then tapers off, with the exact shape depending on how large λ is.

A distinctive feature of the Poisson distribution is that its mean and variance are both equal to λ — a single parameter controls the center and the spread simultaneously, unlike the binomial distribution's separate n and p. This link to the binomial distribution isn't coincidental: if you let the number of binomial trials n grow very large while the per-trial probability p shrinks so that n×p stays fixed at λ, the binomial distribution converges to the Poisson distribution with that same λ — historically, this is exactly how the Poisson distribution was first derived, as the limiting case of many rare, independent chances for an event.

P(X=k)=λkeλk!P(X=k) = \frac{\lambda^{k} e^{-\lambda}}{k!}
λ — the average rate of events per interval, λ ≥ 0 · k — the specific number of events being evaluated, a non-negative integer · e — Euler's number, ≈ 2.71828 · k! — k factorial, the product of all whole numbers from 1 up to k (0! = 1 by definition).
  • Enter the average number of events per interval into Rate / average number of events (λ) — this can be any non-negative number, not just a whole number.
  • Enter the specific event count you want the probability for into Number of events (k) — always a non-negative whole number.
  • Read P(X = k) — probability of exactly k events — the resulting Poisson probability.
  • To compare several possible counts, run the instrument again with the same λ and a different k each time; there is no built-in cumulative total.
  • λ can be a decimal (like 2.5 events per hour) even though k, the count you're asking about, must be a whole number.

Worked example — exactly 4 events at rate λ = 4

Enter λ = 4 into Rate / average number of events and k = 4 into Number of events — the probability of observing exactly 4 events in an interval where 4 is also the long-run average (for instance, 4 customer arrivals in a period that averages 4 arrivals). The instrument computes 4^4 = 256, e^(−4) ≈ 0.0183156389, and 4! = 24.

P(X = k) reads 0.195366814813, computed as 256 × 0.0183156389 / 24. So even when the count you ask about matches the average rate exactly, the probability of hitting that precise count is still under 20% — there's just under a one-in-five chance of landing on exactly 4 events, with the remaining probability spread across 0, 1, 2, 3, 5, 6 and higher counts.

Questions

What conditions does a Poisson distribution require?

Events must occur independently of one another, at a constant average rate λ over the interval being studied, and two events cannot occur at exactly the same instant (or point, for spatial applications). Call arrivals, radioactive decay counts, and typos scattered through a manuscript are classic examples; situations with clustering, seasonal spikes, or events that trigger each other (like retweets) violate the independence assumption and need a different model.

How is the Poisson distribution related to the binomial distribution?

The Poisson distribution is the limiting case of the binomial distribution as the number of trials n grows very large and the per-trial probability p shrinks toward zero, while their product n×p stays fixed at λ. This site's Binomial Distribution calculator needs both n and p specified separately; the Poisson distribution needs only their product, λ, which is why it is the natural choice when 'number of trials' isn't a meaningful concept — like counting emails received per day rather than successes out of a fixed number of attempts.

How is the Poisson distribution related to the exponential distribution?

They describe the same constant-rate random process from two angles. The Poisson distribution, covered here, gives the probability of a specific count of events within a fixed interval. This site's Exponential Distribution calculator gives the probability distribution of the waiting time between consecutive events. Both share the same rate parameter λ — a Poisson process with rate λ automatically implies exponentially distributed gaps between events with that same λ.

Why are the mean and variance both equal to λ?

It's a structural property of the Poisson distribution's derivation as a limit of the binomial distribution: the binomial variance n×p×(1−p) converges to λ×1 = λ as p shrinks toward 0 (since 1−p approaches 1), matching the mean n×p which converges to λ by construction. In practice, this makes λ do double duty — it's both the expected count and a direct measure of how spread out that count is likely to be, with the standard deviation equal to √λ.

Can λ be a decimal, like 2.5?

Yes — λ is an average rate, and averages are routinely fractional even though any single count of events is always a whole number. A call center might average 2.5 calls per minute even though no individual minute can have exactly 2.5 calls; λ = 2.5 is simply the long-run rate that the PMF uses to compute probabilities for each possible whole-number count k.

What does P(X = 0) tell you?

It's the probability of no events at all in the interval, which simplifies to e^(−λ) since λ^0 = 1 and 0! = 1. At λ = 4, as in the worked example, P(X = 0) = e^(−4) ≈ 0.0183, or about 1.8% — a useful quick check, since P(X = 0) is always the easiest Poisson probability to compute by hand and shrinks rapidly as λ grows.

References