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Poisson Probability Calculator

Find the probability of k events with the Poisson distribution.

How it is calculated

For events that occur independently at an average rate λ per interval, the probability of exactly k events is P(X = k) = e−λ λk / k!. The tool works with the logarithms of the terms, so a large λ such as 800 still gives a correct answer, adds them to get the cumulative P(X ≤ k), and finds P(X ≥ k) = 1 − P(X ≤ k) + P(X = k). All three are shown with five decimals.

λ and k cannot be negative, k is cut to a whole number and must be at most 1000. λ is the average count for the same interval you are asking about; if the rate is 12 per hour and you ask about 30 minutes, use λ = 6. The model assumes events are independent and the rate is constant.

Worked example

λ = 4 and k = 2. e−4 ≈ 0.018316. P(0) = 0.018316, P(1) = 0.018316 × 4 = 0.073263, P(2) = 0.073263 × 4 / 2 = 0.146525. So P(X = 2) = 0.14653 and P(X ≤ 2) = 0.23810. P(X ≥ 2) = 1 − 0.23810 + 0.14653 = 0.90842, as the tool shows.

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Common mistakes

  • Using a rate for one interval and asking about a different interval without rescaling λ.
  • Reading P(X ≥ k) as P(X > k); the tool's "at least k" includes k itself.
  • Applying Poisson to events that cluster (such as aftershocks) or that depend on each other.
  • Entering λ as a probability or percent; it is an average count and can be above 1.

About this calculator

The Poisson distribution gives the probability of a specific number of events k occurring in a fixed interval, given a known average rate λ: P(X=k) = (λᵏ·e^(−λ)) / k!. It applies to events that happen independently and at a constant average rate — this calculator also gives the cumulative probabilities P(X≤k) and P(X≥k) for finding "at most" or "at least" that many events.

Statisticians, actuaries, and operations researchers use the Poisson distribution to model call-center arrivals, website traffic spikes, defects per manufactured unit, radioactive decay counts, and rare-disease incidence rates. It's especially useful when events are individually rare but there are many opportunities for them to occur — the classic textbook example is the number of Prussian cavalry soldiers killed by horse kicks per year, one of the earliest documented applications of the distribution.

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Comments (1)

  • Editorial team

    Set k to 0 to get the chance of no events at all: with an average of 3 calls per hour, P(X = 0) is 0.04979, so a silent hour is rare. Large rates work too, since the terms are computed in logarithms; λ = k = 800 gives P(X ≤ k) of about 0.50940. Keep k at or below 1000. What events are you modelling?

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