Poisson Distribution Calculator
Enter the average rate λ and a count k, and this Poisson distribution calculator gives the chance of exactly, at least, at most or between k events, with the formula worked through.
P(X = 2)
- Probability of exactly 2 events P(X = 2) for a Poisson distribution with λ = 3.
- 22.4%
- Formula
- P(X = k) = λ^k e^-λ / k!
- Worked through
- 3^2 x e^-3 / 2! = 9 x 0.0497871 / 2 = 0.224042
- Probability of not exactly 2 The complement: 1 minus the answer above.
- 77.6%
- Mean and variance Both equal λ. That is the Poisson distribution's signature, and a quick check on real data.
- 3 and 3
- Standard deviation The square root of λ.
- 1.73205
- Most likely count λ is a whole number, so 2 and 3 tie at 22.4% each.
- 2 and 3
| k | P(X = k) | P(X ≤ k) | P(X ≥ k) |
|---|---|---|---|
| 0 | 0.04979 | 0.04979 | 1 |
| 1 | 0.1494 | 0.1991 | 0.9502 |
| 2 | 0.224 | 0.4232 | 0.8009 |
| 3 | 0.224 | 0.6472 | 0.5768 |
| 4 | 0.168 | 0.8153 | 0.3528 |
| 5 | 0.1008 | 0.9161 | 0.1847 |
| 6 | 0.05041 | 0.9665 | 0.08392 |
| 7 | 0.0216 | 0.9881 | 0.03351 |
| 8 | 0.008102 | 0.9962 | 0.0119 |
| 9 | 0.002701 | 0.9989 | 0.003803 |
| 10 | 0.0008102 | 0.9997 | 0.001102 |
| 11 | 0.000221 | 0.9999 | 0.0002923 |
- Table values have four significant figures.
How the Poisson Distribution Calculator works
The Poisson distribution counts events in a fixed stretch of time or space when they happen independently at a steady average rate. Calls to a help desk in an hour, goals in a soccer match, decays from a radioactive sample in a second and typos in a long document are all modeled this way. Its one parameter is λ, the average count for the interval you are asking about.
The Poisson formula is P(X = k) = λ^k e^-λ / k!. Say a help desk averages 3 calls an hour. The chance of exactly 2 calls in the next hour is 3^2 x e^-3 / 2! = 9 x 0.049787 / 2 = 22.40%, and the chance of none is e^-3 = 4.98%. With λ = 2, no event happens 13.53% of the time and at least one 86.47%.
Questions about a range add terms together. At most k is the sum from 0 to k. At least k is 1 minus the chance of at most k - 1, so at least 1 is always 1 - e^-λ. Between two counts adds every term from the first to the second, both ends included. For the help desk, at most 4 calls in an hour is 81.53% and at least 5 is 18.47%.
The formula for the Poisson distribution has a property worth knowing: the mean and the variance are both λ, so the standard deviation is the square root of λ. At 3 calls an hour that is about 1.73. The most likely count is the whole part of λ, and when λ is a whole number two counts tie: 2 and 3 calls are each 22.40% likely. If real counts vary much more than their average suggests, the events are probably not independent, and the Poisson model will understate the spread.
The distribution is the limit of the binomial when there are many chances for an event and each one is unlikely, with λ standing for the number of trials times the chance per trial. It assumes the rate holds steady across the interval and that one event does not make the next more or less likely. Siméon Denis Poisson published it in 1837.
The formula as written breaks down for large λ. In double precision e^-λ rounds to zero past about λ = 745, and 171! is too large to store at all. This calculator works in logarithms instead, using Catherine Loader's method, the one R uses, which keeps full precision up to a rate of a billion. For λ = 1,000 the chance of exactly 1,000 events is 1.26%.
Poisson Distribution Calculator FAQ
What is the Poisson formula?
P(X = k) = λ^k e^-λ / k!, where λ is the average number of events in the interval and k is the count you want the chance of. For λ = 3 and k = 2 it is 9 x e^-3 / 2 = 22.40%.
When should you use the Poisson distribution?
For counts of independent events at a steady average rate over a fixed interval: arrivals, calls, defects per batch, accidents per month. It fits poorly when events cluster, or when the rate changes over the interval.
How do I find λ?
Use the average count for the same length of interval you are asking about. A rate of 12 an hour is λ = 3 for a 15-minute window and λ = 24 for two hours.
What is the probability of at least one event?
1 - e^-λ, because the only way to miss is a count of 0, and P(X = 0) = e^-λ. With λ = 2 that is 86.47%.
Why are the mean and variance both λ?
Both come out of the same sum over the formula, and each equals λ. A data set whose variance is far above its mean is overdispersed, which is a sign the Poisson model does not fit it.