Case 1
- Average rate (λ)
- 3
- Number of events (k)
- 2
0.224042
Open with these valuesPoisson Distribution Calculator
0.224042
Result: 0.224042One number describes the whole distribution: the average rate λ is both the mean and the variance. At three calls an hour, exactly two calls and exactly three calls are equally likely — 0.224042 each — and a completely quiet hour still happens 5 % of the time.
Worked examples
Case 2
- Average rate (λ)
- 2.5
- Number of events (k)
- 4
0.133602
Open with these valuesCase 3
- Average rate (λ)
- 5
- Number of events (k)
- 3
0.140374
Open with these valuesHow it's calculated
P(X = k) = λᵏ × e^(−λ) ÷ k!
- StepEnter the average number of events you expect in one interval.
- StepEnter the exact count whose probability you want.
- ResultRead the probability of exactly that many events.
What this number means
The Poisson distribution describes counting: how often something happens in a fixed window when the occasions for it are many and each one is unlikely — calls in an hour, decays in a second, typos on a page. What makes it useful is that a single number, the average rate λ, fixes the whole distribution. The three parts of the formula each do one job. λᵏ weighs k occurrences, e^(−λ) normalises the whole thing so the probabilities of all counts sum to 1, and k! divides the orders back out: the k events are indistinguishable, so counting them as sequences would count each outcome k! times over. At the values set here, λ = 3 and k = 2 give 9 × e^(−3) ÷ 2 = 0.224042. A shortcut hides in the formula: stepping from k to k + 1 multiplies the probability by λ ÷ (k + 1). At λ = 3 that factor is exactly 1 on the step from two to three, which is why both stand at 0.224042 — the distribution has a flat top rather than a peak. What the result does not say is when inside the interval the events fall, and it says nothing about a different interval: λ and k must refer to the same window. Three an hour and two per half hour is not a question this formula answers.
Two assumptions ride along
The events have to arrive independently and at a rate that stays constant across the interval. If they come in bursts, or the rate shifts inside the interval, the number on this page does not apply.
This is P(X = k), not P(X ≤ k)
The result is the probability of exactly k events. At λ = 3 exactly two is 0.224042, while at most two is 0.423190.
One parameter sets both mean and spread
λ is the mean and the variance at once, a built-in property of the distribution. A larger λ therefore means a higher count and a wider distribution together.
k is a whole number, λ need not be
The count k must be 0 or a positive integer, because a fraction of an event cannot be observed. The rate λ may be any positive decimal — 2.5 events per interval is perfectly valid.
Commonly misread
At λ = 3, exactly three events is clearly the most likely outcome.
Exactly two is just as likely — both are 0.224042. And even that peak covers barely more than a fifth of all intervals.
The result gives the probability of k events or fewer.
It holds for exactly k. At λ = 3 and k = 2 that is 0.224042, while at most two would be 0.423190.
The average rate is right, so the Poisson distribution fits.
It also assumes the events are independent and the rate stays constant over the interval. Events that cluster break that assumption, and the figure no longer carries.
Reference table
| λ, k | Case | P(X = k) |
|---|---|---|
| 3, 0 | three an hour, a quiet hour | 0.049787 |
| 3, 2 | three an hour, exactly two | 0.224042 |
| 3, 3 | three an hour, exactly three | 0.224042 |
| 2.5, 4 | rate 2.5, four events | 0.133602 |
| 5, 3 | rate 5, three events | 0.140374 |
Questions
How do I calculate a Poisson probability?
Use P(X = k) = λᵏ × e^(−λ) ÷ k!, where λ is the average rate and k the exact number of events. With λ = 3 and k = 2 that is 9 × e⁻³ ÷ 2 = 0.224042, about a 22.4 % chance of exactly two.
What is the Poisson distribution?
It gives the probability of a given number of independent events in a fixed interval of time or space, when they occur at a constant average rate λ. It models rare, independent events such as calls per hour, decays per second or typos per page.
Why are the mean and the variance equal?
Both are λ; that is a built-in property of the distribution, not a coincidence. The single parameter sets the typical count and the spread at once, so a larger λ means a higher count and a wider distribution.
Can the number of events be a decimal?
No. The count k must be a whole number of 0 or more, because you cannot observe a fraction of an event. The rate λ may be any positive decimal — 2.5 events per interval is perfectly valid.
How does it differ from the binomial distribution?
The binomial counts successes in a fixed number of trials at a fixed success probability; the Poisson counts events in a continuous interval at an average rate. The Poisson is the limit of the binomial when the trials are many and the success probability is small.
Sources and last check
Information, not professional advice.