Poisson Distribution Calculator
Probability of an exact, at most, at least, or between count from the rate lambda, with the mean and standard deviation.
What it does
The Poisson distribution answers a very practical kind of question: how many times will something happen? Calls to a help desk in an hour, defects per batch, particles hitting a detector, accidents at an intersection in a year.
It works whenever events arrive independently, one at a time, at a steady average rate, and you fix the interval up front. The single dial is lambda: the expected count in that interval. Set lambda to 3 and the model describes an interval where three events are expected on average.
The probability of exactly k events is lambda to the k, times e to the minus lambda, divided by k factorial. That combination balances two pulls: more events need rarer arrivals, so the probabilities rise to a peak near lambda and then fall away.
For range questions the tool sums those exact terms. The chance of at most k is the sum from 0 to k; at least k is one minus the sum below k; between two values is the difference of two sums.
Mean and variance both equal lambda, so the spread is the square root of lambda. That is why a bigger interval looks steadier in percentage terms: lambda of 100 has a standard deviation of 10, only 10% of the mean.
How to use
- Enter lambda, the average count for the interval you care about.
- Pick the question: exactly k, at most k, at least k, or between two values.
- Enter the count or counts you are asking about.
- Read the probability as a decimal and a percentage.
- Check the mean and standard deviation to see the typical spread of counts.
- Keep the interval consistent: a rate per hour needs a count for one hour.
Formula
- lambda
- Expected number of events in the interval.
- k
- The count you are asking about.
- e
- Euler's number, about 2.71828.
The mean and the variance of the distribution are both lambda, which makes the standard deviation the square root of lambda. The formula assumes the rate is constant across the whole interval and that the events do not cluster or repel each other. When arrivals bunch up, real counts show more spread than the model predicts.
Worked example
Example: exactly 2 events with average 3
Lambda of 3, asking for exactly 2 (defaults).
- lambda^k = 3^2 = 9.
- e^(-lambda) = e^(-3) = 0.049787.
- k! = 2! = 2.
- P(X = 2) = 9 x 0.049787 / 2 = 0.2240.
- So there is about a 22.4% chance of exactly two events when three are expected.
Reference table
| Events k | P(X = k) | P(X <= k) |
|---|---|---|
| 0 | 0.0498 | 0.0498 |
| 1 | 0.1494 | 0.1991 |
| 2 | 0.2240 | 0.4232 |
| 3 | 0.2240 | 0.6472 |
| 4 | 0.1680 | 0.8153 |
| 5 | 0.1008 | 0.9161 |
| 6 | 0.0504 | 0.9665 |
Limits
- Counts only, in a fixed interval, with a single constant rate.
- Real arrival rates often vary by time of day or season, which breaks the constant-rate assumption.
- Events that cluster, like contagion or aftershocks, need models such as the negative binomial.
- Very large lambda with exact counts still works here, but the numbers get tiny; use ranges instead.
Common mistakes
- Mixing units: a per-hour lambda with a per-day count.
- Asking at least k but computing at most k, which reports the complement of the intended chance.
- Forgetting that lambda is an average, so it can be a decimal like 0.5.
- Treating the result as a guarantee for the next interval rather than a long-run proportion.
- Using Poisson for a fixed number of trials; that is the binomial's job.
Try it yourself
Quick checks. Answers sit tucked under each question.
Lambda 2, chance of zero events?
e^(-2) = 0.1353, about 13.5%.
Lambda 4, chance of exactly 4?
0.1954, about 19.5%.
Lambda 3, chance of at most 2?
0.0498 + 0.1494 + 0.2240 = 0.4232, about 42.3%.
Lambda 1, chance of at least 1?
1 - 0.3679 = 0.6321, about 63.2%.
Questions
When should I use the Poisson distribution?
For counts of independent events in a fixed interval at a steady rate: calls per hour, defects per batch, arrivals per minute. The interval and the rate must match.
What does lambda mean exactly?
Lambda is the average count for the interval you chose. It can be any positive number, whole or decimal.
Why is the variance equal to the mean?
That is a defining property of the Poisson distribution. It means the standard deviation is the square root of lambda, so counts spread out more when rates are higher.
How is this different from the binomial?
Binomial counts successes in a fixed number of trials, each with a probability p. Poisson counts events in continuous time or space, dialed by the rate lambda.
Can I use it for events that can repeat at the same spot?
Yes, as long as arrivals are independent and the rate stays constant. If events trigger more events, the counts spread more than Poisson predicts.
Is there a connection to waiting times?
Yes. If counts follow a Poisson with rate lambda, the gaps between events follow an exponential distribution with mean 1/lambda.
Sources
- NIST Engineering Statistics Handbook: Poisson distribution
- R manual: Poisson distribution functions (density, mean and variance equal to lambda)