Compute exact, at-most, and at-least Poisson probabilities
This free calculator computes the Poisson probability distribution. Enter the average rate \( \lambda \) and target count \( x \) to get:
Enter \( \lambda > 0 \) and a non-negative integer \( x \). Both the mean and the variance of the distribution equal \( \lambda \).
If event \( A \) occurs at an average rate \( \lambda \) per unit interval, then the probability of observing exactly \( x \) events in an interval of the same length is:
\[ P(X = x) = \frac{e^{-\lambda}\, \lambda^x}{x!} \]Mean and variance. For a Poisson distribution: \( \mu = \lambda \) and \( \sigma^2 = \lambda \), so the standard deviation is \( \sqrt{\lambda} \).
The default inputs match the following scenario. Compare the manual calculation below with the outputs from the calculator.
Problem. An event occurs at an average rate of 4 times per 24-hour period. Find the probability that the event occurs:
Setup. \( \lambda = 4 \), and the interval matches (24 hours).
(a) Exactly 5:
\[ P(X = 5) = \frac{e^{-4} \cdot 4^5}{5!} = \frac{0.0183156 \cdot 1024}{120} \approx 0.156293 \](b) At most 5: sum \( P(0) + P(1) + \dots + P(5) \):
| x | P(X = x) |
|---|---|
| 0 | 0.0183156 |
| 1 | 0.0732626 |
| 2 | 0.1465251 |
| 3 | 0.1953668 |
| 4 | 0.1953668 |
| 5 | 0.1562934 |
Sum:
\[ P(X \le 5) \approx 0.785130 \](c) At least 5: use the complement \( P(X \ge 5) = 1 - P(X \le 4) \):
\[ P(X \le 4) \approx 0.628837 \] \[ P(X \ge 5) = 1 - 0.628837 \approx 0.371163 \]Note. \( P(X \ge 5) \) includes \( x = 5 \) itself, so we subtract \( P(X \le 4) \) — not \( P(X \le 5) \). Using the latter would give \( P(X > 5) \), which is a different event.
A call center receives an average of 6 calls per hour. What is the probability of receiving exactly 8 calls in a given hour?
\( \lambda = 6 \), \( x = 8 \):
\[ P(X = 8) = \frac{e^{-6} \cdot 6^8}{8!} \approx 0.103258 \]A website receives an average of 2.5 visitors per minute. What is the probability of receiving at most 3 visitors in a given minute?
\( \lambda = 2.5 \), \( x \le 3 \):
\[ P(X \le 3) = \sum_{x=0}^{3} \frac{e^{-2.5} \cdot 2.5^x}{x!} \approx 0.757576 \]A radioactive source emits an average of 12 particles per second. What are the mean, variance, and standard deviation of the count in one second?
For a Poisson distribution with \( \lambda = 12 \):
\(P(X = x) = \dfrac{e^{-\lambda} \lambda^x}{x!}\), where \(\lambda\) is the average rate per unit interval and \(x\) is the number of events. Requires \(\lambda > 0\) and \(x\) a non-negative integer.
Both equal \(\lambda\): \(\mu = \lambda\) and \(\sigma^2 = \lambda\). The standard deviation is \(\sqrt{\lambda}\).
When events occur independently at a constant average rate \(\lambda\) over a fixed interval. Common cases: call arrivals, radioactive decays, typos per page, accidents per day. Also used to approximate the binomial when \(n\) is large and \(p\) is small, with \(\lambda = np\).
The calculator uses log-space arithmetic for numerical stability up to \(\lambda \approx 1000\) and \(x \approx 1000\). Beyond that, floating-point limits apply.
Last updated: October 20, 2026