Poisson Distribution Calculator

Calculate Poisson Probabilities

Poisson Distribution Formula

Formula

\[ P(X = k) = \frac{e^{-\lambda}\,\lambda^k}{k!} \\ \[ P(X \le k) = \sum_{j=0}^{k} \frac{e^{-\lambda}\,\lambda^j}{j!} \\ \[ P(X \ge k) = 1 - \sum_{j=0}^{k-1} \frac{e^{-\lambda}\,\lambda^j}{j!} \]

Where:

\(\lambda\) — the average (mean) number of events per interval; must be positive
\(k\) — the number of events; a whole number, 0 or larger
\(e \approx 2.71828\) — Euler's number
\(k!\) — the factorial of \(k\)

The first formula gives the probability of observing exactly \(k\) events; the two sums give the cumulative probabilities \(P(X \le k)\) and \(P(X \ge k)\).

Poisson Distribution – Calculation Example

Example: a call center receives an average of \(\lambda = 3\) calls per hour. What is the probability of receiving exactly \(k = 5\) calls in the next hour?

\[ P(X = 5) = \frac{e^{-3} \cdot 3^5}{5!} = \frac{0.0497871 \cdot 243}{120} = \frac{12.0981}{120} \approx 0.100819 \]

Result: \(P(X = 5) \approx 0.1008\), or about 10.08%. With \(\lambda = 3\), the mean and variance are both 3 and the standard deviation is \(\sqrt{3} \approx 1.732\).

How to use the calculator

Select the Probability type: exactly k events, a range between two counts, at most k events, or at least k events.
Enter the Average rate (λ) — the mean number of events per interval (per hour, per page, per batch, etc.). It must be greater than 0.
Enter the Number of events (k) — a whole number, 0 or larger. In range mode this value acts as the lower bound k₁.
If you chose the range mode, also fill in the Upper event count (k₂), which must be greater than or equal to k.
Read the Requested probability at the top of the results, then check \(P(X = k)\), the cumulative \(P(X \le k)\), the mean, and the variance below it.

What is the Poisson distribution?

The Poisson distribution models the number of times a random event occurs within a fixed interval of time, space, or volume, when events happen independently at a constant average rate. It is named after French mathematician Siméon Denis Poisson and is one of the most widely used discrete probability distributions in statistics.

A single parameter, \(\lambda\), fully describes the distribution: it is both the mean and the variance, so \(\mu = \sigma^2 = \lambda\). Classic applications include the number of customer arrivals per hour, typos per page, radioactive decays per second, website visits per minute, or insurance claims per month. If you know the historical average count for the interval, the Poisson formula lets you estimate the probability of any specific outcome.

The most likely outcome is near \(\lambda\) itself, and probabilities taper off on both sides. For example, with \(\lambda = 3\), getting 2, 3, or 4 events is fairly likely, while 0 or 8 events is uncommon but not impossible. The calculator's neighbouring-probabilities table shows this spread around your chosen k, which helps you judge whether an observed count is typical or unusual. Cumulative answers (at most / at least) are often more useful than exact probabilities: they tell you the chance of hitting a threshold, such as "more than 10 arrivals" or "no more than 1 defect".

Assumptions, limitations, and tips

Independence: one event must not influence the probability of another. Arrivals triggered by the same cause (e.g. a viral post spiking visits) violate this assumption.
Constant rate: \(\lambda\) must stay the same across the interval. If the rate varies by hour or season, compute each period separately or use a different model.
Count data only: k counts whole events in a fixed interval. The Poisson distribution does not describe continuous quantities like durations or distances.
Scaling λ: if you double the interval, double \(\lambda\). A rate of 3 calls per hour becomes \(\lambda = 6\) for a two-hour window or \(\lambda = 0.05\) per minute.
Large λ: as \(\lambda\) grows (roughly above 20), the Poisson distribution becomes nearly symmetric and closely approximates a normal distribution with mean \(\lambda\) and variance \(\lambda\).
Complements: remember that \(P(X \ge k) = 1 - P(X \le k-1)\). The "at least" answer already accounts for this, so \(P(X \ge 0)\) is always 100%.

Frequently Asked Questions

How do I calculate a Poisson probability by hand without a calculator?

Compute \(P(X = k) = e^{-\lambda}\lambda^k / k!\) step by step: find \(e^{-\lambda}\), raise \(\lambda\) to the power \(k\), multiply them, and divide by \(k!\). For example, with \(\lambda = 2\) and \(k = 3\): \(e^{-2} \approx 0.1353\), \(2^3 = 8\), \(3! = 6\), so \(P(X = 3) = 0.1353 \times 8 / 6 \approx 0.1804\). For cumulative probabilities, add up the individual terms from 0 up to \(k\).

When should I use the Poisson distribution instead of the binomial distribution?

Use Poisson when you count events over a continuous interval (time, area, volume) and know only the average rate. Use binomial when you have a fixed number of independent trials, each with the same success probability. The Poisson is actually the limit of the binomial when the number of trials is very large and the success probability is very small — a common rule of thumb is that Poisson works well as an approximation when \(n \ge 50\) and \(p \le 0.1\).

Can λ be a decimal or fractional number?

Yes, \(\lambda\) can be any positive number, including decimals — only \(k\) must be a whole number. For instance, if a rare error occurs on average 0.7 times per report, \(\lambda = 0.7\) is perfectly valid. A fractional \(\lambda\) simply means events are uncommon in the interval you chose. The only restrictions are \(\lambda > 0\) and \(k\) being an integer of 0 or larger.

What is the probability of zero events, P(X = 0)?

Since \(0! = 1\) and \(\lambda^0 = 1\), the formula simplifies to \(P(X = 0) = e^{-\lambda}\). With \(\lambda = 3\), that's \(e^{-3} \approx 0.0498\), or about a 5% chance of no events. This figure is useful in reliability work — for example, estimating the chance that no machine failure occurs during a shift.

How do I estimate λ from historical data?

Divide the total number of observed events by the number of intervals. If a shop served 150 customers over 30 one-hour periods, then \(\lambda = 150 / 30 = 5\) customers per hour. The longer your observation window, the more stable this estimate becomes. Keep in mind that \(\lambda\) estimated from limited data carries uncertainty, so treat resulting probabilities as approximations rather than exact values.

Why does my manual calculation give 0 or a huge wrong number for large k?

For large \(k\), both \(\lambda^k\) and \(k!\) become astronomically large and can overflow a calculator or spreadsheet, even though their ratio is small. The workaround is to work with logarithms: \(\ln P = -\lambda + k\ln\lambda - \ln(k!)\), then exponentiate at the end. Another common mistake is entering \(k\) as a non-integer or forgetting that \(e^{-\lambda}\) must be multiplied, not added.

What is the most likely number of events (the mode)?

The most likely outcome is \(k = \lfloor\lambda\rfloor\) — the largest whole number not exceeding \(\lambda\). If \(\lambda\) is itself a whole number, then both \(k = \lambda\) and \(k = \lambda - 1\) are equally likely. For example, with \(\lambda = 4\), both 3 and 4 events have the highest probability; with \(\lambda = 4.5\), the single most likely count is 4.