Poisson Distribution Calculator — Step-by-Step Solutions

The Poisson Distribution Calculator helps you compute probabilities for events that occur at a known average rate. Whether you need the exact probability P(X=k), cumulative probabilities, or the mean and variance of the distribution, this tool handles it all.

Simply enter the average rate (lambda) and the number of events (k) to get instant results with step-by-step working and a visual bar chart of the probability distribution.

Lambda (average rate)

k (number of events)

Your calculations will appear here

How to Use the Poisson Distribution Calculator

Enter lambda, the average number of events in your interval (a rate such as 3.5 calls per hour). It must be zero or more.
Choose a mode: P(X = k) for the chance of exactly k events, Cumulative P(X <= k) for at most k events, Mean & Variance, or Distribution Chart, which draws the bar chart and highlights the most likely count.
For the first two modes, enter k, the number of events, as a whole number. The cumulative mode also shows P(X > k), which is the chance of more than k events.
Match the interval to lambda. If the average is 6 per hour and you care about 15 minutes, use lambda = 1.5.

Formula and Theory

The Poisson distribution models the number of times an event occurs in a fixed interval of time or space when events happen independently at a constant average rate. Typical uses are calls arriving at a switchboard, defects per metre of cable, emails per hour, or typos per page.

The probability of exactly k events is P(X = k) = e^(-lambda) x lambda^k / k!, where lambda is the average count. The mean and the variance both equal lambda, so the standard deviation is sqrt(lambda). Counts that vary much more than the mean suggest the Poisson model does not fit.

Cumulative probabilities add the individual terms: P(X <= k) = P(0) + P(1) + ... + P(k). For "at least" questions use the complement, P(X >= k) = 1 - P(X <= k - 1).

The distribution is a good approximation to the binomial when n is large and p is small, with lambda = np. The model assumes events do not cluster, so a system with bursts needs a different one.

Worked Examples

Example 1: Exactly k events

Problem: A help desk receives 3.5 calls per hour on average. What is the probability of exactly 4 calls in an hour?

Solution: P(X = 4) = e^(-3.5) x 3.5^4 / 4!. e^(-3.5) = 0.030197, 3.5^4 = 150.0625 and 4! = 24. So P = 0.030197 x 150.0625 / 24 = 0.1888.

Answer: P(X = 4) = 0.1888 (18.88%)

Example 2: At most k events

Problem: A page has on average 2 typing errors. What is the probability of at most 3 errors?

Solution: P(0) = e^(-2) = 0.1353. P(1) = 2 e^(-2) = 0.2707. P(2) = 2^2/2 x e^(-2) = 0.2707. P(3) = 2^3/6 x e^(-2) = 0.1804. Sum = 0.1353 + 0.2707 + 0.2707 + 0.1804 = 0.8571.

Answer: P(X <= 3) = 0.8571

Example 3: At least k events using the complement

Problem: A call centre averages 6 calls per hour. What is the probability of 10 or more calls in an hour?

Solution: 10 or more is the complement of 9 or fewer. P(X <= 9) with lambda = 6 is 0.9161, so P(X >= 10) = 1 - 0.9161 = 0.0839.

Answer: P(X >= 10) = 0.0839 (8.39%)

Example 4: At least one event in a short interval

Problem: A shop averages 2 customers per hour. What is the chance of at least one customer in a 15-minute period?

Solution: Scale the rate to the interval: lambda = 2 x 0.25 = 0.5. P(X = 0) = e^(-0.5) = 0.6065, so P(at least one) = 1 - 0.6065 = 0.3935.

Answer: 0.3935 (39.35%)

Frequently Asked Questions

When should I use the Poisson distribution?

Use it to count independent events in a fixed interval when the average rate is known and constant, and the count has no fixed upper limit. If you have a fixed number of trials with a success chance, use the binomial distribution instead.

What does lambda mean?

Lambda is the expected number of events in your interval. It is both the mean and the variance of the distribution. Change it whenever you change the length of the interval.

How do I find P(X > k) or P(X >= k)?

Use the cumulative mode. P(X > k) is shown directly, and P(X >= k) is 1 minus the cumulative probability for k - 1.

Why are the mean and variance equal?

It is a defining property of the Poisson distribution, and a practical check: if your data have a variance much larger than the mean (overdispersion), a Poisson model understates the spread.

What is the most likely value?

The mode is the largest whole number not above lambda. If lambda is a whole number, both lambda and lambda - 1 are equally likely. The chart mode highlights it.

Can k be a decimal?

No. The number of events is a whole number, so k must be 0, 1, 2, and so on. Lambda, the average rate, can be any non-negative decimal.