Poisson Distribution Calculator

The probability of exactly k events in an interval, when events occur at an average rate λ, is P(X = k) = e^−λ λ^k / k!. For λ = 3 and k = 2 it is e⁻³ × 9 / 2 = 0.2240. The calculator also adds up ranges for at most, at least and between.

Find λ, the average number of events in your interval, then plug it and the number of events k into e^−λ λ^k / k!. For a mean of 4 and k = 6, that is e⁻⁴ × 4096 / 720 = 0.1042. The calculator does the arithmetic and shows the steps.

A Poisson distribution gives the probability of a given number of events in a fixed interval of time, distance, area or volume, when the events occur independently at a constant average rate. Typical examples are calls arriving at a help desk or flaws along a roll of cable.

Lambda (λ) is the average number of events in your interval. It is also the mean and the variance of the distribution. If calls come at 6 per hour and you ask about 30 minutes, λ = 3. It must be positive, and it need not be a whole number.

Multiply the average rate by the length of the interval. A rate of 4 calls per hour over 2 hours gives λ = 8, and over 30 minutes gives λ = 2. The rate table above lists more examples. Keep the units of the rate and the interval consistent.

Events must be counted over a fixed interval, occur independently of each other, occur at a constant average rate and happen one at a time. If events cluster, as aftershocks do, or the rate changes within the interval, the Poisson model fits poorly.

Use 1 minus the probability of none: 1 − e^−λ. For λ = 3 that is 0.9502. For λ = 0.5 it is 0.3935. The 'at least' option in the calculator also handles higher counts, such as at least 5 events.

Add the probabilities for 0, 1, up to k events. With λ = 3 and k = 2, P(X ≤ 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232. Choose the 'at most' option in the calculator and it returns this sum along with the exact and 'at least' values for the same k.

Take 1 minus the sum of the probabilities from 0 to k - 1. For λ = 3 and k = 5, P(X ≥ 5) = 1 − 0.8153 = 0.1847. This is the 'at least' option in the calculator, and it is the usual way to ask whether a busy period is unusual.

Add the exact probabilities from the lower to the upper count, inclusive. With λ = 2.5, P(1 ≤ X ≤ 3) = 0.2052 + 0.2565 + 0.2138 = 0.6755. Choose 'between' in the calculator and fill in both the lower count and the upper count, and it adds the terms for you.

The mean is λ. The expected number of events in the interval is the same number that describes the average rate for that interval. It need not be a whole number, so a mean of 2.5 means 2.5 events on average, even though counts are whole numbers.

The variance is also λ, so the standard deviation is the square root of λ. For λ = 4 the standard deviation is 2. A variance well above the mean in real data suggests clustering and means the Poisson model may not fit.

It is the square root of λ. For λ = 3 it is 1.732, and for λ = 12 it is 3.464. The calculator reports it with every result. Since variance equals the mean, the spread grows with the rate but more slowly.

The binomial counts successes in a fixed number of trials, with a probability on each trial. The Poisson counts events in an interval, with no fixed number of trials. When n is large and p is small, the binomial is close to a Poisson with λ = np.

When n is large and p is small, with np moderate. For n = 100 and p = 0.02, the binomial P(X = 2) is 0.2734 and the Poisson with λ = 2 gives 0.2707. The approximation table above compares the two for several counts.

The Poisson is discrete and skewed for small λ, while the normal is continuous and symmetric. For large λ, such as 20 or more, the Poisson is close to a normal curve with mean λ and standard deviation √λ. The calculator uses the exact Poisson.

They describe the same process from two sides. The Poisson counts how many events occur in a fixed interval, and the exponential distribution describes the waiting time between events. If events arrive at rate λ per interval, the average wait is 1/λ intervals.

It is e^−λ. With a mean of 3, there is a 4.98% chance of no events, and with a mean of 0.5 it is 60.65%. Enter k = 0 and the 'exactly' option in the calculator. The chance of at least one event is the remainder, 1 − e^−λ.

Use the average calls per interval as λ. If calls arrive at 3 per hour, the chance of exactly 2 in an hour is 0.2240, and the chance of 5 or more is 0.1847. The scenario table above lists several questions for this rate. It does not size staffing.

Use the average number of defects per unit as λ, such as flaws per metre of cable. With 0.5 flaws per metre, the chance of a clean metre is e^−0.5 = 0.6065. Scale λ by the length you are checking, as in the rate table.

Count events in a defined period and place, find the average and use it as λ. With 2.5 events per month, the chance of 1 to 3 in a month is 0.6755. Remember the independence rule: events that cause one another break the model.

The mode is the largest whole number not exceeding λ. For λ = 3 both 2 and 3 tie at 0.2240. For λ = 4.5 the mode is 4. The table of probabilities above lets you see where the peak falls for several means.

Yes. A mean of 0.5 means events are rare in the interval: the chance of none is 0.6065, one is 0.3033 and two or more is 0.0902. The calculator accepts any positive λ up to 100,000, so rare-event questions such as failures per thousand hours work as well.

No. In principle any number of events can occur, though the probabilities become tiny. The calculator sums up to the count you ask for, using logarithms so it stays accurate even for large λ. Counts far above the mean, such as 30 events at λ = 3, come out as very small probabilities.

Multiply or divide the rate in proportion to the interval length. 12 emails per day is 1.5 per hour in an 8-hour working day. Poisson rates scale linearly: the mean for a 4-hour window is twice the mean for a 2-hour window.

The counts in separate intervals are independent, so what happened in one interval does not change the next. The waiting time between events, which is exponential, is the memoryless quantity. The Poisson count itself is not memoryless in that sense, because it describes a total over a fixed interval, not a wait.

Up to 100,000. Probabilities are computed with logarithms of factorials so they do not overflow. For means that large, the Poisson is very close to a normal curve, but the calculator still returns the exact value instead of a normal approximation, so the answer matches the formula to the digits shown.

Look up or compute e^−λ, raise λ to the power k and divide by k factorial, then multiply. For λ = 2 and k = 3: e⁻² = 0.1353, 2³ = 8 and 3! = 6, so the probability is 0.1353 × 8 / 6 = 0.1804.

The cumulative probability P(X ≤ k) is the sum of the exact Poisson probabilities from 0 to k. For λ = 3, P(X ≤ 2) is 0.4232. The calculator shows it beside the exact probability and the at-least tail for the same k.

Count the average number of arrivals in the period you care about and use that as λ. If 6 customers arrive per hour on average, λ = 3 for 30 minutes, and the chance of exactly 2 arrivals is 0.2240. Arrivals must be independent for the model to fit.

Yes, it is a standard model for rare events such as failures per thousand hours, when the average count is small. With λ = 0.25 the chance of no failures is 0.7788 and of at least one is 0.2212. Enter a small positive λ in the calculator.

Compare the mean and variance of your counts. In a Poisson distribution they are equal, so a variance far above the mean suggests events cluster, and far below suggests events are too regular. Counts in separate intervals should also look independent.

Subtract the cumulative probability up to k from 1. For λ = 4 and k = 6, P(X > 6) = 1 − 0.8893 = 0.1107. Choose 'more than' and enter k in the calculator, which also shows the exact probability for k.