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Poisson Distribution — Definition, Formula & Examples

The Poisson distribution is a probability distribution that gives the likelihood of a certain number of events occurring in a fixed interval of time or space, when those events happen independently at a known constant average rate.

A discrete random variable XX follows a Poisson distribution with parameter λ>0\lambda > 0, written X∼Pois(λ)X \sim \text{Pois}(\lambda), if its probability mass function is P(X=k)=e−λλkk!P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} for k=0,1,2,…k = 0, 1, 2, \ldots, where λ\lambda represents both the mean and the variance of the distribution.

Key Formula

P(X=k)=e−λ λkk!P(X = k) = \frac{e^{-\lambda}\,\lambda^k}{k!}
Where:
  • kk = Number of events you want the probability for (a non-negative integer)
  • λ\lambda = Average number of events per interval (the rate parameter)
  • ee = Euler's number, approximately 2.71828

How It Works

You use the Poisson distribution when counting how many times something happens in a fixed interval — such as the number of emails you receive per hour or the number of typos on a page. The key assumptions are that events occur independently, the average rate λ\lambda stays constant, and two events cannot happen at exactly the same instant. To find the probability of observing exactly kk events, plug kk and λ\lambda into the formula. As λ\lambda increases, the distribution becomes more symmetric and begins to resemble a normal distribution.

Worked Example

Problem: A call center receives an average of 4 calls per minute. What is the probability of receiving exactly 6 calls in a given minute?
Identify the parameters: The average rate is λ=4\lambda = 4 calls per minute, and we want k=6k = 6.
Substitute into the formula: Apply the Poisson probability mass function.
P(X=6)=e−4⋅466!P(X = 6) = \frac{e^{-4} \cdot 4^6}{6!}
Compute the components: Calculate each piece: e−4≈0.01832e^{-4} \approx 0.01832, 46=40964^6 = 4096, and 6!=7206! = 720.
P(X=6)=0.01832×4096720P(X = 6) = \frac{0.01832 \times 4096}{720}
Final calculation: Multiply and divide to get the probability.
P(X=6)=75.05720≈0.1042P(X = 6) = \frac{75.05}{720} \approx 0.1042
Answer: The probability of receiving exactly 6 calls in one minute is approximately 0.1042, or about 10.4%.

Another Example

Problem: A website averages 2 server errors per day. What is the probability of experiencing zero errors on a particular day?
Set up: Here λ=2\lambda = 2 and k=0k = 0.
Apply the formula: Substitute into the Poisson PMF. Note that 20=12^0 = 1 and 0!=10! = 1.
P(X=0)=e−2⋅200!=e−2P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2}
Evaluate: Compute the numerical value.
P(X=0)≈0.1353P(X = 0) \approx 0.1353
Answer: There is roughly a 13.5% chance of zero server errors on a given day.

Visualization

Why It Matters

The Poisson distribution appears throughout introductory statistics and probability courses, especially in problems involving queuing theory, insurance claims, and radioactive decay. In fields like operations research and epidemiology, it models rare-event counts — for instance, the number of patients arriving at an emergency room per hour. Understanding it also lays the groundwork for the Poisson process, a core topic in stochastic modeling.

Common Mistakes

Mistake: Using the Poisson formula when events are not independent or the rate is not constant.
Correction: Before applying the distribution, verify that events occur independently and that λ\lambda does not change across the interval. If the rate varies (e.g., call volume spikes at noon), you need a different model or must break the interval into subintervals.
Mistake: Confusing λ\lambda for one interval with λ\lambda for a different-sized interval.
Correction: The parameter λ\lambda must match the interval in your problem. If the average is 4 calls per minute and you want the probability over 3 minutes, use λ=12\lambda = 12, not λ=4\lambda = 4.

Related Terms