The Poisson Distribution: Modeling Calls, Arrivals, and Rare Events
October 29, 2026
The Poisson distribution gives the probability of a given number of events occurring in a fixed interval of time or space, when events happen independently at a constant average rate. Typical examples include calls arriving at a help desk, customers entering a shop, typos on a page, accidents at an intersection, or radioactive decays per second. If the average rate is λ events per interval, the probability of exactly k events is P(k) = e^(−λ) × λ^k ÷ k!.
A help desk receives on average 4 calls per hour (λ = 4). The probability of no calls in an hour is e^(−4) = 0.0183, or 1.8%. The probability of exactly 4 calls is e^(−4) × 4⁴ ÷ 24 = 0.1954, or 19.5%.
Skip the math and get your answer instantly:
Open the Permutation & Combination Calculator →Probability Table for λ = 4
The most likely counts are 3 and 4, each at 19.5%. The mean and the variance of a Poisson distribution are both λ, so the standard deviation is √4 = 2. About two thirds of hours would have between 2 and 6 calls. The chance of 8 or more calls in an hour is 1 minus the sum of probabilities for 0 to 7, which equals 5.1%: staffing for the average alone would leave the desk overwhelmed about one hour in twenty.
| Calls in an hour (k) | Probability | Cumulative probability |
|---|---|---|
| 0 | 1.83% | 1.83% |
| 1 | 7.33% | 9.16% |
| 2 | 14.65% | 23.81% |
| 3 | 19.54% | 43.35% |
| 4 | 19.54% | 62.89% |
| 5 | 15.63% | 78.51% |
| 6 | 10.42% | 88.93% |
| 7 | 5.95% | 94.89% |
| 8 or more | 5.11% | 100% |
Scaling the Rate
If the rate is 4 per hour, then for a 15-minute window λ = 1, and the probability of no calls is e^(−1) = 36.8%. For 2 hours, λ = 8, and the probability of no calls is e^(−8) = 0.03%. Scale λ proportionally to the length of the interval. The time between events in a Poisson process follows an exponential distribution with mean 1/λ: with 4 calls an hour, the average gap is 15 minutes, and the chance of waiting more than 30 minutes for the next call is e^(−2) = 13.5%.
Using It for Planning
Queueing and staffing: call centers use the Poisson model with the Erlang formulas to find how many agents keep waiting times acceptable. Inventory: demand for slow-moving items per week may be approximately Poisson, helping set safety stock. Quality control: defects per square meter of fabric. Insurance and safety: counts of rare claims or accidents. Epidemiology: counts of cases. If the average number of events per day is 0.5, the probability of at least one event in a day is 1 − e^(−0.5) = 39.3%.
When the Model Fails
Events may not be independent (accidents cluster during storms), the rate may vary over time (more calls in the morning), or events can arrive in batches. In these cases the variance exceeds the mean (overdispersion), and other models like the negative binomial are better. Check the data: compute the mean and variance of counts; if they are far from equal, the Poisson is not a good fit. For rare events with many opportunities, such as 1,000 customers each with a 0.4% chance, the Poisson with λ = 4 approximates the binomial well.
Frequently Asked Questions
What is the Poisson probability formula?
P(k) = e^(−λ) λ^k ÷ k!
What are the mean and variance of a Poisson distribution?
Both equal λ.
How do I adjust λ for a different time period?
Multiply the rate by the length of the new interval.
What is the chance of at least one event?
1 − e^(−λ).
When is the Poisson model inappropriate?
When events are not independent or the rate changes over time.