Poisson Distribution Calculator
Enter the average number of events per interval (or a rate and how long the interval is) and the count you are asking about.
P(X = k)
0.180447
P(X ≤ k)
0.857123
P(X ≥ k)
0.323324
P(X < k)
0.676676
P(X > k)
0.142877
Mean λ
2
Standard deviation √λ
1.4142
How it works
The Poisson distribution describes counts of events that happen independently at a steady average rate: calls arriving at a desk, typos on a page, particles hitting a detector, customers walking in. Its single parameter λ is the average number of events in the interval you are asking about. If you know a rate per unit time, λ is that rate multiplied by the length of the interval: 3 calls per hour over a 2-hour shift is λ = 6.
P(X = k) = e^(−λ) λ^k ÷ k!. The calculator builds the terms one after another from P(0) = e^(−λ) using P(k + 1) = P(k) × λ ÷ (k + 1), carried in logarithms so that a large λ does not underflow, and adds them up for the cumulative probabilities. The upper tail is summed directly rather than taken as 1 minus the lower tail, so small tail probabilities keep their precision.
A Poisson distribution has mean λ and variance λ, so its standard deviation is √λ. It is also the limit of the binomial distribution when there are many trials, each with a small chance of success, with np held at λ.
Formula
λ = rate × interval length (when built from a rate) P(X = k) = e^(−λ) · λ^k ÷ k! P(k + 1) = P(k) · λ ÷ (k + 1) (how the terms are accumulated) P(X ≤ k) = Σ P(X = j) for j = 0 … k P(X ≥ k) = Σ P(X = j) for j ≥ k μ = λ σ = √λ
Example
A help desk averages λ = 2 calls per minute. Exactly 3 calls in a minute: e⁻² × 2³ ÷ 3! = 0.135335 × 8 ÷ 6 = 0.180447. At most 3 calls: e⁻² × (1 + 2 + 2 + 1.3333) = 0.857123, the value a Poisson table lists as 0.8571. At least 3 calls is 1 − P(X ≤ 2) = 1 − 0.676676 = 0.323324. The standard deviation is √2 = 1.4142.
Built from a rate: 3 arrivals per hour over a 2-hour window is λ = 6, and exactly 4 arrivals has probability e⁻⁶ × 6⁴ ÷ 4! = 0.133853, with at most 4 arrivals 0.285057.
Assumptions and limitations
- Events occur independently at a constant average rate, and two cannot occur at exactly the same instant. Rush hours, seasonality, or one event making another more likely (contagion) all break the model.
- λ must be the average for the same interval you are asking about. A rate of 3 per hour over 20 minutes is λ = 1, not 3; use the rate-and-interval option to avoid the slip.
- λ is limited to 1,000,000. Probabilities are rounded to six decimal places for display; a value below 0.0000005 shows as 0.
Frequently asked questions
When should I use Poisson instead of binomial?
Use the binomial when there is a fixed number of trials each with a known success probability (10 coin flips). Use the Poisson when events occur over a continuous stretch of time or space at a known average rate with no fixed maximum (calls per hour). When the binomial's n is large and p small, the two agree closely with λ = np.
What is the probability of no events at all?
Set k = 0: P(X = 0) = e^(−λ). With λ = 2 that is 0.135335, so a quiet minute with no calls happens about 13.5% of the time.
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