Poisson Distribution Calculator

Work out how likely it is to observe a given number of events in a fixed window of time or space — arrivals, calls, defects, accidents — when you only know the long-run average rate.

Before You Calculate

  • State the rate λ and the count k over the same window: if λ is events per hour, k must be a count for one hour.
  • λ does not need to be a whole number — an average of 2.7 support tickets per day is a perfectly valid rate.
  • Check that events occur independently and one at a time; scheduled or clustered arrivals break the model.

What the Tool Returns

P(X = k): The probability of observing exactly k events in the window.

P(X ≤ k) and P(X ≥ k): Cumulative probabilities for planning questions such as "what are the odds of two or more failures this month?"

Distribution summary: The mean λ, the variance λ (a defining Poisson property), and the standard deviation √λ.

When the Poisson Model Fits

The Poisson distribution describes counts of events that happen randomly but at a stable average rate. Three conditions need to hold:

  • Independence: One event occurring does not make another more or less likely. A machine breakdown that triggers a cascade of further breakdowns violates this.
  • Constant rate: The average number of events per unit of time or space stays the same across the window. Lunchtime rushes at a restaurant mean the hourly rate is not constant across the day.
  • No simultaneous events: Events arrive one at a time. Group arrivals — a bus unloading thirty customers at once — should be modeled as batch arrivals instead.

Unlike the binomial distribution, there is no fixed number of trials: the count in a Poisson process has no hard upper limit, only rapidly shrinking probabilities for large values.

The Poisson Probability Formula

The calculator evaluates the probability mass function directly:

P(X = k) = (λ^k × e^(−λ)) / k!

μ = λ

σ² = λ

σ = √λ

Where:

  • λ = average number of events per interval (the rate)
  • k = number of events you are asking about
  • e ≈ 2.71828, the base of the natural logarithm
  • k! = k factorial, the product 1 × 2 × … × k

The equality of mean and variance is the signature of the Poisson distribution. If your observed counts have a variance far larger than their mean — overdispersion — a negative binomial model usually fits better.

Rescaling the Rate to Your Window

A convenient property of Poisson processes is that the rate scales linearly with the size of the window. If a website averages 12 signups per hour, then over a 15-minute window λ = 12 × (15/60) = 3, and over an 8-hour shift λ = 96. Always rescale λ before entering it, rather than rescaling the answer afterward.

The same logic works for space instead of time: flaws per square meter of fabric, potholes per mile of road, or typos per page. For example, with an average of 1 typo per page, the probability that a given page is typo-free is P(X = 0) = e^(−1) ≈ 0.367879 — about 37%.

Real Applications

Poisson counts appear across operations, science, and risk management.

  • Call centers and queues: Staffing models start from the distribution of arrivals per interval to predict wait times.
  • Reliability engineering: The number of component failures in a maintenance period drives spare-part inventory decisions.
  • Insurance: Claim counts per policy year are the classic actuarial application, feeding directly into premium calculations.
  • Epidemiology and biology: Case counts in a region, mutations per genome segment, and radioactive decays per second all follow Poisson models when the underlying rate is stable.

Worked Example: Custom Cake Orders

A bakery receives an average of λ = 4 custom cake orders per day. What is the probability of receiving exactly 2 orders tomorrow?

  1. Write the formula: P(X = 2) = (4² × e⁻⁴) / 2!
  2. Evaluate each piece: 4² = 16, e⁻⁴ ≈ 0.018316, and 2! = 2.
  3. Combine: P(X = 2) = (16 × 0.018316) / 2 ≈ 0.146525, a 14.7% chance.

The cumulative outputs answer the planning questions. P(X ≤ 2) = P(0) + P(1) + P(2) ≈ 0.018316 + 0.073263 + 0.146525 ≈ 0.238103, so there is roughly a 24% chance of a slow day with two orders or fewer. Conversely, P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − 0.091578 ≈ 0.908422 — the bakery can count on at least two orders about 91% of the time.

The distribution summary is immediate: mean = 4 orders, variance = 4, and standard deviation √4 = 2. Entering λ = 4 and k = 2 in the calculator reproduces all of these values.

Frequently Asked Questions

How is the Poisson distribution different from the binomial?

The binomial counts successes out of a fixed number of trials, so the count is capped at n. The Poisson counts events in a fixed interval with no upper bound, using only the average rate lambda. When trials are numerous and the success probability is small, the binomial converges to a Poisson with lambda = np, which is why the Poisson is often called the law of rare events.

Can lambda be a decimal like 2.5?

Yes. Lambda is an average, not a count, so any positive value is valid. Only k, the number of events you ask about, must be a whole number, because you cannot observe half an event.

What is overdispersion and why does it matter?

A Poisson variable has variance equal to its mean. Real count data often shows variance well above the mean - for example when events cluster or the rate drifts over time. That mismatch, called overdispersion, makes Poisson probabilities too optimistic in the tails; a negative binomial model is the usual remedy.

How does the Poisson relate to the exponential distribution?

They describe the same process from two angles. The Poisson counts how many events land in a fixed window, while the exponential measures the waiting time between consecutive events. If counts per hour are Poisson with rate lambda, the gaps between events are exponential with mean 1/lambda hours.

When can I approximate a Poisson with a normal distribution?

Once lambda is reasonably large - a common threshold is lambda >= 20 - the Poisson is nearly symmetric and a normal curve with mean lambda and standard deviation sqrt(lambda) approximates it well. For small lambda the distribution is strongly right-skewed and the normal approximation misstates tail probabilities.