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Poisson Distribution, Explained

After reading this you can compute the probability of exactly, at most, or at least a given number of rare events, and know when the Poisson model fits your data and when it does not.

What the Poisson distribution describes

The Poisson distribution counts how many times an event happens in a fixed interval of time, space, or volume, when the events occur independently at a constant average rate. The only input is that rate, written \lambda (lambda), the mean number of events per interval.

Suppose a call center receives on average 3 calls per minute. How likely is it that exactly 2 calls arrive in the next minute? With \lambda = 3 and k = 2, the answer is about 0.2240, or 22.40%. That single number is the hook: you never observe "3 calls" as a smooth quantity, you observe whole counts, and the Poisson distribution puts a probability on each possible count.

The support is every non-negative integer: 0, 1, 2, 3, and so on with no upper limit. Larger counts get smaller probabilities, but none is exactly zero.

When to use it, and when not to

Reach for the Poisson model when you are counting occurrences and four conditions roughly hold:

  • Events happen one at a time, not in simultaneous clumps.
  • The average rate is stable over the interval you study.
  • Events in non-overlapping intervals are independent.
  • There is no fixed ceiling on how many can occur.

Typical fits: typos per page, radioactive decays per second, customers arriving per hour, defects per square meter of fabric.

If events cluster (one accident triggers several) or the rate swings by time of day, the real variance exceeds \lambda. That is called overdispersion. In that case the Poisson model understates the chance of extreme counts, and a negative binomial model often fits better.

When you count successes in a fixed number of trials n, use the binomial distribution instead. The Poisson is the limiting case of the binomial when n is large and the success probability p is small, with \lambda = np.

The formula and why it looks the way it does

The probability of exactly k events is the probability mass function (PMF):

P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}

Here \lambda is the mean rate, k is the count you ask about (a non-negative integer), e \approx 2.71828, and k! is the factorial of k (so 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24).

The three pieces each play a role. The term \lambda^k grows as you allow more events. The factorial k! in the denominator penalizes large counts, because the events are interchangeable and you divide out their orderings. The factor e^{-\lambda} normalizes everything so the probabilities over all k sum to exactly 1.

A useful fact: the mean and the variance are both equal to \lambda. So the standard deviation is \sqrt{\lambda}. With \lambda = 3, the spread around the mean of 3 is about \sqrt{3} \approx 1.732.

A worked example you can reproduce

Three calls per minute, asking about two

Use the demo defaults: \lambda = 3 and k = 2.

  1. Compute e^{-\lambda} = e^{-3} \approx 0.049787.
  2. Compute \lambda^k = 3^2 = 9.
  3. Compute k! = 2! = 2.
  4. Combine: P(X = 2) = \frac{0.049787 \times 9}{2} = \frac{0.448084}{2} \approx 0.2240.

So there is a 22.40% chance of exactly 2 calls in a minute. For the cumulative probability of at most 2 calls, add k = 0, 1, and 2:

Building the CDF at \lambda = 3
kP(X = k)Running P(X ≤ k)
00.049790.04979
10.149360.19915
20.224040.42319
30.224040.64723

So P(X \le 2) \approx 0.4232. The at-least-3 tail is the complement: P(X \ge 3) = 1 - 0.4232 = 0.5768. Notice that k = 2 and k = 3 share the same probability, 0.2240. That happens whenever \lambda is a whole number: the mode sits at both \lambda - 1 and \lambda.

The distribution peaks at k = 2 and k = 3, then falls off. Bars for k above 8 are tiny but not zero.

See how the rate reshapes the distribution

The single most instructive move is changing \lambda. Small rates give a lopsided distribution piled near zero. Large rates spread out and grow roughly symmetric, approaching a normal shape.

At \lambda = 0.5 the most likely count is 0, with P(X = 0) about 0.6065. At \lambda = 3 the peak sits at 2 and 3. At \lambda = 12 the bars form a near-symmetric mound centered on 12 with standard deviation about 3.46, close to a normal curve.

Reading and interpreting the three probabilities

Every Poisson question is one of three shapes. Match your wording to the right one.

Exactly k: P(X = k)
The single PMF value. Use it for "what is the chance of precisely 2 defects?"
At most k: P(X ≤ k)
The CDF, a sum from 0 up to k. Use it for "no more than 2".
At least k: P(X ≥ k)
The upper tail, equal to 1 - P(X \le k-1). Use it for "3 or more".

The boundary word is where people slip. "More than 2" means P(X \ge 3), not P(X \ge 2). "Fewer than 2" means P(X \le 1). Because the variable is discrete, off-by-one errors change the answer by a full bar. Here that bar can be worth 22 percentage points.

To get the at-least tail, it is safer to subtract the CDF than to add an infinite series. For P(X \ge 3) at \lambda = 3, compute 1 - P(X \le 2) = 1 - 0.4232 = 0.5768.

Scaling the interval and other common mistakes

The rate \lambda is tied to a specific interval length. If you observe 3 calls per minute but want the count over 5 minutes, use \lambda = 3 \times 5 = 15. Rates add across time. Forgetting to rescale is the most frequent error.

Two more to watch:

  • Confusing rate with probability. \lambda can exceed 1; it is a mean count, not a probability. A rate of 3 is fine. A "probability" of 3 is not.
  • Using Poisson when there is a fixed maximum. If only 40 machines exist and you count failures among them, the count cannot exceed 40, so the binomial fits better. Poisson is for counts with no natural ceiling.

The waiting time between Poisson events follows a different law. If arrivals are Poisson with rate \lambda, the gap between consecutive arrivals follows an exponential distribution with mean 1/\lambda. The Poisson counts events; the exponential times them.

Related calculators

These tools cover neighboring models and follow-up steps.

Frequently asked questions

What is a good value of lambda to use?

Use the observed average count per interval. If a warehouse ships 47 orders across 10 days, the daily rate is \lambda = 4.7. Match the interval of \lambda to the interval of your question.

Can the Poisson distribution give a probability for k = 0?

Yes. P(X = 0) = e^{-\lambda}. At \lambda = 3 that is 0.04979, about a 5% chance of no events. As \lambda rises, the chance of zero shrinks fast.

Why do the mean and variance both equal lambda?

It falls out of the math of the PMF. One consequence you can check: at \lambda = 4 the mean is 4 and the standard deviation is \sqrt{4} = 2, so most counts land between 2 and 6.

When can I approximate a Poisson with a normal distribution?

When \lambda is roughly 10 or more, a normal with mean \lambda and standard deviation \sqrt{\lambda} is close. For \lambda = 20, use a normal with mean 20 and standard deviation \sqrt{20} \approx 4.472, and apply a continuity correction of 0.5.

Is Poisson the same as binomial with large n?

It is the limit. With n = 1000 trials and p = 0.003, the binomial count is nearly identical to a Poisson with \lambda = np = 3. The Poisson formula is simpler because it needs only the single number \lambda.

Try it yourself in the tool

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