Poisson Distribution — Formula, Mean & Examples

The Poisson distribution gives the probability of getting $0, 1, 2, \dots$ events in a fixed interval when events happen independently and the average rate stays roughly constant. If you know the average number of calls, defects, or arrivals in one interval, the Poisson model helps you find the chance of an exact count.

The key condition is the model choice, not the algebra. If independence or a roughly constant rate is not reasonable, the Poisson formula can look correct and still answer the wrong question.

Poisson distribution formula

If $X$ follows a Poisson distribution with parameter $\lambda > 0$, then for any whole number $k \ge 0$,

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

Here, $k$ is the exact number of events you want, and $\lambda$ is the expected number of events in the interval you chose.

For a Poisson model, the mean and variance are both equal to $\lambda$:

$$\text{mean} = \text{variance} = \lambda$$

That does not mean every real data set has matching mean and variance. It means the Poisson model predicts that relationship.

What $\lambda$ means in plain language

$\lambda$ is the average count for one specific interval. The interval might be one hour, one square meter, one page, or one kilometer, but you have to define it clearly.

If a shop gets an average of $3$ calls per hour, then $\lambda = 3$ for a one-hour interval. For a two-hour interval, you would use $\lambda = 6$ only if the same average rate still makes sense over those two hours.

This is one of the easiest ways to make a mistake. When the interval changes, $\lambda$ usually changes too.

Worked example: exactly 2 calls in 1 hour

Suppose a small shop receives an average of $3$ customer calls per hour. If call arrivals are reasonably independent and the average rate is stable, what is the probability of getting exactly $2$ calls in the next hour?

Here, $\lambda = 3$ and $k = 2$, so:

$$P(X = 2) = \frac{e^{-3}3^2}{2!}$$

Simplify step by step:

$$P(X = 2) = \frac{9e^{-3}}{2}$$

Using $e^{-3} \approx 0.0498$,

$$P(X = 2) \approx \frac{9(0.0498)}{2} \approx 0.224$$

So the probability is about $0.224$, or $22.4\%$. In context, that means getting exactly $2$ calls in the next hour is a fairly ordinary outcome, not a rare surprise.

When a Poisson model makes sense

Use a Poisson model when all of these are reasonably true:

• You are counting occurrences, not measuring a continuous value like time or height.
• The count is taken over a fixed interval such as one hour or one page.
• The average rate is roughly constant over that interval.
• One event does not directly make another event much more or less likely.

This is why the Poisson distribution appears in queueing, reliability, traffic flow, telecommunications, and quality control. It works best for count data with a stable rate, not for situations with strong clustering or sharp time-of-day effects.

Common mistakes in Poisson problems

Using Poisson for non-count data

The Poisson distribution is for $0, 1, 2, 3, \dots$. It does not model continuous measurements like height, time, or temperature.

Forgetting to rescale $\lambda$

If $\lambda = 3$ per hour, that does not mean $\lambda = 3$ per $30$ minutes. For half an hour, the matching parameter would be $\lambda = 1.5$ if the same average rate holds.

Thinking "rare event" is the full rule

"Rare" helps intuition, but it is not the full rule. The real question is whether a fixed interval, a roughly constant average rate, and approximate independence are reasonable.

Treating mean equals variance as a law of nature

For a Poisson model, the mean and variance are both $\lambda$. Real data do not always behave that cleanly, so the equality is a model property, not a law of nature.

Poisson vs. binomial

Use a Poisson model when you count how many events happen in an interval and there is no fixed number of trials built into the setup.

Use a binomial model when you already have a fixed number of trials, each with the same success probability. For example, counting defective bulbs in a sample of $20$ tested bulbs is binomial because the number of trials is fixed at $20$.

Try a similar problem

Try your own version with an average of $5$ deliveries per day. Find the probability of getting exactly $4$ deliveries tomorrow, then change the interval to half a day and decide what happens to $\lambda$ before you calculate.