Poisson Distribution — Formula, Mean & Examples

Bottom line: reach for the Poisson distribution when you are counting how many independent events fall in a fixed interval at a roughly constant average rate — and reach for the binomial instead when the number of trials is fixed in advance.

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

Here $k$ is the exact number of events you want, and $\lambda$ is the expected number of events in the interval you chose. The hard part is never the algebra — it is the model choice. If independence or a stable rate is not reasonable, this formula can look right and still answer the wrong question.

Poisson vs Binomial at a Glance

Counting defective bulbs in a sample of $20$ tested bulbs is binomial because the number of trials is fixed at $20$. Counting calls in an hour, with no built-in trial count, is Poisson.

What $\lambda$ Means, and When to Use Each

$\lambda$ is the average count for one specific interval — one hour, one square meter, one page, one kilometer — but you must define that interval clearly. If a shop gets an average of $3$ calls per hour, then $\lambda = 3$ for a one-hour interval; for two hours you would use $\lambda = 6$, but only if the same rate still holds.

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 over a fixed interval such as one hour or one page.
• The average rate is roughly constant over that interval.
• One event does not strongly change the chance of another.

For a Poisson model, mean and variance are both $\lambda$. That is a property the model predicts, not a guarantee that real data will match.

Worked Example: Exactly 2 Calls in 1 Hour

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

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

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

So the probability is about $0.224$, or $22.4\%$ — a fairly ordinary outcome, not a rare surprise. In context, getting exactly $2$ calls in the next hour is well within normal variation. This is why Poisson appears in queueing, reliability, traffic flow, telecommunications, and quality control: stable-rate count data, not strong clustering or sharp time-of-day effects.

Practice and Rescaling Check

Try your own version: a courier averages $5$ deliveries per day. Find the probability of exactly $4$ deliveries tomorrow with $\lambda = 5$ and $k = 4$ in

Then change the interval to half a day and decide what $\lambda$ becomes before you calculate. If the same average rate holds, halving the interval halves the parameter to $\lambda = 2.5$ — the single most common slip in Poisson problems is leaving $\lambda$ unchanged when the interval changes.

Confusion Points

• Using Poisson for non-count data. It models $0, 1, 2, 3, \dots$, not continuous measurements like height, time, or temperature.
• Forgetting to rescale $\lambda$. $\lambda = 3$ per hour is not $\lambda = 3$ per $30$ minutes; for half an hour it would be $\lambda = 1.5$ if the same rate holds.
• Thinking "rare event" is the whole rule. "Rare" aids intuition, but the real test is a fixed interval, a roughly constant rate, and approximate independence.
• Treating mean = variance as a law of nature. Both equal $\lambda$ for the model; real data do not always cooperate.