The Poisson distribution: calculating probabilities

By the end of this lesson

By the end you can recognise when a Poisson model applies, write down its parameter λ for the interval in the question, and calculate P(X = r), P(X ≤ r) and P(X ≥ r).

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$X$	the random variable: the number of events in the interval
$r$	one particular count, such as $0$ , $1$ or $2$
$P(X = r)$	"the probability that $X$ equals $r$ "
$\lambda$	"lambda": the mean (average) number of events in the interval
$\mathrm{Po}(\lambda)$	the Poisson distribution with mean $\lambda$
$e$	Euler's number, $e \approx 2.718$ ; use the $e^x$ key on your calculator
$r!$	" $r$ factorial": $r \times (r-1) \times \cdots \times 1$ , with $0! = 1$
$\sum_{k=0}^{r}$	"add up the terms for $k = 0, 1, \dots, r$ "

An event here is one occurrence of the thing being counted (one call, one flaw). An interval is the stretch of time, length, area or volume you count over.

Definition

Poisson distribution

A discrete random variable takes separate values such as $0, 1, 2, \dots$ (here, counts). Its probability function is the formula giving $P(X = r)$ for each value. A discrete random variable $X$ has a Poisson distribution with parameter λ > 0, written

$X \sim \mathrm{Po}(\lambda),$

if its probability function is

$P(X = r) = \frac{e^{-\lambda}\lambda^{r}}{r!} \qquad r = 0, 1, 2, \dots$

Here λ is the mean number of events in the interval being considered. The interval may be a length of time, an area, a volume, or a number of items.

When is a Poisson model appropriate?

A Poisson model fits random events when all of the following hold:

events happen singly (two cannot occur at exactly the same point);
events happen independently of one another;
events occur at a constant average rate throughout the interval.

Typical settings: radioactive decays in a second, misprints on a page, calls arriving at a switchboard, flaws in a length of cable. The events must be countable and the interval fixed in advance. If instead you have a fixed number of trials with a constant probability of success, that is binomial, not Poisson.

The formula, term by term

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

$e^{-\lambda}$ is a constant for the whole distribution, so it is often factored out when you sum several terms.
$r!$ grows very quickly, which is why the probabilities fall away for large $r$ .
For $r = 0$ : $\lambda^0 = 1$ and $0! = 1$ , so $P(X = 0) = e^{-\lambda}$ .
$P(X \le r)$ adds the terms from $0$ up to $r$ . Because the probabilities over all $r$ add to 1, the complement gives $P(X \ge r)$ :

$P(X \le r) = \sum_{k=0}^{r} P(X = k), \qquad P(X \ge r) = 1 - P(X \le r-1).$

For "at least one" questions it is almost always quicker to subtract from 1: $P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-\lambda}$ .

Worked example: Example 1: faults in a cable, λ = 3.2

A cable has flaws at a constant average rate of 3.2 per kilometre. A one-kilometre length is inspected. Find the probability that it contains (a) no flaws, (b) exactly two flaws, (c) at most two flaws.

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