For X Po(3), find P(X = 0). [2 marks]

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How do the Poisson and geometric distributions model counts and waiting times, and when does each apply?

The Poisson distribution and its conditions, mean and variance, the sum of independent Poisson variables, the geometric distribution and its mean, and the Poisson approximation to the binomial.

A focused answer to the OCR A-Level Further Mathematics A Statistics option content on the Poisson and geometric distributions, covering the Poisson model and its conditions, its mean and variance both equal to lambda, the sum of independent Poisson variables, the geometric distribution for the number of trials to the first success and its mean, and the Poisson approximation to the binomial.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The Poisson distribution models the number of events in a fixed interval when they occur independently at a constant average rate $\lambda$ : $\mathrm{P}(X = x) = e^{-\lambda}\dfrac{\lambda^x}{x!}$ , with the distinctive property $\mathrm{E}(X) = \mathrm{Var}(X) = \lambda$ . The sum of independent Poisson variables is Poisson with the summed means. The geometric distribution models the number of trials up to and including the first success in independent trials with success probability $p$ : $\mathrm{P}(X = x) = (1 - p)^{x-1} p$ , with mean $\mathrm{E}(X) = \dfrac{1}{p}$ . When a binomial has large $n$ and small $p$ , it is well approximated by a Poisson with $\lambda = np$ .

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What this dot point is asking
The Poisson distribution
Mean equals variance, and sums
The geometric distribution
The Poisson approximation to the binomial
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What this dot point is asking

OCR's Statistics option wants you to recognise when the Poisson distribution applies (events occurring independently at a constant average rate), use its probability formula and the fact that its mean and variance are both $\lambda$ , add independent Poisson variables, recognise and use the geometric distribution for the number of trials to the first success (with mean $\tfrac{1}{p}$ ), and use the Poisson as an approximation to the binomial when $n$ is large and $p$ is small.

The Poisson distribution

The Poisson distribution counts random events, such as calls to a switchboard or particles detected, that happen independently at a steady average rate. It is defined by a single parameter, the mean rate $\lambda$ .

Mean equals variance, and sums

A hallmark of the Poisson distribution is that the mean and the variance are equal, which is often used to test whether data are plausibly Poisson. A useful structural property is that independent Poisson variables add to give another Poisson.

The geometric distribution

The geometric distribution models how many independent trials are needed to get the first success, when each trial succeeds with probability $p$ . The probability decreases geometrically with the number of trials.

The mean $\tfrac{1}{p}$ is intuitive: if a trial succeeds one time in five ( $p = 0.2$ ), you expect five trials on average to reach the first success.

The Poisson approximation to the binomial

When a binomial has many trials and a small success probability, the binomial probabilities are close to Poisson probabilities with the same mean. This is far quicker than evaluating large binomial coefficients.

The Poisson and geometric distributions build on the discrete random variable framework and feed into the chi-squared goodness-of-fit test, where a Poisson model can be tested against data.

Try this

Q1. For $X \sim \text{Po}(3)$ , find $\mathrm{P}(X = 0)$ . [2 marks]

Cue. $e^{-3}\dfrac{3^0}{0!} = e^{-3} \approx 0.0498$ .

Q2. A geometric variable has $p = 0.25$ . State its mean. [1 mark]

Cue. $\mathrm{E}(X) = \dfrac{1}{p} = 4$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20195 marksThe number of emails arriving at an inbox in an hour follows a Poisson distribution with mean

4

. Find the probability that exactly

2

emails arrive in a given hour, and state the variance of the distribution.

Show worked answer →

Poisson probability $\mathrm{P}(X = x) = e^{-\lambda}\dfrac{\lambda^x}{x!}$ with $\lambda = 4$ (M1).

For $x = 2$ (M1): $\mathrm{P}(X = 2) = e^{-4}\dfrac{4^2}{2!} = e^{-4}\dfrac{16}{2} = 8e^{-4}$ (A1).

Evaluate (A1): $\approx 8 \times 0.0183 = 0.147$ (to 3 sf).

The variance of a Poisson distribution equals its mean, so $\mathrm{Var}(X) = 4$ (A1).

Markers reward the Poisson formula, substituting $x = 2$ , the value, and stating the variance equals the mean.

OCR 20226 marksA biased coin shows heads with probability

0.2

. Let

X

be the number of tosses up to and including the first head. State the distribution of

X

, find

\mathrm{P}(X = 3)

and find

\mathrm{E}(X)

Show worked answer →

$X$ follows a geometric distribution with $p = 0.2$ (the number of trials to the first success) (M1).

$\mathrm{P}(X = x) = (1 - p)^{x-1} p$ (M1). For $x = 3$ : $\mathrm{P}(X = 3) = (0.8)^2(0.2) = 0.64 \times 0.2 = 0.128$ (A1, A1).

The mean of a geometric distribution is $\mathrm{E}(X) = \dfrac{1}{p}$ (M1): $= \dfrac{1}{0.2} = 5$ (A1).

Markers reward identifying the geometric distribution, the probability formula, the value $0.128$ , and the mean $\tfrac{1}{p} = 5$ .

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Sources & how we know this

OCR A Level Further Mathematics A (H245) specification — OCR (2017)