What is not scaling λ for the interval?

If the interval changes, λ must change in proportion before you calculate any probability.

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When is the Poisson distribution the right model, and how do you calculate and combine Poisson probabilities?

The Poisson distribution as a model for random events, its mean and variance, calculating probabilities, the sum of independent Poisson variables, and the Poisson approximation to the binomial.

A focused answer to the AQA A-Level Further Mathematics Poisson distribution content, covering the Poisson distribution as a model for random events, its mean and variance, calculating probabilities, the sum of independent Poisson variables, and the Poisson approximation to the binomial.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

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What this dot point is asking
Conditions and the probability function
Combining and approximating

What this dot point is asking

AQA wants you to recognise when the Poisson distribution is an appropriate model, state its conditions, use the probability formula and tables, know that its mean and variance are both equal to $\lambda$ , add independent Poisson variables, and use the Poisson approximation to the binomial when $n$ is large and $p$ is small.

Conditions and the probability function

The Poisson model applies to events that occur singly (one at a time), independently of one another, and at a constant average rate, within a fixed interval of time or space. Each of these conditions can fail in a real situation, and questions often ask you to judge whether the model is reasonable: clustered events (such as accidents that cause further accidents) break independence, and a rate that varies through the day breaks the constant-rate assumption. When the conditions hold, the distribution is fully specified by the single parameter $\lambda$ , the mean number of events in the chosen interval.

The equality of mean and variance is a signature property used both ways. It lets you predict the variance once you know the rate, and it provides a quick check on whether real data could be Poisson: if the sample mean and sample variance are far apart, a Poisson model is doubtful, which links directly to goodness of fit testing.

Calculate Poisson probabilities

A switchboard receives calls at an average of $3$ per minute, modelled by $X \sim \text{Po}(3)$ . Find $P(X = 2)$ and $P(X \geq 1)$ .

Step 1: Apply the probability formula for $P(X = 2)$

The Poisson probability formula $P(X = x) = e^{-\lambda}\frac{\lambda^x}{x!}$ gives us the probability of exactly $x$ events. With $\lambda = 3$ and $x = 2$ , substitute directly.

P(X = 2) = e^{-3}\frac{3^2}{2!} = e^{-3}\frac{9}{2} = 4.5\, e^{-3}.

Step 2: Evaluate the exact probability

Using $e^{-3} = 0.049787$ :

P(X = 2) = 4.5 \times 0.049787 = 0.224 \text{ (to 3 significant figures)}.

Step 3: Use the complement to find $P(X \geq 1)$

Summing $P(X = 1) + P(X = 2) + \cdots$ would require infinitely many terms. It is far more efficient to use the complement: $P(X \geq 1) = 1 - P(X = 0)$ . The only term we need to calculate is the probability of zero calls.

P(X = 0) = e^{-3}\frac{3^0}{0!} = e^{-3} = 0.0498.

Step 4: State the final result

P(X \geq 1) = 1 - 0.0498 = 0.950 \text{ (to 3 significant figures)}.

Final answer: $P(X = 2) = 0.224$ and $P(X \geq 1) = 0.950$ .

Combining and approximating

The additive property is what makes scaling the interval work. A rate of $3$ events per minute is the sum of independent contributions, so over two minutes the parameter doubles to $\text{Po}(6)$ , and over thirty seconds it halves to $\text{Po}(1.5)$ . Always scale $\lambda$ in proportion to the interval before calculating any probability, because the formula uses the mean for the actual interval in question.

The Poisson approximation to the binomial rests on the same conditions in disguise: when $n$ is large and $p$ is small, the binomial events are rare, roughly independent, and occur at a near-constant rate, which is exactly the Poisson setting. Setting $\lambda = np$ matches the means, and the approximation is good when $n$ is large (often $n > 50$ ) and $p$ is small (often $p < 0.1$ ). It is valuable because the Poisson formula avoids the large factorials of a binomial calculation with big $n$ .

Exam-style practice questions

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

AQA 20197 marksCalls arrive at a help desk at an average rate of

3

per ten-minute period, modelled by a Poisson distribution. Find the probability that (a) exactly

2

calls arrive in a ten-minute period, and (b) more than

1

call arrives in a five-minute period.

Show worked answer →

Part (a). Here $X \sim \text{Po}(3)$ . $P(X = 2) = e^{-3}\frac{3^2}{2!} = e^{-3}\frac{9}{2} = 4.5 e^{-3}$ .

Since $e^{-3} = 0.049787$ , $P(X = 2) = 4.5 \times 0.049787 = 0.224$ (to 3 significant figures).

Part (b). For a five-minute period the rate halves, so $Y \sim \text{Po}(1.5)$ .

$P(Y > 1) = 1 - P(Y = 0) - P(Y = 1)$ . Now $P(Y = 0) = e^{-1.5} = 0.22313$ and $P(Y = 1) = e^{-1.5}\frac{1.5^1}{1!} = 1.5 \times 0.22313 = 0.33470$ .

So $P(Y > 1) = 1 - 0.22313 - 0.33470 = 0.442$ (to 3 significant figures).

Markers reward the formula in (a), scaling the mean to $1.5$ in (b), and computing $P(Y > 1)$ as the complement of $P(Y \leq 1)$ .

AQA 20216 marksA factory produces components,

0.5\%

of which are defective. A batch of

400

components is examined. Justify the use of a Poisson approximation to the binomial, state the approximating distribution, and use it to find the probability that the batch contains at most

1

defective component.

Show worked answer →

The number of defectives is binomial $\text{B}(400, 0.005)$ . A Poisson approximation is justified when $n$ is large and $p$ is small, which holds here ( $n = 400$ , $p = 0.005$ ), so the binomial is well approximated by $\text{Po}(np)$ .

Here $np = 400 \times 0.005 = 2$ , so use $X \sim \text{Po}(2)$ .

$P(X \leq 1) = P(X = 0) + P(X = 1) = e^{-2} + e^{-2}\frac{2^1}{1!} = e^{-2}(1 + 2) = 3e^{-2}$ .

Since $e^{-2} = 0.13534$ , $P(X \leq 1) = 3 \times 0.13534 = 0.406$ (to 3 significant figures).

Markers reward justifying the approximation (large $n$ , small $p$ ), stating $\text{Po}(2)$ , and computing $P(X \leq 1)$ correctly.

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Sources & how we know this

AQA A-level Further Mathematics (7367) specification — AQA (2017)