Paper 6. Calls to a helpline, flaws in a roll of cloth, particles hitting a detector: when events happen at random at a steady average rate, the number in a fixed interval follows the Poisson distribution, and one number, the mean rate, describes it completely.
If events occur singly, independently and at random at a constant mean rate, the number \(X\) in a fixed interval satisfies \(X \sim \text{Po}(\lambda)\) with \(\text{P}(X = r) = e^{-\lambda}\dfrac{\lambda^{r}}{r!}\) for \(r = 0, 1, 2, \ldots\). The mean and the variance are both \(\lambda\), and \(\lambda\) scales in proportion to the length of the interval.
Fig. 1: the Poisson distribution with mean 3, drawn to scale. The teal bars add to P(X ≤ 2) = 0.423; the bars for r = 2 and r = 3 are equal because P(X = 3) = P(X = 2) × 3/3
Section 01
When the Poisson model fits.
The Poisson distribution counts events in a fixed interval of time, length, area or volume. It is the right model when the events behave like scattered random points rather than a scheduled pattern.
Singly: events cannot happen at exactly the same instant or place.
Independently: one event does not make another more or less likely.
At random, at a constant mean rate: the average number per unit interval does not change (no rush hour, no busy season).
No upper limit: \(X\) can take any value \(0, 1, 2, \ldots\), unlike the binomial, which stops at \(n\).
A quick data check: because the mean and variance of \(\text{Po}(\lambda)\) are both \(\lambda\), sample data with mean close to variance support a Poisson model.
Worked example Is a Poisson model suitable?
In 100 one-minute intervals, the numbers of cars passing a point were recorded. The sample mean was 1.96 and the sample variance was 1.82. A second road, next to a set of traffic lights, gave mean 2.0 and variance 0.4. Comment on each.
For a Poisson distribution the mean equals the variance.
First road: \(1.96\) and \(1.82\) are close, so a Poisson model with \(\lambda \approx 1.96\) is reasonable.
Second road: the variance \(0.4\) is much smaller than the mean \(2.0\).
Traffic lights release cars in bunches at regular times, so the cars do not arrive independently at random.
Answer: Poisson is suitable for the first road but not for the second (variance far below the mean; arrivals are not independent).
Examiner trap
Do not just recite "independent and random". Cambridge wants the condition in context: "the calls arrive independently of each other" or "the mean rate of calls is constant during the day". A condition with no reference to the situation usually earns nothing.
Section 02
The formula and the mean rate.
For \(X \sim \text{Po}(\lambda)\), \(\text{P}(X = r) = e^{-\lambda}\dfrac{\lambda^{r}}{r!}\). The only parameter is \(\lambda\), the mean number in the interval you are asked about. If the interval changes, scale \(\lambda\) first.
Worked example Single values and cumulative probabilities
Calls arrive at a helpline at random at a mean rate of 3.2 per 10 minutes. Find the probability that in a 10-minute period there are (a) exactly 2 calls (b) at most 2 calls.
Let \(X\) be the number of calls in 10 minutes, so \(X \sim \text{Po}(3.2)\).
"More than 2" means \(X \ge 3\), so it is \(1 - \text{P}(X \le 2)\), not \(1 - \text{P}(X \le 3)\). Write the inequality in symbols before you calculate; most lost marks here are a boundary slipped by one.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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\(X \sim \text{Po}(4)\). What are the mean and variance of \(X\)?
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Which of these is NOT a condition for a Poisson model?
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Emails arrive at random at a mean rate of 2 per hour. \(X\) is the number of emails in 3 hours. What is the distribution of \(X\)?
Quick check+10 XP
For \(X \sim \text{Po}(\lambda)\), what is \(\text{P}(X = 0)\)?
Section 03
Working backwards and two-stage problems.
Some questions give a probability and ask for \(\lambda\); others use a Poisson probability as the \(p\) in a binomial over several intervals.
Worked example Finding the mean from a probability
Faults on a production line occur at random at a mean rate of 1.2 per day. Find the probability that, in a 5-day week, there are exactly 2 days with no faults.
For one day, \(\text{P}(\text{no faults}) = e^{-1.2} = 0.3012\).
The 5 days are independent, so the number of fault-free days is \(B(5, 0.3012)\).
In a two-stage question, do not add the rates to \(\lambda = 6\) for the week and find \(\text{P}(X = 0)\): that is the probability of no faults all week, a different event. Identify what is being counted at each stage.
Stage 2 · Exam
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