MathematicsCore22 min read

The Poisson Distribution

Counting rare events when there is no fixed number of trials

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01

When the binomial has nothing to count

Definition

Poisson distribution — The distribution of the number of events occurring in a fixed interval, when events happen independently at a constant average rate λ.

The binomial distribution needs a fixed number of trials. Ask how many phone calls a switchboard receives in an hour, or how many flaws appear in a metre of cable, and there is no n at all — there is no natural list of trials that either succeeded or failed.

What these situations share instead is a constant average rate of occurrence over a continuous interval of time, length or area. The Poisson distribution models exactly that.

X ~ Po(λ)P(X = r) = e^(−λ) · λʳ / r!r = 0, 1, 2, …E(X) = λVar(X) = λthe mean and the variance are equal — the signature of a Poisson
λ
the mean number per intervalmust match the interval in the question
r
the countany non-negative integer, with no upper limit
e^(−λ)
the normalising factorwhat makes the probabilities total 1

At small λ the distribution is crushed against zero and heavily skewed. Push λ past about 10 and it becomes visibly bell-shaped — which is precisely why the normal approximation is allowed there and not before.

02

The conditions, and the interval trap

Three conditions must hold, and questions frequently ask you to state them. Events must occur independently, at a constant average rate, and singly — not in clusters arriving together.

The most common calculation error has nothing to do with the formula. λ must match the interval the question asks about. If calls arrive at 6 per hour and you are asked about a 20-minute period, then λ = 2, not 6. Scaling λ is the first line of the answer, not an afterthought.

Given rateQuestion asks aboutλ to use
5 per hourone hour5
5 per hour30 minutes2.5
5 per hour3 hours15
2 per metre4.5 metres9
0.4 per page20 pages8

Scale λ before anything else

Write the scaled λ on its own line at the start. It costs a second, it is often worth a mark in itself, and it prevents the whole question being marked wrong for an otherwise perfect method. If the interval doubles, λ doubles — the rate is per unit, so it scales linearly.

03

Adding Poissons, and approximating a binomial

Two independent Poisson variables add in the simplest possible way: the sum is Poisson again, with the two means added. This is why a switchboard receiving 3 calls an hour on one line and 5 on another can be treated as a single Po(8).

The Poisson also rescues the binomial when n is large and p is small. Computing ⁵⁰⁰C₃(0.004)³(0.996)⁴⁹⁷ is unpleasant; approximating with Po(np) is not. The usual conditions are n > 50 and np < 5 — large n so events are spread thin, small p so they are rare.

if X ~ Po(a) and Y ~ Po(b) independently,X + Y ~ Po(a + b)binomial → Poisson:B(n, p) ≈ Po(np)when n is large (> 50) and p is small (np < 5)Poisson → normal:Po(λ) ≈ N(λ, λ)when λ > 15, with a continuity correctiononly the means add — never add the standard deviations
a + b
the combined meanvalid only if the two are independent
np
the approximating meanthe binomial mean carried across
N(λ, λ)
the normal approximationmean and variance both λ
Worked example

Faults occur in a cable at an average rate of 0.6 per metre. Find the probability that a 5-metre length contains (a) exactly 2 faults, (b) at least 2 faults.

  1. Scale the mean: λ = 0.6 × 5 = 3 for a 5-metre length.The rate is per metre and the interval is five metres, so λ multiplies by 5. This is the step that most often goes wrong.
  2. (a) P(X = 2) = e⁻³ · 3² / 2! = 0.049787 × 9 / 2.Direct substitution into the Poisson formula with r = 2.
  3. = 0.224.To 3 significant figures.
  4. (b) P(X ⩾ 2) = 1 − P(X = 0) − P(X = 1).The complement is far shorter than summing an infinite tail — and the tail really is infinite here, so it is the only practical route.
  5. P(X = 0) = e⁻³ = 0.0498; P(X = 1) = e⁻³ × 3 = 0.1494.The r = 0 term is just e^(−λ), since λ⁰ = 1 and 0! = 1.
  6. P(X ⩾ 2) = 1 − 0.0498 − 0.1494 = 0.801."At least 2" includes 2, so only 0 and 1 are removed.

(a) 0.224; (b) 0.801

Recognising which distribution to use

  1. A fixed number of trials, each success or failure → binomial.
  2. A rate over time, length or area with no fixed n → Poisson.
  3. Mean equal to variance in the data → a strong hint the model is Poisson.
  4. Large n, small p, and awkward arithmetic → approximate the binomial by Po(np).
  5. λ above about 15 → approximate the Poisson by N(λ, λ), with a continuity correction.
04

Recognising a Poisson in a worded question

Questions rarely say which distribution to use. They describe a situation, and the phrasing itself carries the answer if you know what to listen for.

The giveaway is that a Poisson situation has no natural denominator. "Three calls per hour" gives no number of trials — you cannot say how many calls did not happen. A binomial situation always has one: twenty components tested, fifty coin tosses, a fixed n.

A second clue is the interval. Poisson questions describe events spread over time, length, area or volume, and the interval can be rescaled at will. Binomial questions describe a fixed batch that cannot be resized.

WordingDistributionWhy
"4 flaws per metre of cable"Poissona rate; no count of non-flaws
"of 50 components, 3% are faulty"binomialfixed n = 50
"calls arrive at 6 per hour"Poissona rate over time
"20 patients, each 10% likely"binomialfixed n = 20
"accidents per month at a junction"Poissona rate over time

The mean-equals-variance check

If a question gives you sample data rather than a model, compare the mean with the variance. Roughly equal supports Poisson. A variance much larger than the mean means the events cluster rather than occurring independently, and a variance much smaller means they are more evenly spaced than chance would give — neither is Poisson, and saying so is often the mark.

Practice questions

5 questions · 17 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

2 · 4 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State two conditions needed for a Poisson model to be appropriate.
Model answer

Any two of: events occur independently of one another; they occur at a constant average rate; and they occur singly rather than in clusters.

Examiner tip. One mark each. "Randomly" on its own is too vague to earn a mark.

SQ2[2 marks]
A student is told a data set has mean 4.1 and variance 4.0. Explain why this suggests a Poisson model, and state what a variance of 12 would have suggested instead.
Model answer

For a Poisson distribution the mean and variance are equal, both λ. A mean of 4.1 and variance of 4.0 are close enough to support that model. A variance of 12 would be far larger than the mean, so the events would not be occurring independently at a constant rate — they would be clustered, and a Poisson model would be inappropriate.

Examiner tip. The mean-equals-variance property is the standard way of testing a Poisson model, and is worth remembering as a discriminating fact.

Solved numericals

2 · 7 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[3 marks]
Calls arrive at a switchboard at an average of 9 per hour. Find the probability that exactly 4 calls arrive in a 20-minute period.
Full working
  1. 20 minutes is one third of an hour, so λ = 9/3 = 3Scaling the mean to the stated interval is a mark in its own right.[1]
  2. P(X = 4) = e⁻³ × 3⁴ / 4!Substituting into the Poisson formula.[1]
  3. = 0.049787 × 81 / 24 = 0.168To 3 significant figures.[1]

0.168

N2[4 marks]
A book has misprints at an average rate of 0.4 per page. Find the probability that a 10-page chapter contains at least 3 misprints.
Full working
  1. λ = 0.4 × 10 = 4Scaling to the ten-page interval.[1]
  2. P(X ⩾ 3) = 1 − P(0) − P(1) − P(2)The complement, since the upper tail is unbounded.[1]
  3. P(0) = e⁻⁴ = 0.0183; P(1) = 4e⁻⁴ = 0.0733; P(2) = 8e⁻⁴ = 0.1465P(2) uses 4²/2! = 8, not 16.[1]
  4. P(X ⩾ 3) = 1 − 0.2381 = 0.762"At least 3" excludes only 0, 1 and 2.[1]

0.762

Exam questions

1 · 6 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
A factory produces items of which 0.5% are defective. A random sample of 400 is taken.
(a) State the exact distribution of the number of defectives.
(b) Explain why a Poisson approximation is appropriate, and state its parameter.
(c) Use the approximation to find the probability of at most 1 defective.
(d) Find the probability of exactly 2 defectives.
Mark scheme
  1. (a) X ~ B(400, 0.005)A fixed number of trials with constant probability, so binomial exactly.[1]
  2. (b) n = 400 is large (> 50) and np = 2 is small (< 5), so Po(2) is appropriateBoth conditions must be quoted with their values, not just named.[1]
  3. λ = np = 400 × 0.005 = 2The approximating mean is the binomial mean.[1]
  4. (c) P(X ⩽ 1) = P(0) + P(1) = e⁻² + 2e⁻²"At most 1" means 0 or 1.[1]
  5. = 0.1353 + 0.2707 = 0.406To 3 significant figures.[1]
  6. (d) P(X = 2) = e⁻² × 2²/2! = 0.1353 × 2 = 0.2712²/2! = 2, so this equals P(X = 1) — a coincidence that occurs whenever λ = 2.[1]

(a) B(400, 0.005); (b) Po(2); (c) 0.406; (d) 0.271