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Mathematics HL / Extra · Legacy and extension

27 Poisson distribution: legacy extension

20-minute guided session≈ 7 min reading incl. reveals32 bank questions

Start with the exam

What you need to be able to do

This is a legacy extension, not compulsory current AA HL material for the reader’s 2027 target. Use it for deliberate extension practice or when a model is supplied. Rate scaling, conditional probability and nested binomial counts are transferable. Prioritise the current probability chapters before spending revision time memorising the legacy Poisson formula.

Your 20-minute study route

Spend 3 minutes reading the overview and diagram, 7 minutes learning the question patterns, 5 minutes writing out the worked example, and 5 minutes on a bank question or recall check. Take longer when you need to; the full bank is here for later practice.

Legacy / extension material

This collection contains content outside the current course, or a mixture of current and legacy skills. Use the compatibility note for each original question to decide which parts to practise for your exam.

0 min30 min60 minHalf-hour mean: 1Half-hour mean: 1Dots illustrate events, not guaranteed counts.
The expected count scales with interval length.

Recognise → set up → write

How the questions are set

QUESTION PATTERN 01

Scale the parameter

Recognise it: A Poisson model changes duration or area.

  1. For X~Po(λ), use P(X=r)=e^(−λ)λ^r/r!. Its mean and variance both equal λ.
  2. At constant rate c per unit, exposure t has mean ct. Convert time or area units before multiplying.
  3. Translate integer tails carefully: P(X≥r)=1−P(X≤r−1). For independent disjoint intervals, multiply probabilities of the stated interval events.
What to put on the page

State λ with its exposure, then the exact probability expression or cumulative event.

Common mark loss

The standard deviation is √λ, not λ. A daily mean is not a weekly mean.

QUESTION PATTERN 02

Recover the rate

Recognise it: A zero mass or relationship between count probabilities is given.

  1. P(X=0)=e^(−λ), so λ=−ln P(X=0). A legitimate mean must be nonnegative.
  2. For adjacent masses use P(X=r+1)/P(X=r)=λ/(r+1), obtained by cancelling the complete formulas.
  3. A nonlinear equation may have several positive roots. Check the original probability equation and all extra conditions before selecting one.
What to put on the page

Show the logarithmic equation or the cancellation of factorials and powers.

Common mark loss

For integer λ>0 there are two modes, λ−1 and λ; blindly rounding the mean is insufficient.

QUESTION PATTERN 03

Condition and count qualifying intervals

Recognise it: A later part concerns qualifying days or capacity-limited revenue.

  1. Define events A and B and write P(A∩B)/P(B). Nested events are not independent just because counts on different days are independent.
  2. If one interval qualifies with probability q, the count among n independent identical intervals is binomial B(n,q), with mean nq.
  3. For capped income, sum low-count payoffs individually and combine the tail probability with the constant capped payoff.
What to put on the page

Show the conditioning denominator and define any outer count variable.

Common mark loss

Expected arrivals and expected occupied capacity differ when arrivals can exceed capacity.

Write it before you reveal it

A worked exam-style example

Original teaching example, created for this lesson.

Original extension example: a supplied model gives two arrivals per hour. Find the chance of at least one in 30 minutes and the expected number of qualifying intervals among ten independent half-hours.

Reveal the worked solution
  1. The half-hour parameter is λ=2×1/2=1.
  2. q=P(X≥1)=1−e^(−1).
  3. For Y~B(10,q), E(Y)=10q=10(1−e^(−1)).

Answer: Probability ≈0.632; expected qualifying count ≈6.32.

Check: An expectation may be noninteger although every observed interval count is integer.

From your own collection

Apply it to past-paper questions

These questions anchor the patterns above. Attempt one without the answer, then compare your setup and conclusion with the reasoning. The linked reader contains the source answer and the question-specific syllabus review.

Q992 · 2018 · PAPER 2 · 5 MARKS

Question 992

Several linked steps in Poisson mean scaled by area, Conditional probability with an integer threshold.

Compatibility: Poisson distributions are a legacy Mathematics HL requirement, outside current AA HL. Conditional probability itself is current; supply Poisson probabilities or the mass-function/CDF rule for extension practice.

View the original question

Original page images preserve the equations and diagrams. Write your response before revealing the reasoning.

Original Mathematics question 992, page 1251
Reveal the revision reasoning

The reserve mean is λ=3.2×56=179.2, so the supplied model is X∼Poisson(179.2).

Because X>190 implies X≥168, the conditional probability is P(X≥191)/P(X≥168).

Use upper tails 1−P(X≤190) and 1−P(X≤167), respectively. Their ratio is approximately 0.245331.

Editorial revision reasoning. Check the original answer for its marking detail.

Q989 · 2017 · PAPER 1 · 7 MARKS

Question 989

Requires linking Legacy Poisson zero mass, Logarithms and exact probability.

Compatibility: (a)-(b): The probability formula and Poisson parameter scaling are legacy. The subsequent logarithm and complement algebra are current.

View the original question

Original page images preserve the equations and diagrams. Write your response before revealing the reasoning.

Original Mathematics question 989, page 1246
Reveal the revision reasoning
  1. Since P(X>0)=3/4, P(X=0)=1/4=exp(−m).
  2. Take logarithms to obtain m=ln 4.
  3. Y has mean 2m=ln 16, so its zero mass is 1/16 and its one mass isln 16/16.
  4. Subtract these from one to obtain(15−ln 16)/16.

Editorial revision reasoning. Check the original answer for its marking detail.

Q991 · 2006 · PAPER 1 · 6 MARKS

Question 991

Direct application of Legacy Poisson tail, Current conditional probability.

Compatibility: (a): Poisson modelling is outside current AA HL examination content. (b): Conditional probability of nested events is current, but its input probabilities come from the legacy Poisson model.

View the original question

Original page images preserve the equations and diagrams. Write your response before revealing the reasoning.

Original Mathematics question 991, page 1249
Reveal the revision reasoning
  1. For Poisson mean 1.5, use P(X=k)=exp(−1.5)1.5^k/k!.
  2. More than two is 1−[P(0)+P(1)+P(2)], approximately 0.191.
  3. At least one has probability 1−exp(−1.5).
  4. Every outcome with more than two also has at least one, so divide the first probability by the second, giving approximately 0.246.

Editorial revision reasoning. Check the original answer for its marking detail.

All 32 questions in this topic

Close your notes first

Can you retrieve the method?

1. Is Poisson compulsory here?

Check answer 1

No, it is labelled legacy extension.

2. Mean over three hours at hourly rate c?

Check answer 2

3c.

3. Probability of at least one?

Check answer 3

1−e^(−λ).

Before you move on

  1. Identify legacy facts separately from current probability operations.
  2. Record the exposure attached to each λ.
  3. Check the integer boundary before using a cumulative probability.

Your revision note

Record one missed cue, one corrected step and one question to retry. Saved only in this browser.