Scale event rates to the required exposure
Convert time, length or area into consistent units before forming the expected count λ.
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.1. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Po(λ) counts non-negative integers over a stated exposure, with finite λ ≥ 0. Mean and variance are each λ; proofs are not required. The homogeneous process examples stipulate constant rate, independent counts in disjoint exposures, and negligible multiple events in sufficiently small exposures; real applications need evidence. A binomial-to-Poisson approximation uses large n and small p, with n > 50 and np < 5 approximately as the syllabus guide. A Poisson-to-normal approximation uses λ > 15 approximately, matching N(λ,λ) and applying continuity correction. These guides are not universal error guarantees. Exact and approximate calculations are labelled separately; numerical normal probabilities use unrounded z and can differ slightly from rounded-table working. Independent Poisson sums and linear combinations, sample means and hypothesis testing belong to later Statistics 2 sections. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Convert minutes, seconds and hours.
- Define a homogeneous Poisson count.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many hours is twenty minutes?
What you will learn
- Calculate λ as rate multiplied by consistent exposure.
- Compare parameters from different rate and exposure pairs.
Match units before multiplying
At six per hour, twenty minutes is one third of an hour. Thus λ = 6/3 = 2 and X ~ Po(2). The resulting parameter is an expected number of events, without a remaining time unit. Six times twenty would incorrectly mix hours and minutes.
The same reasoning applies to length or area under a stipulated homogeneous spatial model. Four defects per metre over 1.5 metres gives λ = 6. Constant rate and independent disjoint counts remain assumptions requiring evidence in real applications.
Use the whole observation
At two per minute, ninety seconds is 1.5 minutes and λ = 3. At six per hour, thirty minutes is half an hour and also gives λ = 3. Both counts have Po(3) probabilities under their respective assumptions, despite different rates.
Do not reuse a window parameter when duration changes or extend an old constant rate into a period where it no longer applies. For a valid parameter, P(X = 0) = e⁻λ; increasing exposure at a positive fixed rate reduces the zero-count probability.
Pause and explain
At six per hour, what is λ for twenty minutes?
Worked example
At six per hour, find the count model over twenty minutes and its zero probability.
Show the worked solution
- Convert to 1/3 hour.
- λ = 6/3 = 2, so X ~ Po(2).
- P(X = 0) = e⁻² ≈ 0.1353.
Answer Po(2); P(0) ≈ 0.1353.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
At six per hour, find the twenty-minute count model and P(X = 0).
Give me a hint
Convert the duration to hours before finding its parameter.
Compare my reasoning
- Twenty minutes is 1/3 hour.
- λ = 6 × 1/3 = 2.
- X ~ Po(2) and P(0) = e⁻² ≈ 0.1353.
Po(2); P(0) ≈ 0.1353.
Look for these in your work
- Use consistent units.
- Use this window’s parameter.
A stipulated spatial Poisson model has four defects per metre. Find the count model and exactly-two probability on 1.5 metres.
Give me a hint
Use rate times length, then the Poisson formula.
Compare my reasoning
- λ = 4 × 1.5 = 6.
- X ~ Po(6).
- P(2) = e⁻⁶6²/2! = 18e⁻⁶ ≈ 0.0446.
Po(6); P(2) ≈ 0.0446.
Look for these in your work
- Find the length-specific parameter.
- Include the factorial.
Compare two per minute over ninety seconds with six per hour over thirty minutes, assuming both respective homogeneous models.
Give me a hint
Convert each exposure into its own rate’s unit.
Compare my reasoning
- Ninety seconds is 1.5 minutes, giving λ = 3.
- Thirty minutes is 0.5 hours, also giving λ = 3.
- Both count distributions are Po(3) under the assumptions.
Both parameters are 3, giving the same count probabilities.
Look for these in your work
- Show both conversions.
- Compare parameters, not bare rates.
Common mistakes
- Convert exposure to the rate’s unit.
- Rate is not λ for every window.
- Check that the constant rate still applies.
What can you explain now?
Does double exposure give double λ at fixed rate?
Compare with the explanation
Yes, under the homogeneous model.
It doubles the expectation, not necessarily the observed count.
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Make it stick
Convert exposure units, check the rate assumption, then multiply to find the count parameter.
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