Infer a Poisson parameter from a probability or ratio
Solve zero-count logarithmic equations and consecutive-mass ratios, checking exposure units and zero-denominator cases.
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
- Solve exponentials with natural logarithms.
- Use consecutive Poisson masses.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If e⁻λ = q, 0 < q ≤ 1, what is λ?
What you will learn
- Infer λ from a zero probability or positive mass ratio.
- Convert the inferred window parameter to a rate.
Isolate the zero-count parameter
P(0) = q with 0 < q ≤ 1 gives e⁻λ = q and λ = −ln q. Thus q = e⁻² implies λ = 2; q = 1 implies λ = 0. Probability zero has no finite λ, and values outside [0,1] are invalid.
If P(0) = 0.20 over thirty minutes, λ = ln 5 ≈ 1.6094 for that half-hour. A constant hourly rate is λ/0.5 = 2ln 5 ≈ 3.2189 per hour. The inferred parameter belongs to its specified exposure.
Check before dividing
For λ > 0, P(2)/P(1) = λ/2. Thus P(2) = 2P(1) gives λ = 4 under the positive-parameter condition. Cancel the common exponential and parameter factors carefully.
The restriction matters: at λ = 0 both masses vanish, so the original equality also holds but their ratio is undefined. A single other mass equation can have multiple parameter solutions because the mass can rise and fall with λ; do not assume uniqueness without checking admissible branches.
Pause and explain
Why require λ > 0 before dividing P(2) by P(1)?
Worked example
For λ > 0, P(2) = 2P(1). Find λ.
Show the worked solution
- P(1) = λe⁻λ > 0, so division is valid.
- The ratio gives λ/2 = 2.
- Solve λ = 4, consistent with the restriction.
Answer λ = 4 for λ > 0.
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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.
Find λ if P(0) = e⁻² exactly.
Give me a hint
Take natural logarithms of the zero-count formula.
Compare my reasoning
- e⁻λ = e⁻².
- Logarithms give −λ = −2.
- Thus λ = 2, which is admissible.
λ = 2.
Look for these in your work
- Use the zero-count formula.
- Check the parameter range.
For λ > 0, P(2) = 2P(1). Find λ and explain the restriction.
Give me a hint
Check that the denominator is positive before taking the ratio.
Compare my reasoning
- P(1) = λe⁻λ > 0.
- P(2)/P(1) = λ/2 = 2 gives λ = 4.
- At λ = 0 both masses vanish; the equality alone holds but the ratio is undefined.
λ = 4 for λ > 0; without the restriction λ = 0 also satisfies the equality.
Look for these in your work
- Cancel the factorial ratio correctly.
- Identify the zero-denominator case.
A thirty-minute homogeneous Poisson count has P(0) = 0.20. Infer its parameter and hourly rate.
Give me a hint
The logarithm first gives the parameter for the actual observation.
Compare my reasoning
- λ = −ln 0.20 = ln 5 ≈ 1.6094.
- The exposure is half an hour.
- Rate = λ/0.5 = 2ln 5 ≈ 3.2189 per hour.
Half-hour λ ≈ 1.6094; hourly rate ≈ 3.2189.
Look for these in your work
- Infer the exposure-specific parameter.
- State the correct rate unit.
Common mistakes
- Use −ln q for P(0) = q.
- Scale a window parameter before quoting a rate.
- Do not divide by a zero positive-count mass.
What can you explain now?
What gives P(0) = 1?
Compare with the explanation
λ = 0.
This is the degenerate distribution concentrated at zero.
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Make it stick
Solve the logarithm or valid ratio, check zero cases and admissibility, then scale the exposure correctly.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.