Calculate exact Poisson count probabilities
Use the probability mass formula with factorials, support checks and the zero-parameter edge case.
Section 6.1 · The Poisson distribution
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
- Evaluate factorials including 0! = 1.
- Find the exposure-specific parameter.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 0! in the mass formula?
What you will learn
- Calculate e⁻λ λʳ/r! for integer r ≥ 0.
- Handle zero counts, excluded values and λ = 0.
Evaluate one count
For λ > 0 and r = 0,1,2,…, P(X = r) = e⁻λ λʳ/r!. For Po(3), P(2) = e⁻³9/2 = 4.5e⁻³ ≈ 0.2240. This is a discrete mass, unlike a continuous density height. Do not add a binomial nCr factor.
At zero count, use 0! = 1 and λ⁰ = 1, giving e⁻λ. Negative and non-integer counts have probability zero. For λ = 0, state the degenerate distribution directly: P(0) = 1 and P(r) = 0 for r ≥ 1, avoiding an ambiguous 0⁰ expression.
Compare adjacent masses
For positive λ, P(r + 1)/P(r) = λ/(r + 1). For Po(4), masses at three and four are equal because the ratio is 4/4 = 1. Both equal (64/6)e⁻⁴ ≈ 0.1954.
The Po(3) bars show counts zero through ten only. The support continues beyond the drawing; a truncated display is not the complete distribution. Retain full precision until the final probability and check it belongs to [0,1].
Pause and explain
Which expression gives P(2) for Po(3)?
Worked example
Calculate P(X = 2) for Po(3).
Show the worked solution
- Identify λ = 3 and r = 2.
- Substitute e⁻³3²/2!.
- Simplify to 4.5e⁻³ ≈ 0.2240.
Answer 4.5e⁻³ ≈ 0.2240.
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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.
Calculate P(2) for Po(3).
Give me a hint
Write the exponential, power and factorial before simplifying.
Compare my reasoning
- λ = 3 and r = 2.
- P(2) = e⁻³3²/2! = 4.5e⁻³.
- Evaluate ≈ 0.2240, within [0,1].
4.5e⁻³ ≈ 0.2240.
Look for these in your work
- Use the count factorial.
- Round only at the end.
Find P(0) for Po(2), then state zero and positive masses for Po(0).
Give me a hint
Handle zero parameter directly as a count always equal to zero.
Compare my reasoning
- Po(2) gives e⁻²2⁰/0! = e⁻² ≈ 0.1353.
- Po(0) assigns probability one to zero.
- Every positive count then has probability zero.
Po(2): P(0) ≈ 0.1353; Po(0): P(0) = 1, P(r ≥ 1) = 0.
Look for these in your work
- Use 0! = 1.
- Avoid ambiguous 0⁰ in the degenerate case.
For Po(4), calculate P(3) and P(4) and explain their equality.
Give me a hint
Compare consecutive masses with λ/(r + 1).
Compare my reasoning
- P(3) = (64/6)e⁻⁴.
- P(4) = (256/24)e⁻⁴, the same expression.
- Their ratio is 4/(3 + 1) = 1; both are ≈ 0.1954.
P(3) = P(4) ≈ 0.1954.
Look for these in your work
- Evaluate both masses.
- Explain the consecutive-mass ratio.
Common mistakes
- Divide by r!, not r.
- Do not add a binomial order factor.
- State Po(0) directly.
What can you explain now?
What is P(0) for Po(2)?
Compare with the explanation
e⁻² ≈ 0.1353.
The zero power and zero factorial are each one for positive λ.
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Make it stick
Identify parameter and count, substitute every factor, then check support and probability range.
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.