Calculate Poisson tails and inclusive intervals
Translate integer wording and use a finite sum or complement to handle the infinite upper support.
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
- Calculate Poisson masses.
- Use complementary probabilities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which counts are excluded by X ≥ 2?
What you will learn
- Calculate cumulative, tail and inclusive interval probabilities.
- Distinguish more than from at least at an integer boundary.
Complement an upper tail
For Po(2), at least two is X ≥ 2, whose complement is X ≤ 1. Thus P(X ≥ 2) = 1 − e⁻²(1 + 2) ≈ 0.5940. Sum the finite excluded counts instead of truncating an infinite upper tail without accounting for the remainder.
For Po(4), more than two excludes counts zero, one and two: 1 − e⁻⁴(1 + 4 + 8) ≈ 0.7619. At least two instead excludes only zero and one, giving 1 − 5e⁻⁴ ≈ 0.9084. Discrete inclusive and strict events differ by a boundary mass.
Include both interval endpoints
For Po(3), 1 ≤ X ≤ 3 sums masses at one, two and three: e⁻³(3 + 9/2 + 27/6) = 12e⁻³ ≈ 0.5974. Equivalently subtract P(X ≤ 0) from P(X ≤ 3); subtracting P(X ≤ 1) would wrongly remove count one.
For integers 1 ≤ a ≤ b, P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1). The Po(2) diagram shows counts zero through eight only; the highlighted at-least-two event also includes the unshown upper tail.
Pause and explain
Which expression gives P(X > 2) for Po(4)?
Worked example
Find P(X ≥ 2) for Po(2).
Show the worked solution
- The complement contains zero and one.
- Their sum is 3e⁻².
- Subtract from one: 1 − 3e⁻² ≈ 0.5940.
Answer 1 − 3e⁻² ≈ 0.5940.
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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.
For Po(2), find the probability of at least two events.
Give me a hint
List the finite lower complement first.
Compare my reasoning
- The event is X ≥ 2.
- P(X ≤ 1) = 3e⁻².
- Its complement is 1 − 3e⁻² ≈ 0.5940.
≈ 0.5940.
Look for these in your work
- Include count two.
- Account for the entire upper tail.
For Po(3), find P(1 ≤ X ≤ 3).
Give me a hint
Sum the integer masses including both endpoints.
Compare my reasoning
- P(1) = 3e⁻³.
- P(2) = P(3) = 4.5e⁻³.
- Their sum is 12e⁻³ ≈ 0.5974.
12e⁻³ ≈ 0.5974.
Look for these in your work
- Include both endpoints.
- Add mutually exclusive count masses.
Compare P(X > 2) and P(X ≥ 2) for Po(4).
Give me a hint
Write both complements to isolate the count-two mass.
Compare my reasoning
- P(X > 2) = 1 − 13e⁻⁴ ≈ 0.7619.
- P(X ≥ 2) = 1 − 5e⁻⁴ ≈ 0.9084.
- Their difference is P(2) = 8e⁻⁴, included only by the second event.
More than two ≈ 0.7619; at least two ≈ 0.9084.
Look for these in your work
- Translate strict and inclusive wording.
- Explain the boundary mass difference.
Common mistakes
- At least and more than have different endpoints.
- A visible tail may be truncated.
- Use a − 1 when subtracting below an inclusive lower endpoint.
What can you explain now?
What is the complement of X > 2?
Compare with the explanation
X ≤ 2.
The excluded boundary count belongs to the complement.
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Make it stick
List the integer event, preserve endpoints and use a finite complement for an infinite upper tail.
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.