Condition Poisson probabilities on a known count event
Intersect count events and divide by a positive-probability condition to work within a restricted sample space.
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
- Use P(A given B) = P(A and B)/P(B).
- Calculate Poisson tails and intervals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which denominator is used for P(X = 1 | X ≥ 1)?
What you will learn
- Calculate conditional probabilities from Poisson masses.
- Find the intersection before dividing by the condition.
Restrict possible counts
For Po(2), P(X = 1 | X ≥ 1) has numerator P(X = 1) = 2e⁻² because the intersection remains count one. The denominator is 1 − e⁻², giving approximately 0.3130. An unconditional mass is not the conditional answer.
The denominator must be positive. For Po(0), conditioning on X ≥ 1 is undefined because that event has zero probability. Nested events about the same count are not automatically independent.
Remove values contradicted by the condition
For Po(3), P(X ≤ 2 | X ≥ 1) intersects to {1,2}. Divide 7.5e⁻³ by 1 − e⁻³ to obtain approximately 0.3930. Including count zero in the numerator would contradict the known condition.
For Po(4), P(X = 2 | X ≤ 2) = 8e⁻⁴/[e⁻⁴(1 + 4 + 8)] = 8/13 ≈ 0.6154. Cancel the common exponential only after forming both probabilities. Check the result lies in [0,1].
Pause and explain
For Po(4), what is P(X = 2 | X ≤ 2)?
Worked example
For Po(2), find P(X = 1 | X ≥ 1).
Show the worked solution
- The intersection is count one, mass 2e⁻².
- The condition has probability 1 − e⁻² > 0.
- Divide to obtain approximately 0.3130.
Answer 2e⁻²/(1 − e⁻²) ≈ 0.3130.
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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.
For Po(2), calculate P(X = 1 | X ≥ 1).
Give me a hint
Compute the positive-count denominator by complement.
Compare my reasoning
- The numerator is 2e⁻².
- The denominator is 1 − e⁻².
- Their ratio is approximately 0.3130.
≈ 0.3130.
Look for these in your work
- Identify the condition.
- Verify its probability is positive.
For Po(3), calculate P(X ≤ 2 | X ≥ 1).
Give me a hint
Find the common count values before summing masses.
Compare my reasoning
- The intersection is counts one and two.
- Its mass is 7.5e⁻³.
- Divide by 1 − e⁻³ to get approximately 0.3930.
≈ 0.3930.
Look for these in your work
- Exclude zero from the intersection.
- Use the whole condition probability.
For Po(4), calculate P(X = 2 | X ≤ 2) and explain why independence cannot be assumed.
Give me a hint
Form both masses before cancelling their common factors.
Compare my reasoning
- P(2) = 8e⁻⁴.
- P(X ≤ 2) = 13e⁻⁴, hence the ratio is 8/13 ≈ 0.6154.
- Count two implies the condition; knowing it changes the mass, so the events are dependent.
8/13 ≈ 0.6154; the nested events are dependent.
Look for these in your work
- Form the complete condition.
- Explain its effect on the sample space.
Common mistakes
- Use the known condition as denominator.
- Intersect before summing.
- Zero-probability conditions are undefined.
What can you explain now?
Does X ≤ 2 given X ≥ 1 include zero in the numerator?
Compare with the explanation
No; the intersection is {1,2}.
The condition excludes zero from the restricted support.
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Make it stick
Intersect the events, calculate a positive condition probability, then divide within the restricted space.
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