Use independent disjoint intervals and reject overlap products
Calculate independent-window joint probabilities and split overlapping observations into separate pieces.
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
- Scale time-window parameters.
- Distinguish independence from overlapping information.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At six per hour, what is λ for ten minutes?
What you will learn
- Multiply probabilities for justified independent disjoint windows.
- Analyse overlap through shared and unshared counts.
Use the stipulated independence
At six per hour, each ten-minute window has λ = 1. For two disjoint ten-minute windows, P(first zero, second one) = e⁻¹ × e⁻¹ = e⁻² ≈ 0.1353, using the stipulated independent counts.
For twenty minutes followed by a separate ten minutes, parameters are two and one. P(first one, second zero) = 2e⁻²e⁻¹ = 2e⁻³ ≈ 0.0996. Independent windows need not have equal duration or count probabilities.
Split the shared observation
Let A count minute 0–20 and B count minute 10–30. Three disjoint ten-minute pieces have independent Po(1) counts. A = 0 and B = 1 requires piece counts 0,0,1, giving e⁻³ ≈ 0.0498.
The marginal product P(A = 0)P(B = 1) = e⁻² × 2e⁻² = 2e⁻⁴ ≈ 0.0366 is wrong. Knowing A = 0 fixes the shared middle count at zero and changes B’s possible sources. General independent Poisson sums belong to section 6.2; here use the explicit disjoint-piece mechanism.
Pause and explain
Why is P(A = 0)P(B = 1) invalid for these overlapping windows?
Worked example
At six per hour, find P(A = 0,B = 1) for A over 0–20 minutes and B over 10–30.
Show the worked solution
- Split into 0–10, 10–20 and 20–30, each with λ = 1.
- The event needs piece counts 0,0,1.
- Their independent product is e⁻³ ≈ 0.0498.
Answer e⁻³ ≈ 0.0498.
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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 chance of zero in one ten-minute window and one in another disjoint ten-minute window.
Give me a hint
Both parameters are one and disjoint counts are independent by assumption.
Compare my reasoning
- Both λ values equal one.
- Required masses are e⁻¹ and e⁻¹.
- Multiply to obtain e⁻² ≈ 0.1353.
e⁻² ≈ 0.1353.
Look for these in your work
- State independence.
- Use window-specific parameters.
At six per hour, find the chance of one in twenty minutes then zero in the following ten minutes.
Give me a hint
Scale the different disjoint durations separately.
Compare my reasoning
- First λ = 2, mass at one 2e⁻².
- Second λ = 1, mass at zero e⁻¹.
- Product = 2e⁻³ ≈ 0.0996.
2e⁻³ ≈ 0.0996.
Look for these in your work
- Calculate both parameters.
- Justify the product.
For A over 0–20 and B over 10–30 minutes at six per hour, calculate P(A = 0,B = 1) and the incorrect marginal product.
Give me a hint
Split the observation into three independent ten-minute pieces.
Compare my reasoning
- The required piece counts are 0,0,1.
- Correct product e⁻³ ≈ 0.0498.
- Marginal product 2e⁻⁴ ≈ 0.0366 differs because the observations overlap.
Correct ≈ 0.0498; incorrect marginal product ≈ 0.0366.
Look for these in your work
- Translate the joint event into pieces.
- Explain dependence through the shared count.
Common mistakes
- Overlapping marginal events are dependent here.
- Each piece needs its own parameter.
- Independence must be justified by the model.
What can you explain now?
What is shared by windows 0–20 and 10–30?
Compare with the explanation
The 10–20 count.
That common contribution causes dependence.
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Make it stick
Draw the exposures, split overlaps, and justify independence before multiplying probabilities.
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.