Choose an exposure using a Poisson probability target
Solve exponential probability inequalities and check neighbouring allowed durations before reporting a minimum or maximum.
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
- Solve logarithmic inequalities.
- Scale a positive rate with exposure.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At a positive fixed rate, how does P(no events) change with duration?
What you will learn
- Solve minimum-arrival or maximum-idle duration targets.
- Verify whole-unit thresholds without early rounding.
Decide which direction the target requires
At three per hour over t hours, P(at least one) = 1 − e⁻³ᵗ. Requiring at least 0.90 gives e⁻³ᵗ ≤ 0.10, hence t ≥ ln 10/3. The minimum continuous duration is about 0.767528 hours or 46.0517 minutes.
At two per minute, P(no events) ≥ 0.50 instead gives e⁻²ᵗ ≥ 0.50. Dividing the logarithmic inequality by minus two reverses its direction: t ≤ ln 2/2 minutes, about 20.7944 seconds. This is a maximum idle duration, since longer exposure reduces P(0).
Respect the allowed time unit
At six per hour, m minutes gives λ = m/10. An at-least-one probability of at least 0.90 requires m ≥ 10ln 10 ≈ 23.02585. If m must be whole, its minimum is 24, not the nearest rounded value 23.
At 23 minutes, 1 − e⁻²⋅³ ≈ 0.899741 fails; at 24, 1 − e⁻²⋅⁴ ≈ 0.909282 passes. Round only after solving the inequality. These probabilities are conditional on the ideal process, not empirical guarantees of an event by a deadline.
Pause and explain
If minimum whole m must satisfy m ≥ 23.02585, what is m?
Worked example
At six per hour, find the minimum whole minutes for P(at least one) ≥ 0.90.
Show the worked solution
- Set 1 − e⁻ᵐ⁄¹⁰ ≥ 0.90.
- Solve m ≥ 10ln 10 ≈ 23.02585.
- The minimum whole m is 24; verify 23 fails and 24 passes.
Answer 24 minutes under the model.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 three per hour, find the minimum continuous duration with P(at least one) ≥ 0.90.
Give me a hint
Use the zero-count complement and solve for hours.
Compare my reasoning
- e⁻³ᵗ ≤ 0.10.
- Taking logs and dividing by −3 gives t ≥ ln 10/3.
- The minimum is ≈ 0.767528 hours = 46.0517 minutes.
t ≥ ln 10/3 hours; minimum ≈ 46.0517 minutes.
Look for these in your work
- Preserve the inequality direction.
- State continuous duration and units.
At two per minute, find the maximum continuous duration for P(0) ≥ 0.50.
Give me a hint
Longer exposure decreases P(0), so expect an upper bound.
Compare my reasoning
- e⁻²ᵗ ≥ 1/2 gives −2t ≥ −ln 2.
- Divide by −2 to get t ≤ ln 2/2 minutes.
- The maximum is ≈ 0.346574 minutes = 20.7944 seconds.
Maximum ≈ 20.7944 seconds.
Look for these in your work
- Reverse order for negative division.
- Interpret the bound as a maximum.
At six per hour, choose the minimum whole minutes for chance at least 0.90 of an event and check its preceding minute.
Give me a hint
Solve continuously before comparing adjacent permitted times.
Compare my reasoning
- m ≥ 10ln 10 ≈ 23.02585.
- At 23 minutes the probability ≈ 0.899741 < 0.90.
- At 24 it is ≈ 0.909282 ≥ 0.90; thus 24 is minimal.
24 minutes; 23 fails and 24 passes.
Look for these in your work
- Respect whole-minute constraints.
- Verify the adjacent candidate.
Common mistakes
- Negative division reverses an inequality.
- Idle and arrival targets give different bound directions.
- Round thresholds according to allowed units.
What can you explain now?
Why does 23 minutes fail a minimum of 23.02585?
Compare with the explanation
It is below the required lower bound.
The preceding probability check confirms the failure.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Write the target, solve without early rounding, then check its direction, units and neighbouring allowed value.
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.