Interpret geometric expectation and choose attempt thresholds
Distinguish average waiting time from a probability guarantee and find the first integer deadline meeting a stated target.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.4. A probability table needs non-negative probabilities summing to one. E(X) is a probability-weighted mean; Var(X) = E(X²) − [E(X)]², with squared units. A binomial model B(n,p) counts successes in a fixed non-negative integer number n of independent two-outcome trials with a constant success probability p. A geometric model Geo(p) counts the trial of the first success, starting at 1, with independent trials and a constant 0 < p ≤ 1. Formula proofs are not required. The expectation of the geometric distribution is required; its variance is not required by section 5.4. Normal approximations belong to section 5.5 and are not used here. Exact model probabilities do not establish real-world independence or constant rates without evidence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate geometric tail probabilities.
- Use logarithms and check inequality directions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is E(T) for Geo(1/5) under the first-success convention?
What you will learn
- Use E(T) = 1/p under the first-success convention.
- Find and verify an integer attempt threshold for a specified success probability.
An average is not a guaranteed deadline
For T ~ Geo(p) counting the first-success trial from one, E(T) = 1/p. At p = 1/5 the expected waiting time is five trials. A particular experiment may finish on trial one or much later; the expectation is the long-run average under repeated independent experiments with the same model. Formula proofs and geometric variance are not required by section 5.4.
If the expected first-success trial is exactly four, then p = 1/4. The chance of success by trial two is 1 − (3/4)² = 7/16. The chance by trial four is 175/256, about 68.36%, rather than 100% or 90%. The alternative failure-count variable T − 1 has mean 1/p − 1, so its convention would change an inferred probability.
Verify both sides of an integer threshold
With p = 1/4, a success-by-m probability of at least 90% requires 1 − (3/4)ᵐ ≥ 0.90, or (3/4)ᵐ ≤ 0.10. Taking logarithms and dividing by the negative ln(3/4) reverses the inequality: m ≥ ln(0.10)/ln(3/4), approximately 8.004. Therefore the first possible integer is nine, not eight.
Check the neighbours with unrounded values: by eight trials the probability is 0.8998870849609375, still below 0.90; by nine it is 0.9249153137207031, above the target. Rounding the eight-trial probability to 0.90 before comparison would choose the wrong threshold. For 0 < p < 1, no finite number of attempts guarantees success with probability one, even though eventual success has probability one under the ideal infinite-trial model.
Pause and explain
For Geo(1/4), what is the minimum number of trials giving at least a 90% chance of success?
Worked example
T ~ Geo(1/4). Find the minimum integer m with P(T ≤ m) ≥ 0.90.
Show the worked solution
- Require (3/4)ᵐ ≤ 0.10.
- Divide logarithms carefully to get m ≥ ln(0.10)/ln(3/4) ≈ 8.004, so test m = 9.
- The eight-trial probability is 0.8998870849609375 and the nine-trial probability is 0.9249153137207031.
Answer Minimum m = 9.
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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.
T ~ Geo(1/5). Find E(T) and explain what it means.
Give me a hint
Take the reciprocal of p and interpret it over repeated experiments.
Compare my reasoning
- E(T) = 1/(1/5) = 5 trials.
- The variable includes the successful trial, rather than counting only failures.
- Five is the long-run mean waiting time under the model, not a guaranteed trial in one experiment.
E(T) = 5 trials; an average, not a guaranteed finishing time.
Look for these in your work
- Use the first-success convention.
- Distinguish expectation from a probability deadline.
A geometric first-success variable has exact expectation 4. Find p and P(T ≤ 2).
Give me a hint
Infer p from the reciprocal mean, then complement two failures.
Compare my reasoning
- 1/p = 4 gives p = 1/4.
- P(T > 2) = (3/4)² = 9/16.
- P(T ≤ 2) = 1 − 9/16 = 7/16.
p = 1/4; P(T ≤ 2) = 7/16.
Look for these in your work
- Infer p using the correct variable convention.
- Use a cumulative event rather than exactly two trials.
T ~ Geo(1/4). Find the minimum integer m giving P(T ≤ m) ≥ 0.90 and explain why the mean is insufficient.
Give me a hint
Solve the failure-tail inequality and compare the candidate with its preceding integer.
Compare my reasoning
- (3/4)ᵐ ≤ 0.10 gives m ≥ ln(0.10)/ln(3/4) ≈ 8.004.
- At m = 8 the success probability is 0.8998870849609375; at m = 9 it is 0.9249153137207031.
- Choose m = 9. The mean is four trials, but success by four is only 175/256, so it does not meet the 90% target.
Minimum m = 9; the mean of four trials is not a 90% deadline.
Look for these in your work
- Reverse the inequality when dividing by a negative logarithm.
- Check the unrounded probability at both neighbouring integers.
Common mistakes
- Expected waiting time is not a guaranteed completion time.
- The first-success mean differs from the mean number of failures.
- Do not round a near-target probability before choosing the integer threshold.
What can you explain now?
Does E(T) = 4 mean success is certain by trial four?
Compare with the explanation
No; for p = 1/4 that probability is only 175/256.
An average trial number and a cumulative probability answer different questions.
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Make it stick
Use 1/p for the mean first-success trial; use cumulative probabilities and neighbouring integers for a reliability threshold.
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.