Use geometric tails and conditions after early failures
Translate waiting-time boundaries into a number of failures and condition on the observed early part of the experiment.
Section 5.4 · Discrete random variables
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 first-success probabilities.
- Use complements and conditional probability with a positive denominator.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does T > 4 mean?
What you will learn
- Calculate P(T > m), P(T ≥ r) and P(T ≤ m).
- Use conditional probability after a run of initial failures.
Count how many trials must fail
For T ~ Geo(p), T > m means no success in trials 1 through m, so P(T > m) = (1 − p)ᵐ for non-negative integer m. By contrast, T ≥ r requires only r − 1 initial failures, giving (1 − p)ʳ⁻¹. The one-trial boundary difference matters.
At least one success by trial m is T ≤ m, with probability 1 − (1 − p)ᵐ. For p = 1/4, P(T > 4) = (3/4)⁴ = 81/256, while P(T ≤ 4) = 175/256. For p = 1/5, P(T ≥ 3) = (4/5)² = 0.64, whereas P(T > 3) = (4/5)³ = 0.512.
Condition on the known initial failures
Suppose p = 1/4 and the first two trials failed, so T > 2. First success on trial five has joint probability (3/4)⁴(1/4). Divide by P(T > 2) = (3/4)² to get P(T = 5 | T > 2) = (3/4)²(1/4) = 9/64. After the observed failures, two further failures and then success are still required.
Similarly, P(T > 5 | T > 2) = (3/4)⁵/(3/4)² = (3/4)³ = 27/64. Three additional trials must fail. This unchanged future probability follows from the stipulated independent constant-p model; it does not mean success becomes ‘due’ after failures. If the given event has probability zero, conditioning is undefined, so handle the p = 1 endpoint separately.
Pause and explain
For Geo(1/5), which expression gives P(T ≥ 3)?
Worked example
T ~ Geo(1/4). Find P(T > 4) and P(T ≤ 4).
Show the worked solution
- T > 4 requires all four initial trials to fail.
- Its probability is (3/4)⁴ = 81/256.
- The complementary chance of success by trial four is 1 − 81/256 = 175/256.
Answer P(T > 4) = 81/256; P(T ≤ 4) = 175/256.
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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.
T ~ Geo(1/4). Find P(T > 4) and P(T ≤ 4).
Give me a hint
Use four failures and then the complementary event.
Compare my reasoning
- P(T > 4) = (3/4)⁴ = 81/256.
- T ≤ 4 and T > 4 are complementary.
- P(T ≤ 4) = 1 − 81/256 = 175/256.
81/256 and 175/256.
Look for these in your work
- Do not add a final success to a tail.
- Check the two probabilities sum to one.
T ~ Geo(1/5). Find P(T ≥ 3) and P(T > 3).
Give me a hint
Count the initial failures separately for each boundary.
Compare my reasoning
- T ≥ 3 requires two failures, giving (4/5)² = 0.64.
- T > 3 requires three failures, giving (4/5)³ = 0.512.
- The difference 0.128 is exactly P(T = 3).
P(T ≥ 3) = 0.64; P(T > 3) = 0.512.
Look for these in your work
- Preserve the inclusive and strict boundaries.
- Explain the one-trial difference.
T ~ Geo(1/4). Given T > 2, find P(T = 5 | T > 2) and P(T > 5 | T > 2).
Give me a hint
Divide the relevant joint event by the positive probability of the first two failures.
Compare my reasoning
- P(T > 2) = (3/4)²; the event T = 5 is contained within it.
- P(T = 5 | T > 2) = (3/4)⁴(1/4)/(3/4)² = 9/64.
- P(T > 5 | T > 2) = (3/4)⁵/(3/4)² = 27/64.
9/64 and 27/64.
Look for these in your work
- Use a positive conditional denominator.
- Count additional failures after the known initial run.
Common mistakes
- T ≥ r and T > r have different failure counts.
- A tail event does not need a final success factor.
- Independent failures do not make the next success more likely or guaranteed.
What can you explain now?
For T ≥ 3, how many initial failures are required?
Compare with the explanation
Two: trials one and two must fail.
Success on trial three is allowed by the inclusive lower boundary.
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Make it stick
Translate the boundary into initial failures, use the complement for success by a deadline, and condition on exactly what is known.
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.