Calculate geometric first-success probabilities
Use the single required sequence of failures followed by success, without a binomial order factor.
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
- Define a geometric first-success variable.
- Multiply probabilities along independent paths.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many failures precede first success on trial five?
What you will learn
- Calculate P(T = r) = (1 − p)ʳ⁻¹p.
- Explain why the first-success event has exactly one success placement.
Failures must come before the success
For T ~ Geo(p), the event T = r requires r − 1 failures followed by one success. Thus P(T = r) = (1 − p)ʳ⁻¹p for integer r ≥ 1. The successful trial is fixed at the final position; there is no choice of success positions and no combination factor.
For p = 1/4 and r = 3, the only permitted string is failure, failure, success, with probability (3/4)²(1/4) = 9/64. A string containing a success on trial one or two would have stopped earlier. Exactly one success among three fixed trials is a different binomial event and would allow three positions.
Use the variable’s support
For p = 1/5, first success on trial five has four preceding failures, giving (4/5)⁴(1/5) = 256/3125 = 0.08192. Counting five failures would move the success to trial six. For p = 1/3, P(T = 1) = 1/3, while P(T = 4) = (2/3)³(1/3) = 8/81.
T cannot equal zero or a non-integer, so those events have probability zero. At p = 1 the first trial succeeds with certainty, giving probability one at T = 1 and zero for all later values. Address this endpoint directly if a calculator presents an ambiguous zero-power expression. Formula proofs are not required; the event definition provides the needed reasoning.
Pause and explain
For T ~ Geo(1/4), what is P(T = 3)?
Worked example
T ~ Geo(1/4). Find P(T = 3).
Show the worked solution
- First success on trial three requires two failures followed by success.
- Multiply their probabilities: (3/4)(3/4)(1/4).
- The result is 9/64, with no order factor.
Answer P(T = 3) = 9/64.
Try a topic test
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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 = 3).
Give me a hint
Write the unique two-failure, one-success path.
Compare my reasoning
- Trials one and two must fail, each with probability 3/4.
- Trial three must succeed with probability 1/4.
- Multiply (3/4)²(1/4) = 9/64.
9/64.
Look for these in your work
- Use two preceding failures.
- Avoid an extra success-position factor.
T ~ Geo(1/5). Find the probability of first success on trial five.
Give me a hint
The first four trials fail and the fifth succeeds.
Compare my reasoning
- Use four failure factors, each 4/5.
- Multiply by the final success factor 1/5.
- (4/5)⁴(1/5) = 256/3125 = 0.08192.
256/3125 = 0.08192.
Look for these in your work
- Match the exponent to the preceding failures.
- State the exact first-success event.
T ~ Geo(1/3). Compare P(T = 1) and P(T = 4), explaining the exponents.
Give me a hint
The first case needs no preceding failure; the second needs three.
Compare my reasoning
- P(T = 1) = 1/3 because success occurs immediately.
- P(T = 4) = (2/3)³(1/3) = 8/81.
- The failure exponents are zero and three because the successful trial is included in T.
P(T = 1) = 1/3; P(T = 4) = 8/81.
Look for these in your work
- Keep the first-trial convention consistent.
- Explain the number of preceding failures.
Common mistakes
- Use r − 1 failures, not r.
- Do not add a combination factor to the unique first-success path.
- First success on trial r differs from one success among r fixed trials.
What can you explain now?
Why is there no 3C1 factor for first success on trial three?
Compare with the explanation
The only allowed success position is trial three.
Any earlier success changes the first-success trial and leaves the event.
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Make it stick
First success on trial r means exactly r − 1 initial failures and a final success in a single permitted order.
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.