Calculate exact binomial probabilities with the order factor
Multiply one fixed success/failure path, then count all distinct placements of the successes.
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
- Use combinations and independent path products.
- Recognise the binomial assumptions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many placements of two successes among four trials exist?
What you will learn
- Use P(X = r) = nCr pʳ(1 − p)ⁿ⁻ʳ.
- Handle zero successes, all successes and the model’s parameter endpoints.
One path is not the whole event
For X ~ B(4,1/3), one particular sequence containing two successes and two failures has probability (1/3)²(2/3)² = 4/81. The two success positions can be chosen in 4C2 = 6 ways. Hence P(X = 2) = 6 × 4/81 = 8/27.
In general, P(X = r) = nCr pʳ(1 − p)ⁿ⁻ʳ for integer r in 0–n. The combination counts success positions, while the powers give the probability of one arrangement. There is no extra r! factor: successes are outcome classifications, not individually named people being ordered. Outside the support, the probability is zero.
Check endpoint cases directly
Zero successes gives (1 − p)ⁿ, because there is one all-failure path; all n successes gives pⁿ. For n = 5 and p = 0.40, P(X = 0) = 0.60⁵ = 0.07776. For n = 3 and p = 1/4, exactly two successes gives 3(1/4)²(3/4) = 9/64.
If p = 0, the success count is certainly zero. If p = 1, it is certainly n. If n = 0, it is certainly zero for any admissible p. Handle these degenerate cases from the experiment rather than relying on ambiguous calculator displays involving zero to the zeroth power. Formula proofs are not required, but understanding the order factor prevents common errors.
Pause and explain
For B(4,1/3), what is P(X = 2)?
Worked example
X ~ B(4,1/3). Find P(X = 2).
Show the worked solution
- One two-success path has probability (1/3)²(2/3)² = 4/81.
- There are 4C2 = 6 placements of the success positions.
- Multiply 6 × 4/81 = 8/27.
Answer P(X = 2) = 8/27.
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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.
X ~ B(4,1/3). Find P(X = 2).
Give me a hint
Count success placements and multiply by one path probability.
Compare my reasoning
- The path factor is (1/3)²(2/3)² = 4/81.
- The order factor is 4C2 = 6.
- Their product is 24/81 = 8/27.
8/27.
Look for these in your work
- Include all placements.
- Use two success and two failure factors.
X ~ B(5,0.40). Find P(X = 0).
Give me a hint
There is only one all-failure sequence.
Compare my reasoning
- Each failure has probability 0.60.
- All five must fail, giving 0.60⁵.
- The result is 0.07776, exactly 243/3125.
0.07776 = 243/3125.
Look for these in your work
- Use the failure probability.
- Recognise the endpoint event has one path.
X ~ B(3,1/4). Find P(X = 2) and explain why the all-success probability is different.
Give me a hint
Exactly two successes has three placements; three successes has one.
Compare my reasoning
- P(X = 2) = 3C2(1/4)²(3/4) = 9/64.
- P(X = 3) = (1/4)³ = 1/64.
- The events differ in both their failure factor and number of permitted paths.
Exactly two: 9/64; all three: 1/64.
Look for these in your work
- Keep exact and all-success events distinct.
- Explain the appropriate order factor.
Common mistakes
- Include the combination factor for all placements.
- Match exponents to successes and failures.
- Treat probabilities outside the support as zero and endpoint models directly.
What can you explain now?
Why is 4/81 insufficient for exactly two successes in four trials?
Compare with the explanation
It counts one particular arrangement, not all six success-position choices.
The event includes every two-success path under the independent common-p model.
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Make it stick
Choose the success positions, multiply the common path probability, and check the support and endpoint cases.
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