Infer table parameters without overclaiming a unique distribution
Combine moments with normalisation and recognise when a supplied mean leaves several valid tables.
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 exact table moments.
- Solve linear equations and probability bounds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is E(X) for values 0, 2 with probabilities 1 − p, p?
What you will learn
- Infer probabilities from an expectation or variance condition.
- Check whether the given information identifies one distribution or a family.
A mean condition supplies an equation
For values 0, 1, 2 with probabilities k, 2k, 1 − 3k, the mean is 2k + 2(1 − 3k) = 2 − 4k. If it equals 1, then k = 1/4, giving probabilities 1/4, 1/2, 1/4. Their second moment is 3/2 and variance 1/2. Verify the inferred k satisfies 0 ≤ k ≤ 1/3.
For values 0 and 2 with probabilities 1 − p and p, E(X) = 2p. Mean 1.20 gives p = 0.60. The second moment is 4p = 2.40, so variance 2.40 − 1.44 = 0.96. The variable is not itself a success count with values 0 and 1; use its supplied numerical values.
A shared mean does not fix spread
For values 0, 1, 2 with probabilities p, 1 − 2p, p, every valid p in [0,1/2] gives mean 1. The second moment is 1 + 2p, so variance is 2p. Knowing only the mean 1 leaves a family of distributions rather than a unique p.
At p = 0.10 the variance is 0.20; at p = 0.40 it is 0.80, despite the same mean. A supplied variance 0.60 would identify p = 0.30. Count the independent equations and check admissible bounds before claiming uniqueness. An extra moment can constrain a table without proving a particular binomial or geometric model.
Pause and explain
A symmetric table p, 1 − 2p, p on 0, 1, 2 has mean 1. Does the mean determine p?
Worked example
X takes 0, 1, 2 with probabilities p, 1 − 2p, p and variance 0.60. Find p and the mean.
Show the worked solution
- The mean is 1 for every admissible p.
- E(X²) = 1 + 2p, so variance = 2p.
- 2p = 0.60 gives p = 0.30, within [0,1/2].
Answer p = 0.30; mean 1.
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.
X takes 0, 1, 2 with probabilities k, 2k, 1 − 3k. Its mean is 1. Find k and variance.
Give me a hint
Write the weighted mean, solve for k and then calculate the second moment.
Compare my reasoning
- Mean 2 − 4k = 1 gives k = 1/4.
- The table is 1/4, 1/2, 1/4 with second moment 3/2.
- Variance is 3/2 − 1 = 1/2; all probabilities are valid.
k = 1/4; variance 1/2.
Look for these in your work
- Use the moment condition as an equation.
- Verify the inferred table before calculating spread.
X takes 0, 1, 2 with probabilities p, 1 − 2p, p. Show that mean 1 does not determine p; compare p = 0.10 and 0.40.
Give me a hint
Calculate the mean and variance as functions of p.
Compare my reasoning
- E(X) = 1 − 2p + 2p = 1 for every p in [0,1/2].
- E(X²) = 1 + 2p, so variance = 2p.
- p = 0.10 gives variance 0.20; p = 0.40 gives 0.80, both with mean 1.
p is not unique; variances 0.20 and 0.80 share mean 1.
Look for these in your work
- State the valid parameter range.
- Provide two distinct admissible examples.
X takes 0 or 2 points with probabilities 1 − p and p. Its mean is 1.20 points. Find p and variance.
Give me a hint
Use the value two, rather than treating X as a zero–one indicator.
Compare my reasoning
- 2p = 1.20 gives p = 0.60.
- E(X²) = 4p = 2.40.
- Variance = 2.40 − 1.20² = 0.96 points squared.
p = 0.60; variance 0.96 points squared.
Look for these in your work
- Use the stated numerical support.
- Check probability bounds and squared units.
Common mistakes
- A mean may leave several admissible tables.
- Moment equations do not automatically establish a named distribution.
- Check the inferred probabilities against their bounds.
What can you explain now?
Can equal means accompany different variances?
Compare with the explanation
Yes; p = 0.10 and 0.40 in the symmetric table give variance 0.20 and 0.80.
The mean fixes centre here while the free parameter changes spread.
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Make it stick
Turn supplied moments into equations, check bounds, and distinguish a uniquely identified table from a valid family.
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.