HL AA · Statistics and Probability
Updated October 2026
Quick answer: For a discrete random variable, the probabilities add to 1, E(X) = Σ x P(X = x) and Var(X) = E(X²) − [E(X)]², where E(X²) = Σ x² P(X = x). For a linear change, E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X).
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Practise HL AA Statistics and Probability →The random variable X has P(X = x) = kx for x = 1, 2, 3, 4. Find k, E(X) and Var(X), and hence E(2X + 3) and Var(2X + 3).
Answer: k = 1/10, E(X) = 3, Var(X) = 1, E(2X + 3) = 9, Var(2X + 3) = 4.
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5 quick questions on discrete random variables. No sign-in needed.
| \(x\) | 10 | 20 | 50 |
|---|---|---|---|
| \(P(X = x)\) | \(\frac{7}{10}\) | \(\frac{1}{5}\) | \(\frac{1}{10}\) |
| \(x\) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| \(P(X = x)\) | \(\frac{1}{10}\) | \(\frac{1}{5}\) | \(\frac{3}{10}\) | \(\frac{2}{5}\) |
| \(x\) | 0 | 1 | 2 |
|---|---|---|---|
| \(P(X = x)\) | \(\frac{1}{4}\) | \(\frac{1}{2}\) | \(\frac{1}{4}\) |
| \(x\) | 2 | 5 | 10 |
|---|---|---|---|
| \(P(X = x)\) | \(\frac{1}{2}\) | \(\frac{3}{10}\) | \(\frac{1}{5}\) |
| \(x\) | -1 | 0 | 3 |
|---|---|---|---|
| \(P(X = x)\) | \(\frac{2}{5}\) | \(\frac{2}{5}\) | \(\frac{1}{5}\) |
Stretch yourself: original probability problems at Senior level (ages 16 to 18), with hints and full solutions: Problem S33 · Problem S106 · Problem S34 · Problem S107. All 20 →
Find E(X) = Σ x P(X = x) and E(X²) = Σ x² P(X = x), then Var(X) = E(X²) − [E(X)]². Check that the variance is not negative: if it is, there is an arithmetic slip.
On the Advanced Random Variables questions page: 52 exam-style HL AA questions with full mark schemes, plus free step-by-step worked examples.
In HL AA, Paper 1 is non-calculator and the other papers allow a GDC, so learn the exact method as well as the calculator one.
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