Discrete Random Variables (AA HL)
A discrete random variable \(X\) takes a countable set of values, each with its own probability. This page covers the two calculations examiners ask for most - finding a missing probability and finding the expected value \(E(X)\) - plus how expectation and variance behave under a linear transformation. It's part of the broader Probability Distributions topic.
12 questions on this sub-topic.
Expectation and variance
Covered under IB syllabus reference SL4.7: the concept of discrete random variables and their probability distributions, and \(E(X)\) as a measure of the average long-run outcome. Both formulas below are in the formula booklet.
Expected value
\(E(X)=\sum x\,P(X=x)\)
Multiply each value by its probability and add the results. \(E(X)\) doesn't have to equal any value \(X\) can actually take.
Variance
\(\text{Var}(X)=E(X^2)-[E(X)]^2\)
Find \(E(X^2)=\sum x^2 P(X=x)\) first, then subtract the square of \(E(X)\).
Need the full syllabus wording and formula-booklet reference table? See Probability Distributions.
Worked examples
A discrete random variable \(X\) takes the values shown in the table.
| \(x\) | 1 | 2 | 3 |
|---|---|---|---|
| \(P(X=x)\) | 0.2 | 0.5 | ? |
Find \(P(3)\).
Worked solution
Probabilities sum to 1: \(P(3) = 1 - 0.2 - 0.5\) M1
\(= 0.3.\) A1
A spinner has the payouts and probabilities shown in the table.
| Payout | $10 | $2 | $0 |
|---|---|---|---|
| Probability | 0.1 | 0.4 | 0.5 |
Find the expected payout.
Worked solution
\(P($0) = 1 - 0.1 - 0.4\) M1
\(= 0.5.\) A1
\(E = 10(0.1) + 2(0.4) + 0(0.5)\) M1
\(= 1 + 0.8\) A1
\(= $1.80.\) A1
A discrete variable has \(P(X=x) = k(x+1)\) for \(x = 0, 1, 2, 3.\)
| \(x\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| \(P(X=x)\) | \(k\) | \(2k\) | \(3k\) | \(4k\) |
(a) Find \(k\).
(b) Find \(E(X)\).
Worked solution
(a) \(\sum P = k(1+2+3+4) = 10k = 1 \Rightarrow k\) M1
\(= 0.1.\) A1
(b) \(E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4)\) M1
\(= 2.\) A1
Common mistakes
- Forgetting probabilities must sum to 1. This is the standard way to find an unknown constant \(k\) or a missing probability in a discrete distribution - if the values you've written down don't sum to 1, something upstream is wrong.
- Treating \(E(X)\) as the most likely outcome. \(E(X)\) is a long-run average, not the value with the highest probability - it can easily land between two values \(X\) never actually takes.
- Applying \(\text{Var}(aX+b)=a^2\text{Var}(X)\) incorrectly. A shift by \(b\) changes \(E(X)\) but has no effect on the spread, so \(b\) drops out of the variance formula entirely - only the scale factor \(a\) survives, and it gets squared.
Ready to practise properly?
12 discrete-random-variable questions, marked instantly like the real exam.
Quick answers
How do you find a missing probability in a discrete distribution?
Use the fact that all the probabilities in a discrete probability distribution must sum to 1, then solve for the unknown value.
What is the expected value of a discrete random variable?
\(E(X) = \sum x\,P(X=x)\), summed over every value \(X\) can take. It is the long-run average outcome, not necessarily a value \(X\) actually takes. GDC guidance for computing it is covered on the Probability Distributions page.