Discrete Random Variables (AA SL)
A discrete random variable takes a finite (or countable) list of values, each with its own probability. Once you have that probability distribution, you can find its expected value - the long-run average outcome if the experiment were repeated many times. This page covers the setup and the \(E(X)\) formula, with worked examples and the mistakes that lose marks. It's part of the broader Random Variables & Distributions topic.
14 questions on this sub-topic.
The key formula
Covered under IB syllabus reference SL4.7: the concept of discrete random variables and their probability distributions, and the expected value (mean) for discrete data.
Expected value
\(E(X)=\sum xP(X=x)\)
This formula is in the formula booklet. Multiply each outcome by its own probability, then add every term together.
Valid distributions
\(\sum P(X=x) = 1\)
Every entry in a probability table must lie between 0 and 1, and the whole row must sum to exactly 1 - this is often how you find an unknown \(k\).
Need the full syllabus wording and formula-booklet reference table? See Random Variables & Distributions.
Worked examples
The distribution of \(X\) is given in the table.
| \(x\) | 1 | 2 | 3 |
|---|---|---|---|
| \(P(X=x)\) | 0.2 | 0.5 | \(k\) |
(a) Find \(k\).
(b) Find \(P(X\ge 2)\).
Worked solution
(a) Probabilities sum to 1: \(0.2+0.5+k=1\), so \(k=0.3\). M1
\(k=0.3.\) A1
(b) \(P(X\ge 2)=P(X=2)+P(X=3)=0.5+0.3=0.8.\) A1
A die game pays $5 for a six and costs $1 otherwise.
Find the expected gain per roll.
Worked solution
Roll a six (gain \(+5\)) with probability \(\tfrac16\); roll anything else (gain \(-1\)) with probability \(\tfrac56\). A1
Apply \(E(X)=\sum x\,P(x)\): \(E(X)=5\cdot\tfrac16+(-1)\cdot\tfrac56=\tfrac{5}{6}-\tfrac{5}{6}=0.\) M1A1
The expected gain is \($0\) per roll, so in the long run the game is fair. R1
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
\(= 1 + 0.8 + 0 = $1.80.\) A1
Common mistakes
- Forgetting the probabilities must sum to 1. If a table has an unknown \(k\), the very first step is almost always \(\sum P(X=x)=1\) - skipping this leaves \(k\) unfound and every later part wrong.
- Averaging the outcomes instead of weighting them. \(E(X)\) is not the plain mean of the \(x\)-values - each outcome must be multiplied by its own probability before adding.
- Missing negative payoffs in a game context. When a question involves a cost as well as a prize, the cost is a negative value of \(x\), not something to subtract at the end - it belongs inside the sum.
Ready to practise properly?
15 discrete random variable questions, marked instantly like the real exam.
Quick answers
How do you find the expected value of a discrete random variable?
Multiply each possible value \(x\) by its probability \(P(X=x)\), then add the results: \(E(X)=\sum xP(X=x)\).
How do you know a probability distribution table is valid?
All the probabilities must lie between 0 and 1, and the whole set of probabilities in the table must add up to exactly 1.