The distribution of X is given in the table.x1234P(X=x)0.20.30.4p(a) Find p. (b) Find E(X).

(a) 0.2+0.3+0.4+p = 1, so p = 0.1 (b) E(X) = 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4

Discrete Random Variables (AI HL)

Q: What is the formula for the expected value of a discrete random variable?

E(X) = the sum of x times P(X=x) over every possible value of x.

Q: What does E(X) = 0 mean in a game?

If X is a player's gain and E(X) = 0, the game is fair - in the long run a player expects to neither win nor lose money.

A discrete random variable takes a finite (or countable) set of values, each with its own probability, usually laid out as a table or a formula. This page covers how to find a missing probability, how to calculate the expected value $E(X)$, and what it means when $E(X)=0$ in a game - with worked examples and the mistakes that lose the most marks. It's part of the broader Probability & Distributions topic.

12 questions on this sub-topic.

Practise discrete random variables → Try exam-style questions

The two formulas

Covered under IB syllabus reference SL4.7: the concept of discrete random variables and their probability distributions, and expected value $E(X)$ for discrete data - including that $E(X)=0$ indicates a fair game where $X$ is a player's gain. Neither formula is in the formula booklet.

Expected value

$E(X) = \Sigma xP(X=x)$

Multiply each value of $X$ by its probability, then add. This is the long-run average of $X$.

Variance

$\text{Var}(X) = E(X^2)-[E(X)]^2$

Find $E(X^2) = \Sigma x^2P(X=x)$ first, then subtract the square of $E(X)$.

Need the full syllabus wording and formula-booklet reference table? See Probability & Distributions. For calculator routines, see the parent topic's GDC guidance.

Worked examples

Easy

No calc

[3 marks]

The distribution of $X$ is given in the table.

$x$	1	2	3	4
$P(X=x)$	0.2	0.3	0.4	$p$

(a) Find $p$.
(b) Find $E(X)$.

Worked solution

(a) 0.2+0.3+0.4+p = 1, so p = 0.1 A1

(b) E(X) = 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) = 0.2 + 0.6 + 1.2 + 0.4 M1
= 2.4 A1

A1 P-value M1 E(X) formula A1 Answer

Medium

GDC

[4 marks]

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
$E(X)=5\cdot\tfrac16+(-1)\cdot\tfrac56=\tfrac{5}{6}-\tfrac{5}{6}=0.$ M1 A1
The expected gain is $$0$ per roll, so in the long run the game is fair. R1

A1 Payoffs and probabilities M1 Apply $E(X)=\sum x\,P(x):$ A1 Expectation set-up and value R1 Interpretation

Common mistakes

Using $P(X\le k)$ when the question means $P(X For a discrete variable, "fewer than 3" is \(X\le2$, not $X\le3$ - always translate the wording into an inequality before reaching for binomcdf.

Forgetting the probabilities must sum to 1. Whenever a table has an unknown $p$, the very first line should be an equation setting the sum of all the probabilities equal to $1$ - it's easy to jump straight to $E(X)$ and skip this check.

Missing the sign of a loss in $E(X)$. In a game, a cost or loss must be entered as a negative value of $x$ - writing it as positive by mistake flips the sign of the whole expected value and can turn a fair game into a winning (or losing) one on paper.

Full topicProbability & Distributions RelatedPoisson Distribution RelatedInverse Normal and Parameters FormulasAI HL formula booklet explained PrintAI HL formula sheet

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Quick answers

What is the formula for the expected value of a discrete random variable?

$E(X) = \Sigma xP(X=x)$ - multiply each value of $X$ by its probability and add the results.

What does E(X) = 0 mean in a game?

If $X$ is a player's gain and $E(X) = 0$, the game is fair - in the long run a player expects to neither win nor lose money.

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\(x\)	1	2	3	4
\(P(X=x)\)	0.2	0.3	0.4	\(p\)