SAMath MethodsSyllabus dot point

How do we describe the possible outcomes of a chance experiment and their probabilities?

A discrete random variable assigns a numerical value to each outcome, and its probability distribution lists every value together with its probability.

What a discrete random variable is, how to build and read a probability distribution table, and the two rules every valid distribution must satisfy.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The probability distribution
The two rules of a valid distribution
Reading probabilities from the table
Common errors
Why it matters

What this dot point is asking

A random variable is a rule that attaches a number to each outcome of a chance experiment. It is discrete when its possible values can be listed - typically counts such as the number of heads in three tosses, or the number of defective items in a sample.

The probability distribution

The probability distribution of $X$ lists each value alongside its probability. For tossing a fair coin three times and counting heads, $X\in\{0,1,2,3\}$ :

$x$	0	1	2	3
$P(X=x)$	$\tfrac18$	$\tfrac38$	$\tfrac38$	$\tfrac18$

The two rules of a valid distribution

Every probability distribution must satisfy two conditions:

The second rule is the workhorse of exam questions: you are often given a distribution containing an unknown and must use "the probabilities sum to 1" to solve for it.

Finding an unknown probability

Worked example

The random variable $X$ has the distribution below, where $k$ is a constant.

$x$	1	2	3	4
$P(X=x)$	$0.1$	$2k$	$0.3$	$k$

Find $k$ , then $P(X\ge 3)$ .

The probabilities must sum to 1:

0.1+2k+0.3+k=1 \quad\Rightarrow\quad 0.4+3k=1 \quad\Rightarrow\quad 3k=0.6 \quad\Rightarrow\quad k=0.2.

So $P(X=2)=0.4$ and $P(X=4)=0.2$ . Then

P(X\ge 3)=P(X=3)+P(X=4)=0.3+0.2=0.5.

Reading probabilities from the table

Compound events are found by adding the relevant rows. Watch the inequality carefully:

$P(X>2)=P(X=3)+P(X=4)$ (strictly greater, so $2$ is excluded).
$P(X\ge 2)=P(X=2)+P(X=3)+P(X=4)$ ( $2$ included).
The complement: $P(X\ge 2)=1-P(X=1)$ can be quicker.

Common errors

Why it matters

Discrete random variables are the foundation for expected value, variance, and the binomial distribution that follow in this topic. The "sum to 1" technique and careful inequality reading reappear throughout the probability strand and into Topic 5's continuous distributions.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2017 SACE Stage 24 marksA discrete random variable X takes values -1, 0 and 2 with probabilities a, b and 0.1. (a) Explain why a + b = 0.9. (b) Given E(X) = 0, write a second equation for a and b. (c) Find a and b.

Show worked answer →

(a) Every valid probability distribution must have probabilities that sum to 1. Here a + b + 0.1 = 1, so a + b = 0.9. (1 mark)

(b) The expected value is E(X) = sum of (x times P(X = x)) = (-1)(a) + (0)(b) + (2)(0.1) = -a + 0.2. Setting E(X) = 0 gives the second equation -a + 0.2 = 0, i.e. a = 0.2. (1 mark)

So a = 0.2 and b = 0.7. The key facts being tested are the two defining properties of a discrete distribution: probabilities sum to 1, and the definition of expected value.

2018 SACE Stage 22 marksIn a game, the number X printed on a randomly selected duck takes the values 1, 2, 5 and 10 with probabilities 0.5, 0.2, 0.2 and 0.1 respectively. Calculate the expected value E(X).

Show worked answer →

Use E(X) = sum of (x times P(X = x)) across all four values:

E(X) = (1)(0.5) + (2)(0.2) + (5)(0.2) + (10)(0.1)
= 0.5 + 0.4 + 1.0 + 1.0
= 2.9.

So E(X) = 2.9. Marks: one for setting up the sum of value times probability, one for the correct arithmetic. A quick validity check first - the probabilities 0.5 + 0.2 + 0.2 + 0.1 = 1 - confirms this is a proper distribution before computing the mean.