Discrete random variables and distributions (SACE Stage 2 Mathematical Methods)

Syllabus dot point

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

7 min readUpdated 2 Jun 2026

Quick answer

A discrete random variable $X$ takes a countable set of numerical values, each with a probability. Its probability distribution is the table or rule pairing every value $x$ with $P(X=x)$ . To be valid, all probabilities must satisfy $0\le P(X=x)\le 1$ and they must sum to exactly $1$ .

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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
The cumulative distribution
Common errors
Distributions given by a rule
Why it matters
Exam-style questions

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.

Discrete random variable

A discrete random variable $X$ is a variable whose possible values form a countable set $\{x_1, x_2, \dots\}$ , where each value $x_i$ occurs with a known probability $P(X=x_i)$ .

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:

Conditions for a valid distribution

0\le P(X=x)\le 1 \quad\text{for every } x, \qquad \sum_{\text{all }x} P(X=x)=1.

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

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 $P(X\ge 3)$ .

Step 1: Write the sum-to-one equation

Every valid probability distribution must have all its probabilities add to exactly 1. That rule gives us an equation to solve for the unknown $k$ .

0.1 + 2k + 0.3 + k = 1.

Step 2: Solve for $k$

Collect the constant terms on the left and the $k$ terms together.

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

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

Step 3: Find $P(X \ge 3)$

The event $X \ge 3$ means $X = 3$ or $X = 4$ , so add those two probabilities.

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

Final answer: $k = 0.2$ and $P(X \ge 3) = 0.5$ .

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.

Use the complement for "at least" questions

For "at least one" or "at least two" style questions, it is often faster to compute $1-P(\text{none})$ or $1-P(\text{fewer than})$ than to add many rows. This also reduces arithmetic slips.

The cumulative distribution

Beyond individual probabilities, you can build a running total $P(X \le x)$ , called the cumulative distribution. For the three-toss coin example, the cumulative probabilities are $P(X\le 0)=\tfrac18$ , $P(X\le 1)=\tfrac48$ , $P(X\le 2)=\tfrac78$ and $P(X\le 3)=1$ . Cumulative values make "less than or equal to" questions immediate and let you find any interval probability by subtraction: $P(1 < X \le 3)=P(X\le 3)-P(X\le 1)=1-\tfrac48=\tfrac12$ . Examiners use cumulative phrasing ("no more than two") precisely to test whether you include or exclude the boundary value.

Common errors

Probabilities that do not sum to 1, or muddled inequalities

Two errors recur. First, building a distribution whose probabilities do not total exactly 1 - always check the sum before answering. Second, confusing strict and inclusive inequalities: $P(X>2)$ excludes $x=2$ whereas $P(X\ge 2)$ includes it. Write out the values you are summing rather than working in your head.

Distributions given by a rule

A distribution need not be a table; it can be a formula. If $P(X=x)=\dfrac{x}{10}$ for $x=1,2,3,4$ , you generate the probabilities by substitution: $0.1,0.2,0.3,0.4$ . Always confirm the rule produces a valid distribution by checking each value lies in $[0,1]$ and the total is exactly $1$ . Rule-based distributions are common because they let an examiner embed an unknown constant - for example $P(X=x)=kx$ - which you solve for using $\sum P(X=x)=1$ .

Solving for a constant in a rule

The random variable $X$ has $P(X=x)=kx^2$ for $x=1,2,3$ . Find $k$ and then $P(X=3)$ .

Step 1: Generate the probabilities from the rule

Substitute each value of $x$ into the formula $P(X=x) = kx^2$ to see what each probability looks like in terms of $k$ .

$P(X=1) = k(1)^2 = k$
$P(X=2) = k(2)^2 = 4k$
$P(X=3) = k(3)^2 = 9k$

Step 2: Apply the sum-to-one condition

Because all probabilities must sum to 1, add the three expressions and set the total equal to 1.

k + 4k + 9k = 1 \quad\Rightarrow\quad 14k = 1 \quad\Rightarrow\quad k = \frac{1}{14}.

Step 3: Find $P(X=3)$

Now that $k$ is known, substitute back into the expression for $P(X=3)$ .

P(X=3) = 9k = \frac{9}{14} \approx 0.643.

Final answer: $k = \dfrac{1}{14}$ and $P(X=3) = \dfrac{9}{14} \approx 0.643$ .

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, where the sum becomes an integral.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style4 marks

Calculator-assumed. A 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 $E(X) = \sum x\,P(X=x) = (-1)(a) + (0)(b) + (2)(0.1) = -a + 0.2$ . Setting $E(X) = 0$ gives $-a + 0.2 = 0$ , i.e. $a = 0.2$ . (1 mark)

So $a = 0.2$ and $b = 0.7$ . The two ideas tested are that probabilities sum to 1 and the definition of expected value.

SACE 2022-style2 marks

Calculator-assumed. In 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 $E(X) = \sum x\,P(X=x)$ 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 the sum of value times probability, one for the arithmetic. A quick check that $0.5 + 0.2 + 0.2 + 0.1 = 1$ confirms the distribution is valid before computing the mean.

SACE 2021-style3 marks

Calculator-assumed. The distribution of

X

P(X=x)= \dfrac{x}{10}

for

x = 1, 2, 3, 4

. (a) Verify this is a valid probability distribution. (b) Find

P(X \ge 3)

Show worked answer →

(a) Compute each probability: $P(1) = 0.1$ , $P(2) = 0.2$ , $P(3) = 0.3$ , $P(4) = 0.4$ . Each lies between $0$ and $1$ , and the sum is $0.1 + 0.2 + 0.3 + 0.4 = 1$ , so it is valid. (2 marks)

(b) $P(X \ge 3) = P(3) + P(4) = 0.3 + 0.4 = 0.7$ . (1 mark)

Marks reward checking both validity conditions and reading the inclusive inequality $X \ge 3$ correctly.

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board

What this dot point is asking

The probability distribution

The two rules of a valid distribution

Step 1: Write the sum-to-one equation

Step 2: Solve for kkk

Step 3: Find P(X≥3)P(X \ge 3)P(X≥3)

Reading probabilities from the table

The cumulative distribution

Common errors

Distributions given by a rule

Step 1: Generate the probabilities from the rule

Step 2: Apply the sum-to-one condition

Step 3: Find P(X=3)P(X=3)P(X=3)

Why it matters

Exam-style questions

Practise this

Sources & how we know this

Step 2: Solve for $k$

Step 3: Find $P(X \ge 3)$

Step 3: Find $P(X=3)$