SACE Board SACE 2021-style practice: Calculator-assumed. The distribution of X is P(X=x)= x10 for x = 1, 2, 3, 4. (a) Verify this is a valid probability distribution. (b) Find P(X 3).

(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 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 3 correctly.

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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 4.87 min answerUpdated 2026-06-02

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
The cumulative distribution
Common errors
Distributions given by a rule
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

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.

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

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 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.

SACE 20234 marksCalculator-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 20222 marksCalculator-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 20213 marksCalculator-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.