Unit 3 · Topic 5.1 · Discrete random variables

General discrete random variables

This topic used to be Year 11 work. In the 2025 syllabus it moved into Unit 3, so from 2026 you can expect it on the external exam for the first time. There are only a handful of real QCAA questions on it, so most of the questions below are ones I have written in the style I expect the QCAA to use. Tick off each skill once you can do it without help.

Skills in this topic

General discrete random variables, about 5 hours of class time. Only tick a skill when you could do it in an exam.

Part 1

Discrete probability distributions

A random variable is a quantity whose value you cannot predict, because it depends on the outcome of an experiment. We name it with a capital letter, like $X$, and write the values it takes in lower case, so $P(X=x)$ means "the probability that $X$ takes the value $x$". It is discrete when you can count the possible outcomes, such as the number of heads or the number of books borrowed.

A probability distribution lists every outcome with its probability, usually in a table. Here is one for the number of heads when you toss three coins:

$x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Every discrete probability distribution must satisfy both of these

Each probability is between 0 and 1: $0\le P(X=x)\le1$.
The probabilities add to 1: $\sum P(X=x)=1$.

A distribution is uniform when every outcome has the same probability, like a fair die. Otherwise it is non-uniform, like the coin table above. You can show either kind as a table, a bar chart or a histogram.

Practise

Questions: properties of a distribution

Teacher-written question

Short response · Technology-free · [4 marks]

Two tables are shown. Each is proposed as the probability distribution of a discrete random variable.

Table A

$x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{4}$	$\frac{3}{8}$	$\frac{1}{4}$	$\frac{1}{8}$

Table B

$y$	$0$	$1$	$2$	$3$
$P(Y=y)$	$0.3$	$0.45$	$0.35$	$-0.1$

Determine whether each table represents a discrete probability distribution. Justify your decisions.

Check both rules for each table. You need to state both to get full marks.

Table A: every probability is between 0 and 1, and $\frac{2}{8}+\frac{3}{8}+\frac{2}{8}+\frac{1}{8}=1$. Both rules hold.

Table B: the probabilities do add to 1, which is the trap. But $P(Y=3)=-0.1$, and a probability cannot be negative.

Answer

Table A is a discrete probability distribution. Table B is not.

Teacher-written question

Short response · Technology-free · [5 marks]

The discrete random variable $X$ has the probability function

$$P(X=x)=k(4-x),\qquad x=0,1,2,3$$

(a) Construct a discrete probability distribution table for $P(X=x)$.[1 mark]

(b) Determine the value of $k$.[2 marks]

(a) Substitute each value of $x$ into the function:

$x$	$0$	$1$	$2$	$3$
$P(X=x)$	$4k$	$3k$	$2k$	$k$

(b) The probabilities must add to 1: $4k+3k+2k+k=10k=1$, so $k=0.1$.

Answer

(b) $k=0.1$ · (c) $P(X\ge2)=0.3$

Teacher-written question

Short response · Technology-free · [4 marks]

The discrete random variable $X$ has the probability distribution shown, where $k$ is a constant.

$x$	$1$	$2$	$3$
$P(X=x)$	$k$	$k^{2}$	$5k^{2}$

Determine $P(X>1)$. Evaluate the reasonableness of your solution.

Use the sum rule to find $k$. It gives you a quadratic:

$$k+k^{2}+5k^{2}=1\ \Rightarrow\ 6k^{2}+k-1=0\ \Rightarrow\ (3k-1)(2k+1)=0$$

So $k=\frac{1}{3}$ or $k=-\frac{1}{2}$. Reject $k=-\frac{1}{2}$, because it would make $P(X=1)$ negative.

Then $P(X>1)=P(X=2)+P(X=3)=6k^{2}=\frac{6}{9}=\frac{2}{3}$.

To evaluate reasonableness, check the whole distribution with $k=\frac{1}{3}$: the probabilities are $\frac{1}{3}$, $\frac{1}{9}$ and $\frac{5}{9}$. Each is between 0 and 1, and they add to 1, so the answer is reasonable.

Answer

$P(X>1)=\frac{2}{3}$

Part 2

Building a distribution from data

Sometimes you work out probabilities from the situation, like a fair die. Other times you only have data, and you use relative frequencies instead. Divide each frequency by the total number of trials. The result is a point estimate of the probability:

$$P(X=x)\approx\frac{\text{frequency of }x}{\text{total number of trials}}$$

Data never matches a model perfectly. If a die is fair, 120 rolls will not give exactly 20 of each number. When you judge whether a model is reasonable, look at whether the differences are small enough to be chance.

Watch the estimates settle

Practise

Questions: building a distribution

Teacher-written question

Short response · Technology-free · [4 marks]

In a board game, players roll a fair eight-sided die with faces numbered 1 to 8. Let $X$ represent the number rolled.

(a) Construct a table showing the probability distribution of $X$.[2 marks]

(b) Explain why $X$ is a uniform discrete random variable.[1 mark]

(a) The die is fair, so each of the 8 faces has probability $\frac{1}{8}$:

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X=x)$	$\frac{1}{8}$	$\frac{1}{8}$	$\frac{1}{8}$	$\frac{1}{8}$	$\frac{1}{8}$	$\frac{1}{8}$	$\frac{1}{8}$	$\frac{1}{8}$

(b) Every outcome has the same probability, $\frac{1}{8}$, which is what makes it uniform.

Answer

Teacher-written question

Short response · Technology-active · [4 marks]

A games café suspects a six-sided die used at one of its tables is damaged. The die was rolled 120 times, and the results are displayed below.

(a) Use relative frequencies to estimate the probability of rolling a five with this die.[2 marks]

(b) The café's manager makes the following statement: The results show the die is not reasonably modelled by a uniform discrete random variable. Evaluate the reasonableness of this statement.[2 marks]

(a) Five was rolled 18 times out of 120: $P(X=5)\approx\frac{18}{120}=0.15$.

(b) Compare the data with what a uniform model predicts. If the die were fair, you would expect about $\frac{120}{6}=20$ of each number, a probability of $\frac{1}{6}\approx0.167$.

The frequencies are all between 18 and 22, so the relative frequencies are all between 0.15 and 0.18. They are all close to $\frac{1}{6}$, and differences this small are what you would expect by chance in 120 rolls.

Answer

(a) $0.15$ · (b) The statement is not reasonable. The data is consistent with a uniform model.

Teacher-written question

Short response · Technology-free · [5 marks]

A librarian recorded the number of books borrowed by each of 50 visitors to a library. The results are displayed below.

Let $X$ represent the number of books borrowed by a randomly selected visitor.

(a) Use the data to construct a table of point estimates for the probability distribution of $X$. Express the probabilities as fully simplified fractions.

(b) Determine a point estimate of the probability that a randomly selected visitor borrows at least 3 books.

(a) Read the frequencies off the graph (9, 15, 12, 9, 5) and check they add to 50. Then divide each by 50 and simplify:

$x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{9}{50}$	$\frac{3}{10}$	$\frac{6}{25}$	$\frac{9}{50}$	$\frac{1}{10}$

(b) "At least 3" means 3 or 4: $\frac{9}{50}+\frac{5}{50}=\frac{14}{50}=\frac{7}{25}$.

Answer

(b) $P(X\ge3)\approx\frac{7}{25}$

Part 3

Expected value, variance and standard deviation

For a random variable, the mean is called the expected value, $E(X)$. Multiply each value of $X$ by its probability, then add them all up. The variance measures spread, and the standard deviation is its square root.

Expected value (on your formula sheet)

$$E(X)=\mu=\sum p_{i}x_{i}$$

Variance (the syllabus definition)

$$\text{Var}(X)=\sum p_{i}\left(x_{i}-\mu\right)^{2}$$

Variance (the quicker way, not on the formula sheet)

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

Standard deviation

$$\text{SD}(X)=\sigma=\sqrt{\text{Var}(X)}$$

Both variance formulas give the same answer. I use the quicker one. To find $E\left(X^{2}\right)$, square the $x$-values first, multiply each by its probability, then add. Adding $x^{2}$ and $p\times x$ rows to the table keeps your working tidy and easy to mark.

Watch the variance get built

Try it yourself

The expected value is the balance point

Picture the bars sitting on a beam. The expected value is where you would put the support so the beam balances. Change the probabilities and watch it move. The readouts only appear when your table is a valid distribution.

Practise

Questions: expected value, variance and standard deviation

QCAA sample Unit 1 and 2 assessment (2019), Question 5

Short response · Technology-free · [5 marks]

The following table shows the probability distribution of a discrete random variable $X$.

$x$	$0$	$2$	$5$	$9$
$P(X=x)$	$0.3$	$a$	$2a$	$0.1$

(a) Determine the value of $a$.

(b) Determine $E(X)$.

(a) The probabilities add to 1: $0.3+a+2a+0.1=1$, so $3a=0.6$ and $a=0.2$.

(b) Multiply each value by its probability and add:

$$E(X)=0(0.3)+2(0.2)+5(0.4)+9(0.1)=0+0.4+2+0.9=3.3$$

Answer

(a) $a=0.2$ · (b) $E(X)=3.3$

Teacher-written question

Short response · Technology-free · [5 marks]

The table shows the probability distribution of a discrete random variable $X$.

$x$	$1$	$3$	$5$	$7$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{6}$

(a) Determine $E(X)$.[2 marks]

(b) Determine the standard deviation of $X$. Give your answer in simplest form.[3 marks]

(a) $E(X)=\frac{1}{6}+\frac{3}{3}+\frac{5}{3}+\frac{7}{6}=\frac{1+6+10+7}{6}=4$. The table is symmetric about 4, so that is a good check.

(b) Square the $x$-values first, then find $E\left(X^{2}\right)$:

$$E\left(X^{2}\right)=\frac{1}{6}+\frac{9}{3}+\frac{25}{3}+\frac{49}{6}=\frac{1+18+50+49}{6}=\frac{59}{3}$$

Then $\text{Var}(X)=\frac{59}{3}-4^{2}=\frac{11}{3}$. For simplest form, rationalise the denominator of the square root: $\sqrt{\frac{11}{3}}=\frac{\sqrt{33}}{3}$.

Answer

(a) $E(X)=4$ · (b) $\text{SD}(X)=\dfrac{\sqrt{33}}{3}$

QCAA sample Unit 1 and 2 assessment (2019), Question 3

Short response · Technology-active · [4 marks]

Consider the following income statistics for businesses in a particular suburb.

Year's earnings	Percentage
$40 000 loss	10%
$0	35%
$40 000 profit	45%
$200 000 profit	10%

Determine the expected value and the standard deviation.

Turn the table into a distribution first. A loss is a negative value, and the percentages become probabilities: $x=-40000,\ 0,\ 40000,\ 200000$ with $p=0.1,\ 0.35,\ 0.45,\ 0.1$.

$$E(X)=-4000+0+18000+20000=34000$$

This is Paper 2, so let your calculator's statistics mode do the rest, or work it by hand:

$$E\left(X^{2}\right)=0.1(40000)^{2}+0.45(40000)^{2}+0.1(200000)^{2}=4.88\times10^{9}$$

Then $\text{Var}(X)=4.88\times10^{9}-34000^{2}=3.724\times10^{9}$, and $\sigma=\sqrt{3.724\times10^{9}}\approx61025$.

Answer

Expected value $34 000, standard deviation about $61 025

Teacher-written question

Short response · Technology-active · [4 marks]

The daily profit ($x$) of a weekend market stall depends on weather and foot traffic. Based on past records, the stall's daily profit is modelled by the following distribution, $X$.

$x$	$-150$	$0$	$250$	$600$
$P(X=x)$	$0.10$	$0.2$	$0.5$	$0.2$

Determine the expected value and standard deviation of the stall's daily profit.

$$E(X)=-150(0.1)+0(0.2)+250(0.5)+600(0.2)=-15+125+120=230$$

$$E\left(X^{2}\right)=22500(0.1)+62500(0.5)+360000(0.2)=105500$$

Then $\text{Var}(X)=105500-230^{2}=52600$ and $\sigma=\sqrt{52600}\approx229.35$. Squaring $-150$ gives a positive number, so the loss still adds to the spread.

Answer

Expected profit $230, standard deviation about $229.35

Part 4

Modelling with discrete random variables

In modelling questions you usually have to build the table yourself from the words. Decide what $X$ stands for and write it down. Then the expected value tells you what happens on average over many repeats, such as a long-run gain per game or the average number of extras per order.

Practise

Questions: modelling

Teacher-written question

Short response · Technology-free · [5 marks]

At a school fete, a sideshow game costs $5 to play. The player spins a wheel that awards a prize of $0 with probability $\frac{5}{8}$, $4 with probability $\frac{1}{4}$ and $20 with probability $\frac{1}{8}$.

The game's designer makes the following claim: On average, a player will lose less than $1 per game.

Evaluate the reasonableness of the claim.

Let $X$ be the player's net gain in dollars, which is the prize minus the $5 cost. That gives the distribution:

$x$	$-5$	$-1$	$15$
$P(X=x)$	$\frac{5}{8}$	$\frac{1}{4}$	$\frac{1}{8}$

$$E(X)=-\frac{25}{8}-\frac{2}{8}+\frac{15}{8}=-\frac{12}{8}=-1.5$$

On average a player loses $1.50 per game, which is more than $1. A quick check: the expected prize is $\frac{1}{4}(4)+\frac{1}{8}(20)=3.5$, and $3.5-5=-1.5$.

Answer

The claim is not reasonable. The average loss is $1.50 per game.

Teacher-written question

Short response · Technology-active · [5 marks]

A café tracks the number of extra espresso shots added to each coffee order. In a random sample of orders, 45% had no extra shots, 35% had one, 15% had two and 5% had three. Each extra shot is charged at $0.80.

The café expects to sell 250 coffees next Saturday.

Determine the revenue the café should expect from extra espresso shots next Saturday.

Let $X$ be the number of extra shots in an order, with $x=0,1,2,3$ and probabilities $0.45,\ 0.35,\ 0.15,\ 0.05$. They add to 1, so it is a valid distribution.

$$E(X)=0+0.35+2(0.15)+3(0.05)=0.8$$

So an average order has 0.8 extra shots, worth $0.8\times0.80=0.64$ dollars. Over 250 coffees that is $250\times0.64=160$.

Answer

The café should expect about $160 from extra shots.

Game plan

Your game plan for discrete random variables

1Are there more than two countable outcomes, in a table, a graph or the words? Then it is a discrete random variable. Write down what $X$ stands for.
2Asked to show it is a probability distribution? Check both rules: every probability is between 0 and 1, and they add to 1.
3Got an unknown in the table? Set the sum of the probabilities equal to 1 and solve. Reject any answer that makes a probability negative.
4Working from data? Divide each frequency by the total to get point estimates.
5Expected value: multiply each value by its probability and add.
6Variance: square the $x$-values, find $E\left(X^{2}\right)$ the same way, then use $\text{Var}(X)=E\left(X^{2}\right)-\big[E(X)\big]^{2}$.
7Standard deviation: take the square root of the variance.

Practice

Practice questions

Eight more questions of my own, covering every skill. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to.

Question 1 of 8

Skill 1: probability functions · Technology-free · [2 marks]

The table shows the probability distribution of a discrete random variable $X$.

$x$	$0$	$1$	$2$	$3$
$P(X=x)$	$k$	$2k$	$0.3$	$0.1$

Determine the value of $k$.

1The probabilities must add to 1.

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

2Check the distribution is valid. The probabilities are 0.2, 0.4, 0.3 and 0.1. Each is between 0 and 1, and they add to 1.

Answer

$k=0.2$

Question 2 of 8

Skill 1: probability functions · Technology-free · [3 marks]

The discrete random variable $X$ has probability function $P(X=x)=kx^{2}$ for $x=1,2,3$. Determine the value of $k$, and $P(X\ge2)$.

1Write out each probability by substituting each value of $x$.

$$P(X=1)=k,\qquad P(X=2)=4k,\qquad P(X=3)=9k$$

2They must add to 1.

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

3$X\ge2$ means $X=2$ or $X=3$, so add those two.

$$P(X\ge2)=\frac{4}{14}+\frac{9}{14}=\frac{13}{14}$$

Answer

$k=\frac{1}{14}$ and $P(X\ge2)=\frac{13}{14}$

Question 3 of 8

Skill 2: estimates from data · Technology-free · [3 marks]

A four-sided spinner was spun 200 times. The results are shown.

Number	$1$	$2$	$3$	$4$
Frequency	$46$	$62$	$38$	$54$

Estimate $P(X=2)$ and the probability of spinning an even number. Comment on whether the spinner appears to be fair.

1Each estimate is the frequency divided by the total number of spins, 200.

$$P(X=2)\approx\frac{62}{200}=0.31$$

2"Even" means a 2 or a 4, so add those frequencies first.

$$P(X\text{ is even})\approx\frac{62+54}{200}=\frac{116}{200}=0.58$$

3If the spinner were fair, each probability would be 0.25, so you would expect about 50 of each. The 62 is a little high, but relative frequencies always wobble. The animation in Part 2 shows this. With only 200 spins you cannot be sure the spinner is unfair. More spins would give better estimates.

Answer

$P(X=2)\approx0.31$ and $P(X\text{ is even})\approx0.58$. These are only estimates, so 200 spins is not enough to say the spinner is unfair.

Question 4 of 8

Skill 3: uniform and non-uniform · Technology-free · [3 marks]

A bag contains three red marbles and two blue marbles. Two marbles are drawn at random without replacement. Let $X$ be the number of red marbles drawn. Construct the probability distribution table for $X$.

1Work out each probability. Without replacement, the second draw has only four marbles left.

$$P(X=0)=P(BB)=\frac{2}{5}\times\frac{1}{4}=\frac{1}{10}$$

$$P(X=2)=P(RR)=\frac{3}{5}\times\frac{2}{4}=\frac{3}{10}$$

2One red can happen two ways, red then blue or blue then red.

$$P(X=1)=\frac{3}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{3}{4}=\frac{6}{10}$$

3Put them in a table and check they add to 1.

$x$	$0$	$1$	$2$
$P(X=x)$	$\frac{1}{10}$	$\frac{6}{10}$	$\frac{3}{10}$

The probabilities are not all equal, so this is a non-uniform distribution.

Answer

$P(X=0)=\frac{1}{10}$, $P(X=1)=\frac{6}{10}$, $P(X=2)=\frac{3}{10}$

Question 5 of 8

Skill 3: uniform and non-uniform · Technology-free · [3 marks]

$X$ is the number rolled on a fair ten-sided die with faces numbered 1 to 10. Explain why $X$ is a uniform discrete random variable, then determine $P(X>7)$ and $E(X)$.

1The die is fair, so every one of the 10 outcomes has the same probability, $\frac{1}{10}$. That is what makes it uniform.

2$X>7$ means 8, 9 or 10. That is three outcomes.

$$P(X>7)=\frac{3}{10}$$

3For the expected value, multiply each value by $\frac{1}{10}$ and add.

$$E(X)=\frac{1+2+\dots+10}{10}=\frac{55}{10}=5.5$$

For a uniform distribution the expected value is always halfway between the smallest and largest values. It does not have to be a value $X$ can take.

Answer

Each outcome has probability $\frac{1}{10}$, so $X$ is uniform. $P(X>7)=\frac{3}{10}$ and $E(X)=5.5$.

Question 6 of 8

Skill 4: expected value, variance and standard deviation · Technology-free · [4 marks]

The number of goals, $X$, scored by a hockey team in a match has the probability distribution shown.

$x$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.1$	$0.3$	$0.4$	$0.2$

Determine the mean and standard deviation of $X$.

1Expected value. Multiply each value by its probability and add.

$$E(X)=0(0.1)+1(0.3)+2(0.4)+3(0.2)=1.7$$

2For the variance, this time use $E\left(X^{2}\right)$. Square each value, multiply by its probability and add.

$$E\left(X^{2}\right)=0(0.1)+1(0.3)+4(0.4)+9(0.2)=3.7$$

$$\mathrm{Var}(X)=E\left(X^{2}\right)-\big(E(X)\big)^{2}=3.7-1.7^{2}=0.81$$

3Take the square root for the standard deviation.

$$\sigma=\sqrt{0.81}=0.9$$

This is the distribution from the variance animation in Part 3. The two methods give the same answer, so use whichever you find quicker.

Answer

The mean is 1.7 goals and the standard deviation is 0.9 goals.

Question 7 of 8

Skill 4: expected value, variance and standard deviation · Technology-free · [5 marks]

The discrete random variable $X$ has the probability distribution shown, and $E(X)=2.6$.

$x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$a$	$b$	$0.2$

Determine the values of $a$ and $b$, and the exact standard deviation of $X$.

1There are two unknowns, so you need two equations. The first comes from the probabilities adding to 1.

$$0.1+a+b+0.2=1\quad\Rightarrow\quad a+b=0.7$$

2The second comes from the expected value.

$$1(0.1)+2a+3b+4(0.2)=2.6\quad\Rightarrow\quad 2a+3b=1.7$$

3Solve simultaneously. Substitute $a=0.7-b$ into the second equation: $1.4-2b+3b=1.7$, so $b=0.3$ and $a=0.4$.

4Now the variance, using $E\left(X^{2}\right)$.

$$E\left(X^{2}\right)=1(0.1)+4(0.4)+9(0.3)+16(0.2)=7.6$$

$$\mathrm{Var}(X)=7.6-2.6^{2}=7.6-6.76=0.84$$

5Take the square root. $\sqrt{0.84}=\sqrt{\frac{84}{100}}=\frac{2\sqrt{21}}{10}$.

Answer

$a=0.4$, $b=0.3$, and $\sigma=\frac{\sqrt{21}}{5}\approx0.917$

Question 8 of 8

Skill 5: modelling · Technology-free · [3 marks]

At a school fete, it costs $2 to play a game. The prize, $X$ dollars, has the probability distribution shown.

$x$	$0$	$2$	$10$
$P(X=x)$	$0.6$	$0.3$	$0.1$

The organisers claim: The stall will make a profit of at least $100 if the game is played 200 times. Evaluate the reasonableness of the claim.

1Find the expected prize.

$$E(X)=0(0.6)+2(0.3)+10(0.1)=1.60$$

2The stall takes $2 a game and pays out $1.60 on average, so it expects to keep the difference.

$$\text{expected profit per game}=2-1.60=0.40\text{ dollars}$$

3Scale up to 200 games.

$$200\times0.40=80\text{ dollars}$$

4Compare with the claim and give a reason. The expected profit is less than the claimed amount.

Answer

The expected profit over 200 games is $80, which is less than $100, so the claim is not reasonable.

Questions marked as teacher-written are original Tangent Tuition questions in the style of the external exam. Other past questions and marking-guide steps are from QCAA Mathematical Methods assessments, © State of Queensland (QCAA), licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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