2020 Paper 1, Question 6
If the probability of success in a Bernoulli trial is 0.30, the variance is
$\text{Var}(X)=p(1-p)=0.3\times0.7=0.21$. (A) is $1-p$ and (C) is the mean.
Answer
(D) $0.21$
Unit 3 · Topic 5.2 · Discrete random variables
Bernoulli and binomial distributions
A Bernoulli trial is the simplest random experiment there is: one go, two outcomes. Repeat it a fixed number of times and count the successes, and you have a binomial distribution. Both are discrete random variables, so everything from Topic 5.1 still applies. They are special cases that come with their own shortcut formulas. The QCAA ask about this every year, and most of the marks lost come from reading the wording wrongly. The maths is rarely the problem. Tick off each skill once you can do it without help.
Part 1
Bernoulli distributions
A Bernoulli trial is a single event with exactly two outcomes, success (recorded as 1) and failure (recorded as 0), each with a fixed probability. Everything is controlled by one number, the probability of success $p$.
| $x$ | $0$ | $1$ |
|---|---|---|
| $P(X=x)$ | $1-p$ | $p$ |
Mean
$$E(X)=\mu=p$$
Variance
$$\text{Var}(X)=p(1-p)$$
Standard deviation
$$\text{SD}(X)=\sqrt{p(1-p)}$$
Where this fits
A Bernoulli random variable is a discrete random variable, just like the ones in Topic 5.1. Its table just happens to have only two columns, 0 and 1. That means there are no new rules to learn. The formulas above are the Topic 5.1 ones, worked out once for this table.
$$E(X)=0(1-p)+1(p)=p$$
$$\text{Var}(X)=E\left(X^{2}\right)-\big(E(X)\big)^{2}=\left[0^{2}(1-p)+1^{2}(p)\right]-p^{2}=p(1-p)$$
A binomial random variable is a discrete random variable too. Each trial has only two outcomes, but $X$ counts the successes over $n$ trials, so it can take any of the values $0, 1, 2, \dots, n$. Its probabilities come from a formula instead of a table, and its mean and variance have shortcuts. Underneath, the probabilities still add to 1 and $E(X)$ is still $\sum x\,P(X=x)$.
On their own, Bernoulli questions are short. The exam usually repeats the trial, which takes you into the binomial distribution in Part 2.
Practise
Exam questions: Bernoulli distributions
For the written-answer questions, the solutions show where each mark goes, based on the QCAA marking guides. Many marks allow follow-through: if an earlier step is wrong but you use your wrong answer correctly, you can still get the later marks. That is another reason to show every step.
2024 Paper 2, Question 4
Consider the Bernoulli distribution where the outcomes for rolling a six-sided die are a four and not rolling a four. Determine the variance of the resulting Bernoulli distribution in this scenario.
Success is rolling a four, so $p=\frac{1}{6}$. Then $\text{Var}(X)=\frac{1}{6}\times\frac{5}{6}=\frac{5}{36}=0.13\dot{8}$. (C) is $p$ itself, and (D) is $1-p$.
Answer
(B) $0.13\dot{8}$
2022 Paper 1, Question 12
The probability that a debating team wins a debate can be modelled as a Bernoulli distribution. Given that the probability of winning a debate is $\frac{4}{5}$:
(a) Determine the mean of this distribution.[1 mark]
(b) Determine the variance of this distribution.[1 mark]
(c) Determine the standard deviation of this distribution.[1 mark]
The mean is $p=\frac{4}{5}$.
1 markThe mean.
The variance is $p(1-p)=\frac{4}{5}\times\frac{1}{5}=\frac{4}{25}$.
1 markThe variance.
The standard deviation is $\sqrt{\frac{4}{25}}=\frac{2}{5}$.
1 markThe standard deviation. Follow-through applies if your variance was wrong.
Answer
(a) $\frac{4}{5}$ · (b) $\frac{4}{25}$ · (c) $\frac{2}{5}$
2023 Paper 1, Question 11
Two random samples (A and B) were obtained using two different Bernoulli experiments. Each Bernoulli trial in the random samples was recorded as 1 (for success) or 0 (for failure). The results are shown.
| A | $1$ | $1$ | $1$ | $1$ | $0$ | $1$ | $1$ | $0$ | $1$ | $1$ |
|---|---|---|---|---|---|---|---|---|---|---|
| B | $0$ | $0$ | $1$ | $1$ | $1$ | $0$ | $1$ | $1$ | $0$ | $0$ |
In sample A, for each trial the mean is 0.8 and the variance is 0.16.
(a) Use the sample B results to determine the mean and variance for each trial in sample B.[2 marks]
(b) Compare the variability about the means of samples A and B.[2 marks]
(a) Sample B has 5 successes in 10 trials, so $p=0.5$, and the mean is $p=0.5$.
1 markThe mean for sample B.
The variance is $p(1-p)=0.5\times0.5=0.25$.
1 markThe variance for sample B.
(b) Compare the variances, and say what that means. B's variance (0.25) is bigger than A's (0.16).
1 markSaying that sample B has the larger variance.
So the results in B are more spread out about their mean, and A's results are more consistent.
1 markExplaining that a larger variance means more variability.
Answer
(a) Mean $0.5$, variance $0.25$ · (b) Sample B varies more about its mean than sample A.
Part 2
Is it binomial?
A binomial random variable counts the number of successes when you repeat a Bernoulli trial $n$ times. We write $X\sim\text{Bin}(n,p)$. Before you use it, check for all four flags below. If any one is missing, it is not binomial. Learn the phrases as well as the rule.
1. Only two outcomes
Each trial is a success or a failure, with nothing in between.
"delayed or not delayed" · "hit or miss" · "a four and not a four"
2. A fixed number of trials
You know how many times before you start. This is your $n$.
"if eight seeds are planted" · "in a batch of 20" · "in 10 attempts" · "for three days"
3. The same chance every trial
The word "each" is the giveaway. $p$ never changes.
"each seed … 0.8" · "each customer … 20%" · "one in five … every day"
4. Trials do not affect each other
One result tells you nothing about the next.
"independent of all other sales" · "with replacement"
"Without replacement" is the most common way the QCAA break a binomial. Each draw changes what is left, so $p$ changes and the trials are not independent.
Practise
Exam questions: is it binomial?
2023 Paper 1, Question 3
A bag contains 10 buttons of the same shape and size in different colours: 5 blue, 3 green and 2 red. If 3 buttons are randomly drawn from the bag, which probability can be calculated using the binomial distribution?
Run the flags. Without replacement, $p$ changes after each draw, which rules out (B) and (D). (C) has three categories (green, red and neither), so it is not two outcomes. Only (A) has success as "green", a fixed $p=\frac{3}{10}$ and independent draws.
Answer
(A)
2022 Paper 1, Question 2
A binomial random variable arises from the number of successes in $n$ independent Bernoulli trials. A context not suitable for modelling using a binomial random variable is recording the number of
"Without replacement" breaks flags 3 and 4. After a red marble is drawn, the chance of the next one being red drops from $\frac{5}{9}$ to $\frac{4}{8}$.
Answer
(D)
2023 Paper 1, Question 13
At a certain airport, the departure of one in five international flights is delayed every day. The status of any flight is independent of other flights. One international flight is selected at random each day for three days. Each selection is recorded as either 'delayed' or 'not delayed'.
(a) State two conditions that make this context suitable for modelling using a binomial random variable.[2 marks]
(b) Calculate the probability that at least two of the selected flights were delayed.[3 marks]
(a) Any two of the four flags, tied to the context: there are only two outcomes (delayed or not delayed). There is a fixed number of trials (three days). The probability of a delay is the same each day ($\frac{1}{5}$). The flights are also independent.
1 markOne condition, tied to the context.
1 markA second condition. Two are enough for both marks.
(b) $X\sim\text{Bin}(3,\,0.2)$. "At least two" means 2 or 3:
1 markThe number of trials, 3, and the probability of a delay, 0.2.
$$P(X\ge2)=\binom{3}{2}(0.2)^{2}(0.8)+\binom{3}{3}(0.2)^{3}=0.096+0.008=0.104$$
1 markA suitable method, adding $P(X=2)$ and $P(X=3)$.
1 markThe probability.
Answer
(b) $P(X\ge2)=0.104$
Part 3
Mean, variance and standard deviation
A binomial is just $n$ Bernoulli trials added together, so its formulas are the Bernoulli ones multiplied by $n$.
Mean
$$\mu=E(X)=np$$
Variance
$$\sigma^{2}=\text{Var}(X)=np(1-p)$$
Standard deviation
$$\sigma=\sqrt{np(1-p)}$$
Watch the shape change
Practise
Exam questions: mean, variance and standard deviation
2024 Paper 2, Question 1
The probability of hitting a target in a particular binomial experiment is 0.72. Determine the mean of the number of hits if this experiment is repeated eight times.
$\mu=np=8\times0.72=5.76$. (B) uses $1-p$ instead of $p$.
Answer
(C) $5.76$
2023 Paper 1, Question 15
In a certain game, players throw one water balloon at a target. There is a one in four chance of hitting the target.
(a) State the probabilities of all the possible outcomes for one throw at the target.[2 marks]
(b) Let $H$ be the discrete random variable for one of the possible outcomes. Determine the mean and variance of the distribution of random variable $H$ when 20 players throw a water balloon at the target.[2 marks]
(a) $P(\text{hit})=\frac{1}{4}$ and $P(\text{miss})=\frac{3}{4}$.
1 markThe probability of a hit.
1 markThe probability of a miss.
(b) Let $H$ be the number of hits, so $H\sim\text{Bin}\left(20,\frac{1}{4}\right)$. The mean is $\mu=np=20\times\frac{1}{4}=5$.
1 markThe mean.
The variance is $\sigma^{2}=np(1-p)=20\times\frac{1}{4}\times\frac{3}{4}=\frac{15}{4}$.
1 markThe variance.
Answer
(b) Mean $5$, variance $\frac{15}{4}$
2020 Paper 2, Question 11
A sugar company samples the packets of sugar it produces and finds that 5% of packets are underweight. Consider a batch of 20 packets.
(a) Determine how many packets of sugar the company can expect to be underweight in a batch of 20 packets.[1 mark]
(b) Determine the variance of the batch.[1 mark]
(c) Determine the probability that at most one of the packets of sugar in the batch is underweight.[2 marks]
$X\sim\text{Bin}(20,\,0.05)$. (a) The expected number is $np=20\times0.05=1$.
1 markThe expected number of underweight packets.
(b) The variance is $np(1-p)=20\times0.05\times0.95=0.95$.
1 markThe variance.
(c) "At most one" means $P(X\le1)$. Use the binomial cumulative function on your calculator with $n=20$, $p=0.05$, from 0 to 1: about $0.736$.
1 markShowing your approach. Writing $P(X\le1)$ with $n$ and $p$ is enough, and so are the two formula terms for 0 and 1.
1 markThe probability. 0.736 or 73.58% is fine.
Answer
(a) 1 packet · (b) $0.95$ · (c) $P(X\le1)\approx0.736$
2022 Paper 2, Question 11
A salesperson has a 20% probability of making a sale to each customer who enters the store. Each sale is independent of all other sales.
(a) Determine the mean number of sales on a day where 25 customers enter the store.[2 marks]
(b) Determine the standard deviation of the number of sales on a day where 25 customers enter the store.[2 marks]
(c) Determine the minimum number of customers who would have to enter the store to have an 88% chance or more of making at least one sale.[3 marks]
(a) $X\sim\text{Bin}(25,\,0.2)$, so $\mu=np=25\times0.2=5$ sales.
1 markSubstituting into the formula for the mean.
1 markThe mean.
(b) $\sigma=\sqrt{np(1-p)}=\sqrt{25\times0.2\times0.8}=\sqrt{4}=2$.
1 markSubstituting into the formula for the standard deviation.
1 markThe standard deviation.
(c) Now $n$ is the unknown. "At least one" is fastest as $1-P(X=0)$, and $P(X=0)=0.8^{n}$:
$$1-0.8^{n}\ge0.88\ \Rightarrow\ 0.8^{n}\le0.12$$
1 markThe inequation.
Try values on your calculator: $n=9$ gives $1-0.8^{9}\approx0.866$, which is not enough, and $n=10$ gives $\approx0.893$. You need a whole number of customers, so round up.
1 markFinding the unknown. Solving $0.8^{n}=0.12$ on your calculator gives $n\approx9.5$, which also works.
1 markThe minimum number of customers, 10.
Answer
(a) 5 sales · (b) 2 · (c) 10 customers
Part 4
Binomial probabilities
The probability of exactly $r$ successes in $n$ trials is given by the formula below. The binomial coefficient counts the number of ways the successes can be arranged.
$$P(X=r)=\binom{n}{r}p^{r}(1-p)^{n-r}$$
Watch where the formula comes from
Which probability are they asking for?
Once you know it is binomial, the wording tells you which values of $X$ to include. This is where most marks are lost.
Exactly:
One exact count, often with no comparison word at all.
"exactly two" · "all three days" means $P(X=3)$ · "getting 2 defective"
Low:
"Fewer than" is strict. "At most" includes the number.
"fewer than three" means $P(X<3)$ · "at most two" and "no more than two" mean $P(X\le2)$
High:
"At least" includes the number. "More than" is strict.
"at least two" means $P(X\ge2)$ · "at least one" is fastest as $1-P(X=0)$
Middle:
Two ends. It also hides inside "within one standard deviation".
"between 2 and 5 inclusive" · "within one standard deviation" (find $\mu$ and $\sigma$ first)
Watch for two traps. A number with no comparison word means exactly, so "getting 2 defective shoes" is $P(X=2)$. And "no less than $a$" is $P(X\ge a)$, so do not let the double negative flip you into a less-than. Always ask yourself whether the boundary value is in or out.
Technology-free
Use the formula, adding one term for each value of $X$ you need. If it is quicker, find the probability of the event not happening and subtract from 1. These shortcuts save time on the coefficients:
$$\binom{n}{0}=\binom{n}{n}=1,\qquad \binom{n}{1}=\binom{n}{n-1}=n,\qquad \binom{n}{2}=\frac{n(n-1)}{2}$$
Technology-active
For a single value, $P(X=r)$, use your calculator's binomial probability (BPD) function. For anything else, use the binomial cumulative (BCD) function with the right lower and upper bounds. If the question gives you the probability and asks for $n$ or $p$, set up the equation and solve it.
Try it yourself
Turn the words into a probability
Set $n$ and $p$, then choose a phrase and a number. The bars in black are the values of $X$ that phrase includes. The orange marker sits at the mean, $np$.
The phrase
Practise
Exam questions: binomial probabilities
2023 Paper 2, Question 2
The probability of hitting a bullseye on a standard dartboard is 1 in 1250. What is the probability of hitting a bullseye exactly once in 10 attempts?
Match each part of the formula. $n=10$ trials, so the coefficient is $\binom{10}{1}$. One success, so $p$ has power 1 and $1-p$ has power $10-1=9$.
Answer
(C)
2025 Paper 2, Question 8
The table shows the probability distribution for a random variable $X$ in a Bernoulli experiment. The random variable has only two possible values: 0 represents failure and 1 represents success.
| $x$ | $0$ | $1$ |
|---|---|---|
| $P(X=x)$ | $0.6$ | $0.4$ |
If six Bernoulli experiments are conducted, determine the probability of getting exactly two successes.
Success has probability 0.4, so $X\sim\text{Bin}(6,\,0.4)$. Use BPD, or the formula: $\binom{6}{2}(0.4)^{2}(0.6)^{4}=15\times0.16\times0.1296\approx0.311$.
Answer
(C) $0.311$
2021 Paper 1, Question 9
A basket contains 10 green apples and 30 red apples. Three apples are drawn at random from the basket with replacement. Determine the probability that exactly two of the three apples are green.
With replacement, $p=\frac{10}{40}=\frac{1}{4}$ every time. Then $\binom{3}{2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)=3\times\frac{1}{16}\times\frac{3}{4}=\frac{9}{64}$. Forgetting the coefficient gives (A).
Answer
(B) $\frac{9}{64}$
2020 Paper 1, Question 8
A test includes six multiple choice questions. Each question has four options for the answer. If the answers are guessed, the probability of getting at most two questions correct is represented by
"At most two" includes 2, so you need $P(0)+P(1)+P(2)$, with $p=0.25$. (A) leaves out $X=2$, and (C) and (D) give "more than" instead.
Answer
(B)
2021 Paper 1, Question 6
A random variable $X$ is the number of successes in a Bernoulli experiment with $n$ trials, each with a probability of success $p$ and a probability of failure $q$. The probability distribution table of $X$ is shown.
| $k$ | $0$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|---|
| $P(X=k)$ | $\frac{1}{81}$ | $\frac{8}{81}$ | $\frac{24}{81}$ | $\frac{32}{81}$ | $\frac{16}{81}$ |
Which values of $n$, $p$ and $q$ will generate this probability distribution?
$X$ runs from 0 to 4, so $n=4$. $P(X=0)=q^{4}=\frac{1}{81}$ gives $q=\frac{1}{3}$, so $p=\frac{2}{3}$. Check the other end: $P(X=4)=p^{4}=\frac{16}{81}$, which matches.
Answer
(A) $n=4,\ p=\frac{2}{3},\ q=\frac{1}{3}$
2024 Paper 1, Question 12
Each day over a three-day long weekend, a family spins a pointer on a circular board to decide whether they will spend the day at the beach or bushwalking. The circular board consists of three equal sections.
(a) Determine the probability that the family will spend all three days bushwalking.[1 mark]
(b) Determine the following binomial probabilities, expressed as fully simplified fractions.
(i) Exactly two days will be spent at the beach.[2 marks]
(ii) Fewer than three days will be spent at the beach.[3 marks]
Two of the three sections are Beach, so $P(\text{beach})=\frac{2}{3}$ and $P(\text{bushwalking})=\frac{1}{3}$.
(a) $\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$.
1 markThe probability.
(b)(i) Let $X$ be the number of beach days, $X\sim\text{Bin}\left(3,\frac{2}{3}\right)$. $P(X=2)=\binom{3}{2}\left(\frac{2}{3}\right)^{2}\left(\frac{1}{3}\right)=3\times\frac{4}{9}\times\frac{1}{3}=\frac{4}{9}$.
1 markThe binomial expression for $P(X=2)$.
1 markThe probability as a fully simplified fraction.
(ii) "Fewer than three" is $P(X<3)=P(0)+P(1)+P(2)$. It is quicker to subtract the one case you do not want: $1-P(X=3)=1-\frac{8}{27}=\frac{19}{27}$.
1 markA method. Using $1-P(X=3)$ or adding the three cases both work.
1 markThe binomial expression for the calculation.
1 markThe probability as a fully simplified fraction.
Answer
(a) $\frac{1}{27}$ · (b)(i) $\frac{4}{9}$ · (ii) $\frac{19}{27}$
2021 Paper 1, Question 17
In any five-day working week Leonardo either catches a bus to work or uses another form of transportation. On average, he catches the bus to work on three of the five days. His decision on any given day is independent of his decision on any other day.
Determine the probability that Leonardo catches a bus to work on exactly one day in a given five-day working week.
The hidden step is finding $p$. "On average three of the five days" means the mean is 3, so $np=3$ with $n=5$, giving $p=\frac{3}{5}$.
1 markThe values of $n$ and $p$.
$$P(X=1)=\binom{5}{1}\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)^{4}=5\times\frac{3}{5}\times\frac{16}{625}=\frac{48}{625}$$
1 markThe expression for the probability.
1 markThe probability. Follow-through applies if your $p$ was wrong.
Answer
$\frac{48}{625}$
The full walkthrough of 2021 Paper 1, Question 17 goes through it mark by mark.
2021 Paper 2, Question 6
When seeds of a certain variety of flower are planted, the probability of each seed germinating is 0.8. If eight seeds are planted, what is the probability that at least six seeds will germinate?
$X\sim\text{Bin}(8,\,0.8)$ and you want $P(X\ge6)=P(6)+P(7)+P(8)$. Use BCD with lower bound 6 and upper bound 8: about 0.797.
Answer
(A) $0.797$
2020 Paper 2, Question 7
The records of a shoe manufacturer show that 10% of shoes made are defective. Assuming independence, the probability of getting 2 defective shoes in a batch of 20 is
There is no comparison word, so it means exactly 2. Use BPD with $n=20$, $p=0.1$, $r=2$: about 0.2852. The other options are cumulative traps. (C) is $P(X\le1)$ and (D) is $P(X\ge2)$.
Answer
(B) $0.2852$
Part 5
Modelling with binomial distributions
The harder questions hide one binomial inside another, or make you find $p$ from something else first, such as another distribution. Write down what "success" means and what $n$ is before you calculate anything.
Practise
Exam questions: modelling
2024 Paper 2, Question 17
At a particular orchard, 3% of fruit is bruised during picking. After picking, the fruit is packed into boxes, each containing four pieces of fruit. A grocery shop orders 140 boxes of fruit to sell to their customers.
Determine the expected number of boxes that will contain bruised fruit.
There are two binomials here. First, a single box: it "contains bruised fruit" if at least one of its four pieces is bruised.
$$P(\text{box has bruised fruit})=1-0.97^{4}\approx0.1147$$
1 markThe probability that one box has bruised fruit. 0.11 is fine.
Now each box is a trial with that probability. With 140 boxes, the expected number is $np=140\times0.1147\approx16.06$.
1 markThe expected number as a decimal.
Answer
About 16 boxes
1 markThe number of boxes as a whole number. Rounding up or down is accepted, depending on your decimals.
The full walkthrough of 2024 Paper 2, Question 17 goes through it mark by mark.
2025 Paper 2, Question 15
A tour operator offers day cruises off the Queensland coast. They advertise that on any given day, customers have a 65% probability of seeing at least one whale. The tour operator conducts cruises for 40 consecutive days.
Let $X$ be the binomial random variable representing the outcome (success or failure) over the 40 days. A success is customers seeing at least one whale.
Calculate the probability that the number of days with a success will be within one standard deviation of the mean for $X$.
$X\sim\text{Bin}(40,\,0.65)$, so $\mu=np=40\times0.65=26$.
1 markThe mean.
$\sigma=\sqrt{np(1-p)}=\sqrt{40\times0.65\times0.35}=\sqrt{9.1}\approx3.017$.
1 markThe standard deviation.
Within one standard deviation means $22.98\le X\le29.02$. $X$ is a whole number of days, so that is $23\le X\le29$. Check the boundaries carefully here: 22 and 30 are both outside.
Use BCD with lower bound 23 and upper bound 29: $P(23\le X\le29)\approx0.755$.
1 markThe probability. 75% is fine.
Answer
About $0.755$
The full walkthrough of 2025 Paper 2, Question 15 goes through it mark by mark.
2025 Paper 1, Question 17
In a computer fishing game, a player repeatedly casts their hook into either a blue pond or a red pond. Each fish they catch scores points.
In the blue pond, the probability of catching a fish on a cast is $\frac{2}{3}$ and each fish caught scores 10 points.
In the red pond, the probability of catching a fish on a cast is $\frac{1}{3}$ and each fish caught scores 15 points.
A player has three casts left and needs to score at least 30 points to win. All remaining casts must be in the same pond.
It is claimed that the probability of winning if casting in the blue pond is $\frac{1}{27}$ more than the probability of winning if casting in the red pond. Evaluate the reasonableness of the claim.
First turn "at least 30 points" into a number of fish for each pond. In the blue pond each fish is 10 points, so you need all 3. In the red pond each fish is 15 points, so you need at least 2.
Blue: $X\sim\text{Bin}\left(3,\frac{2}{3}\right)$, and $P(X=3)=\left(\frac{2}{3}\right)^{3}=\frac{8}{27}$.
1 markA method for the blue pond. At least 30 points needs all three casts to catch a fish.
1 markThe probability for the blue pond.
Red: $Y\sim\text{Bin}\left(3,\frac{1}{3}\right)$, and $P(Y\ge2)=\binom{3}{2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)+\left(\frac{1}{3}\right)^{3}=\frac{6}{27}+\frac{1}{27}=\frac{7}{27}$.
1 markA method for the red pond. At least 30 points needs at least two fish.
1 markSubstituting into the binomial terms for 2 and 3 fish.
1 markThe probability for the red pond.
The difference is $\frac{8}{27}-\frac{7}{27}=\frac{1}{27}$, which is exactly what was claimed.
1 markEvaluating the claim. This mark only counts if your working backs up the decision.
Answer
The claim is reasonable. Blue gives $\frac{8}{27}$ and red gives $\frac{7}{27}$.
The full walkthrough of 2025 Paper 1, Question 17 goes through it mark by mark.
2022 Paper 2, Question 18
The intelligence quotient (IQ) of individuals in a population is normally distributed, with a mean of 100 and a standard deviation of 16. Nine individuals are chosen at random from the population.
Determine the probability that no more than two of the individuals have an IQ of at least 120.
This one also needs the normal distribution from Unit 4, so come back to it after Topic 3.2 if you have not done that yet.
Find $p$ first. A "success" is a person with an IQ of at least 120. Using the normal cumulative function with mean 100 and standard deviation 16: $p=P(\text{IQ}\ge120)\approx0.1056$.
1 markThe probability of an IQ of at least 120.
Now $X\sim\text{Bin}(9,\,0.1056)$ and "no more than two" is $P(X\le2)$. Use BCD from 0 to 2: about 0.939.
1 markRecognising that a binomial model fits: 9 people, each one a success or not. Saying Bernoulli trial is accepted too.
1 markThe probability.
Answer
About $0.939$
The full walkthrough of 2022 Paper 2, Question 18 goes through it mark by mark.
Game plan
Your game plan for binomial questions
- 1Check it is binomial: two outcomes, a fixed number of trials, the same $p$ every trial, and independent trials.
- 2Write down $n$, $p$ and $1-p$, and what counts as a success. If $p$ is not given, find it first.
- 3For the mean, variance or standard deviation, use $np$, $np(1-p)$ and $\sqrt{np(1-p)}$.
- 4For a probability, translate the words into an inequality, and decide whether the boundary value is in or out.
- 5Technology-free: use the formula, and consider $1-P(\text{the opposite})$ if it is shorter.
- 6Technology-active: BPD for exactly one value, BCD for everything else. If you are given the probability, set up the equation and solve for the unknown.
Practice
Practice questions
Eight 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
Which of these is best modelled by a Bernoulli random variable?
A Bernoulli random variable is one trial with two outcomes, recorded as 1 or 0. Whether one seed germinates is exactly that.
Options (A) and (D) count successes over many trials, so they are binomial. Option (B) has six outcomes, so it is a general discrete random variable like the ones in Topic 5.1.
Answer
(C)
Question 2 of 8
The outcome of a netball shot can be modelled as a Bernoulli distribution. The probability that the goal shooter scores from a shot is $\dfrac{3}{5}$.
(a) Determine the mean of this distribution.[1 mark]
(b) Determine the standard deviation of this distribution.[2 marks]
1Success is scoring, so $p=\frac{3}{5}$ and $1-p=\frac{2}{5}$.
2(a) The mean of a Bernoulli distribution is $p$.
$$E(X)=p=\frac{3}{5}$$
3(b) Find the variance first, then take the square root.
$$\text{Var}(X)=p(1-p)=\frac{3}{5}\times\frac{2}{5}=\frac{6}{25},\qquad \text{SD}(X)=\sqrt{\frac{6}{25}}=\frac{\sqrt{6}}{5}$$
Answer
(a) $\frac{3}{5}$ · (b) $\frac{\sqrt{6}}{5}$
Question 3 of 8
At a call centre, one in six calls is transferred to a supervisor. The outcome of any call is independent of other calls. Five calls are selected at random, and each is recorded as either "transferred" or "not transferred". State two conditions that make this context suitable for modelling with a binomial random variable.
Any two of these earn the marks, as long as each one is said in the context of the question.
1There is a fixed number of trials. Five calls are selected.
2Each trial has only two outcomes. A call is transferred or it is not.
3The trials are independent. One call does not affect any other.
4The probability of success is the same every time. It is $\frac{1}{6}$ for every call.
Just listing "two outcomes, fixed trials" without the call centre will not get full marks.
Answer
Any two of the four conditions, each tied to the calls
Question 4 of 8
A café finds that 20% of its customers order a toasted sandwich. Each order is independent of all other orders. Determine the mean and standard deviation of the number of toasted sandwiches ordered on a day when 50 customers visit the café.
1It is binomial: a fixed number of customers, two outcomes, independent orders and a constant probability. So $X\sim\text{Bin}(50,\,0.2)$.
2Apply the formulas.
$$E(X)=np=50\times0.2=10$$
$$\text{Var}(X)=np(1-p)=50\times0.2\times0.8=8,\qquad \text{SD}(X)=\sqrt{8}=2\sqrt{2}$$
Answer
Mean 10 sandwiches, standard deviation $2\sqrt{2}$
Question 5 of 8
On any school day, the probability that Mia catches the early bus is $\dfrac{1}{3}$, independent of any other day. Calculate the probability that she catches the early bus on at least three of the next four school days. Express your answer as a fully simplified fraction.
1It is binomial with $n=4$ and $p=\frac{1}{3}$, so $1-p=\frac{2}{3}$.
2"At least three" means 3 or 4, so $P(X\ge3)=P(X=3)+P(X=4)$.
3Apply the formula to each.
$$P(X=3)=\binom{4}{3}\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)^{1}=4\times\frac{1}{27}\times\frac{2}{3}=\frac{8}{81}$$
$$P(X=4)=\binom{4}{4}\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$$
4Add and simplify.
$$P(X\ge3)=\frac{8}{81}+\frac{1}{81}=\frac{9}{81}=\frac{1}{9}$$
Answer
$\frac{1}{9}$
Question 6 of 8
A barista finds that 35% of customers order oat milk. Each order is independent. On a particular morning, 20 customers are served.
(a) Determine the probability that exactly seven customers order oat milk.[1 mark]
(b) Determine the probability that at least nine customers order oat milk.[2 marks]
1It is binomial with $n=20$ and $p=0.35$. On Paper 2, use BPD for "exactly" and BCD for everything else.
2(a) "Exactly seven" means $P(X=7)$.
$$\text{BPD}(x=7,\ n=20,\ p=0.35)\approx0.1844$$
3(b) "At least nine" means $P(X\ge9)=P(9\le X\le20)$.
$$\text{BCD}(\text{lower}=9,\ \text{upper}=20,\ n=20,\ p=0.35)\approx0.2376$$
The upper bound is 20, the number of customers. Leaving it at 9 is a common slip.
Answer
(a) 0.1844 · (b) 0.2376
Question 7 of 8
A telemarketer has a 15% probability of making a sale on each call. Each call is independent of all other calls. Determine the minimum number of calls the telemarketer must make to have a 90% chance or more of making at least one sale.
1It is binomial with $p=0.15$, but $n$ is the unknown.
2"At least one" is quickest as one minus "none". $P(X=0)=0.85^{n}$.
$$1-0.85^{n}\ge0.9\quad\Rightarrow\quad 0.85^{n}\le0.1$$
3Solve on your calculator. $0.85^{n}=0.1$ when $n\approx14.17$.
4$n$ is a number of calls, so round up. Check: $1-0.85^{14}\approx0.897$ is not enough, and $1-0.85^{15}\approx0.913$ is.
Answer
15 calls
Question 8 of 8
A quiz has 12 multiple choice questions, each with four options. A student guesses every answer. The probability that the student gets between 3 and 6 questions correct, inclusive, is closest to
$X\sim\text{Bin}(12,\,0.25)$, and "between 3 and 6 inclusive" means $P(3\le X\le6)$. Use BCD with lower bound 3 and upper bound 6: about 0.595.
Option (A) leaves out 3 and 6, which is what "between" means without "inclusive". Option (C) is $P(X\le6)$ and (D) is $P(X\ge3)$. Each one comes from reading the phrase wrongly.
Answer
(B) 0.595
Keep practising
Ready for something harder?
The IQ question joins the binomial to the normal distribution, which is exactly the kind of link the complex unfamiliar questions like to make. Try it once you have finished Unit 4 Topic 3.2.
Past questions and marking-guide steps are from the 2020–2025 QCAA Mathematical Methods external assessments, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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