Mathematical Methods · Past QCAA questions All simple familiar questions
2024 · Paper 1 · Simple familiar

Question 12

[6 marks] Technology-free
Topic guidesPremiumUnit 3 Topic 5.2 · Is it binomial?Unit 3 Topic 5.2 · Binomial probabilities
Game planPremium3.2 Binomial distributions
The question

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.

BeachBushwalkingBeach
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]
Watch the situation first

Three spins, eight paths, and each path is a product

Watch the spinner, then the tree of all eight ways the weekend can go. Each part of the question picks out some of the paths.

Each spin: two sections are beach, one is bushwalking Three spins: multiply along each path a) All three days bushwalking: one path b) i. Exactly two beach days: three paths b) ii. Fewer than three beach days: all paths but one BeachBushwalkingBeach beach 2 3 bush 1 3 beach, beach, beachbeach, beach, bushbeach, bush, beachbeach, bush, bushbush, beach, beachbush, beach, bushbush, bush, beachbush, bush, bush 8 27 4 27 4 27 2 27 4 27 2 27 2 27 1 27 a)b) i.b) ii. P = 1 3 × 1 3 × 1 3 = 1 27 P(X = 2) = 3 × 4 27 = 4 9 P(X < 3) = 1 − 8 27 = 19 27

Paths with the same number of beach days have the same probability. The binomial coefficient just counts those paths.

Now for the mathematics

Which values of $X$ make each event happen?

Let $X$ be the number of beach days, so $X\sim\text{Bin}\left(3,\frac{2}{3}\right)$. Choose a part, then choose the values of $X$ that make its event happen. The bars you choose are added up, and the strip shows how much of the total probability your event takes.

127 627 1227 827 0123 number of beach days, X 0123

The strip under the graph is the whole probability, 1, split up by the value of $X$. Orange is the part your event takes.

Each chosen bar

No values chosen yet.

Add them up
Or take the other bars away from 1
The event

Values of $X$ in this event

Choose the values of $X$ that make the event happen. Remember that $X$ counts beach days.

That’s it. Bushwalking all three days means no beach days, so the event is $X=0$. It is a single path, $\left(\frac{1}{3}\right)^3=\frac{1}{27}$.

That’s it. Three paths have exactly two beach days, and each has probability $\frac{4}{27}$. So $P(X=2)=\frac{12}{27}$, which simplifies to $\frac{4}{9}$.

That’s it. Fewer than three means $X=0$, 1 or 2. Adding three bars works, but the strip shows the quicker way: everything except $X=3$, so $1-\frac{8}{27}=\frac{19}{27}$.

Not this event yet.

$X$ counts beach days. Three bushwalking days means zero beach days, not three.

“Fewer than three” means strictly less than 3, so $X=3$ is not included.

Not part of this event, so take out .

This event also includes .

QCAA marking guide · 6 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. Part a) is one product. In part b), the last mark of each part is for a fully simplified fraction.

Step 1 · part a)1 mark

One of the three equal sections is bushwalking, so each day it has probability $\frac{1}{3}$. The spins are independent, so multiply for the three days.

$$\begin{aligned}&P(3\text{ bushwalking days})\\&=\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}\end{aligned}$$

$P=\dfrac{1}{27}$

Marker: correctly determines the probability

Step 2 · part b) i.1 mark

Let $X$ be the number of beach days. There are 3 trials and the probability of beach is $\frac{2}{3}$, so $X\sim\text{Bin}\left(3,\frac{2}{3}\right)$. Write the binomial expression for exactly 2 successes.

$$\begin{aligned}&P(X=2)\\&=\binom{3}{2}\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^1\end{aligned}$$

Writing $3\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)$ is also accepted. The 3 counts the orders: beach, beach, bush in any of three arrangements.

Marker: correctly determines the required binomial probability expression

Step 3 · part b) i.1 mark

Evaluate it. Cancel the 3s before you multiply and the fraction comes out already simplified.

$$P(X=2)=3\times\frac{4}{9}\times\frac{1}{3}$$

$P(X=2)=\dfrac{4}{9}$

$\frac{12}{27}$ is not fully simplified, so it doesn’t meet this descriptor. Follow-through applies if your expression in step 2 had a slip.

Marker: determines the probability as a fully simplified fraction

Step 4 · part b) ii.1 mark

“Fewer than three” is $X<3$. The only value left out is $X=3$, so use the complement.

$$P(X<3)=1-P(X=3)$$

Adding $P(X=0)+P(X=1)$ $+\,P(X=2)$ is also accepted. It is just more work.

Marker: correctly determines the required method

Step 5 · part b) ii.1 mark

Write the binomial expression for $X=3$: all three days at the beach.

$$\begin{aligned}&P(X<3)\\&=1-\binom{3}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^0\\&=1-\frac{8}{27}\end{aligned}$$

Equivalent expressions are accepted, and this mark can be implied by your later working. Follow-through applies if your method in step 4 had a slip.

Marker: determines the required binomial expression for the calculation

Step 6 · part b) ii.1 mark

Subtract. 19 is prime, so the fraction is already fully simplified.

$P(X<3)=\dfrac{19}{27}$

Marker: determines the probability as a fully simplified fraction

Putting it all together

Each phrase of the question gave you something

“Each day over a three-day long weekend”
One spin per day, so 3 independent trials. That is $n=3$.
“three equal sections”
Two of the three are beach, so $P(\text{beach})=\frac{2}{3}$ and $P(\text{bushwalking})=\frac{1}{3}$.
“all three days bushwalking”
One path, $\left(\frac{1}{3}\right)^3$. With $X$ counting beach days, it is $X=0$.
“Exactly two days will be spent at the beach”
$X=2$. Three orders, so the expression starts with $\binom{3}{2}=3$.
“Fewer than three days”
$X<3$, which is everything except $X=3$.
“expressed as fully simplified fractions”
Two of the marks depend on it. $\frac{12}{27}$ has to become $\frac{4}{9}$.
What makes this simple familiar

A spinner with equal sections, three trials, and small powers of $\frac{1}{3}$ and $\frac{2}{3}$. Every probability is a fraction over 27 that you can work out by hand.

If I had to guess where you might lose marks, it would be the simplifying. $\frac{12}{27}$ is the right value, but the mark is for $\frac{4}{9}$. In part b) i., leaving out the $\binom{3}{2}$ gives $\frac{4}{27}$, which is only one of the three orders. In part b) ii., don’t include $X=3$. “Fewer than three” is strictly less than three.

Only mark this done when you could do it without help. Reading the solution does not count.

Back to the index

Question wording and marking-guide steps are from the 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

Spot a mistake on this page? Report it