Mathematical Methods · Past QCAA questions All simple familiar questions
2023 · Paper 1 · Simple familiar

Question 15

[4 marks] Technology-free
Topic guidesPremiumUnit 3 Topic 5.2 · Bernoulli distributionsUnit 3 Topic 5.2 · Mean, variance and standard deviation
Game plansPremium3.1 Bernoulli distributions3.2 Binomial distributions
The question

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]
Watch the situation first

One throw, then twenty

Watch one throw split into a hit and a miss. Then 20 throws build a binomial distribution, with its mean and its spread.

One throw has two outcomes: hit or miss One in four hit, so three in four miss 20 players each throw once: H counts the hits The mean of a binomial is np The variance is np(1 − p) Hit Miss 14 34 H ∼ Bin(20, 14 ) E(H) = np = 20 × 14 = 5 Var(H) = np(1 − p) = 5 × 34 = 154 05101520 Number of hits, H μ = 5 σσ

Part a) gives you $p$ and the question gives you $n=20$. After that, the mean and the variance are one line each.

Now for the mathematics

Choose what $H$ counts

The question lets $H$ count either outcome. Choose hits or misses, then change the number of players. The orange marker is the mean, and the blue bracket is one standard deviation each side of it.

00.10.20.30.4 Probability μ σσ
The mean, One standard deviation,
counts the

This is part b) with $H$ counting hits, so $p=\frac{1}{4}$. The mean is 5 hits and the variance is $\frac{15}{4}$. The orange marker sits at 5, where the bars balance.

With $H$ counting misses, $p=\frac{3}{4}$, so the mean is 15. The marking guide accepts 15 for the mean when $H$ represents misses. The variance is still $\frac{15}{4}$: the bars are the mirror image of the hits, so the spread is the same.

The question has 20 players. With a different number, the formulas stay the same but the mean and variance change. Both grow in proportion to the number of players.

QCAA marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. Part a) has a mark for each probability. Part b) has a mark for the mean and one for the variance.

Step 1 · part a)1 mark

One throw has two possible outcomes: a hit or a miss. “A one in four chance of hitting” is the probability of a hit.

$$P(\text{hit})=\frac{1}{4}$$

Marker: correctly states the probability of a ‘hit’

Step 2 · part a)1 mark

A hit and a miss are the only possible outcomes, so their probabilities add to 1.

$$P(\text{miss})=1-\frac{1}{4}=\frac{3}{4}$$

$P(\text{hit})=\dfrac{1}{4}$,$P(\text{miss})=\dfrac{3}{4}$

You don’t have to name the outcomes hit and miss to earn these marks. Other formats are accepted too, such as a list or a table, with or without headings.

Marker: correctly states the probability of a ‘miss’

Step 3 · part b)1 mark

Let $H$ be the number of hits. Each of the 20 players throws once, each throw is a hit or a miss, and the chance of a hit is $\frac{1}{4}$ every time. So $H$ is binomial with $n=20$ and $p=\frac{1}{4}$. The mean of a binomial is $np$.

$$\begin{aligned}&H\sim\text{Bin}\left(20,\ \tfrac{1}{4}\right)\\&E(H)=np=20\times\frac{1}{4}=5\end{aligned}$$

If you let $H$ count misses instead, a mean of 15 is accepted. Follow-through applies if your probabilities from part a) had a slip.

Marker: determines the mean

Step 4 · part b)1 mark

The variance of a binomial is $np(1-p)$. You already have $np=5$, so multiply it by $1-p=\frac{3}{4}$.

$$\begin{aligned}\text{Var}(H)&=np(1-p)\\&=5\times\frac{3}{4}\\&=\frac{15}{4}=3\frac{3}{4}\end{aligned}$$

Mean $=5$,variance $=\dfrac{15}{4}$

Other formats for the variance are accepted, such as $5\times\frac{3}{4}$, $\frac{15}{4}$ or 3.75. Follow-through applies if your mean had a slip. Counting misses gives the same variance, because $20\times\frac{3}{4}\times\frac{1}{4}$ is the same product.

Marker: determines the variance

Putting it all together

Each phrase of the question gave you something

“a one in four chance of hitting the target”
The probability of a hit, $\frac{1}{4}$.
“all the possible outcomes for one throw”
Hit and miss. The probabilities add to 1, so a miss is $\frac{3}{4}$. One mark each.
“Let $H$ be the discrete random variable for one of the possible outcomes”
$H$ counts hits (or misses) across the throws, so it is binomial.
“when 20 players throw a water balloon”
The number of trials, $n=20$.
“the mean and variance”
$np=5$ and $np(1-p)=\frac{15}{4}$. One mark each.
What makes this simple familiar

Part a) is reading one probability from the words and using the fact that the two outcomes add to 1. Part b) is the binomial mean and variance formulas, and the numbers stay as simple fractions.

If I had to guess where you might lose marks, it would be the variance. It is $np(1-p)$, not $np$, so 5 is the mean, not the variance. Don’t give the standard deviation, $\sqrt{\frac{15}{4}}\approx1.94$, when the question asks for the variance. And in part a), state both probabilities. The miss has a mark of its own.

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

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Question wording and marking-guide steps are from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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