Statistics & Probability — Worked Solutions (Year 10 Maths)
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Worked examples for Year 10 Maths Statistics & Probability. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Year 10 Maths — Statistics & Probability. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Box plots and the IQR
Question
The reaction times (in seconds) of $11$ students are: $4, 5, 5, 6, 7, 8, 8, 9, 10, 12, 18$. Find the five-number summary and the interquartile range, then determine whether $18$ is an outlier using the $1.5 \times \text{IQR}$ rule.
Solution
The data is already in order, and there are $11$ values, so the median is the $6$th: $Q_2 = 8$.
Split off the median. The lower half is the first five values $4, 5, 5, 6, 7$, so $Q_1$ is the middle one $= 5$. The upper half is $8, 9, 10, 12, 18$, so $Q_3 = 10$. Minimum $= 4$, maximum $= 18$.
Five-number summary: $4,\ 5,\ 8,\ 10,\ 18$. IQR $= Q_3 - Q_1 = 10 - 5 = 5$.
Outlier test: upper fence $= Q_3 + 1.5 \times \text{IQR} = 10 + 7.5 = 17.5$. Since $18 > 17.5$, yes — $18$ is an outlier.
Let's build the five-number summary step by step. With $11$ values in order, the median sits exactly in the middle at the $6$th position, which is $8$, so $Q_2 = 8$.
To find the quartiles we look at each half excluding that middle value. The lower half is $4, 5, 5, 6, 7$ and its middle value is $5$, so $Q_1 = 5$. The upper half is $8, 9, 10, 12, 18$ with middle value $10$, so $Q_3 = 10$. The smallest and largest values are $4$ and $18$.
The interquartile range measures the spread of the middle half: IQR $= Q_3 - Q_1 = 10 - 5 = 5$.
Now the outlier rule. A value is unusually high if it sits more than $1.5$ IQRs above $Q_3$. That upper fence is $10 + 1.5 \times 5 = 10 + 7.5 = 17.5$. Because $18$ is just past $17.5$, it counts as an outlier — which matches our intuition that one student was much slower than the rest.
$n = 11$, ordered.
- Median $Q_2$ = 6th value $= 8$
- Lower half $4,5,5,6,7$ → $Q_1 = 5$
- Upper half $8,9,10,12,18$ → $Q_3 = 10$
- Min $= 4$, Max $= 18$
Summary: $4, 5, 8, 10, 18$. IQR $= 10 - 5 = 5$.
Outlier: upper fence $= 10 + 1.5(5) = 17.5$; $18 > 17.5$ → outlier.
Where the marks go
- 1 mark: Correct median $Q_2 = 8$
- 1 mark: Correct quartiles $Q_1 = 5$ and $Q_3 = 10$
- 1 mark: Correct IQR $= 5$
- 1 mark: Applies the $1.5 \times \text{IQR}$ rule and concludes $18$ is an outlier
Key idea
Find the median first, then the quartiles from each half; the IQR measures the middle spread, and a value beyond $Q_3 + 1.5 \times \text{IQR}$ is an outlier.
Example 2 — Tree diagrams and conditional probability
Question
A bag contains $3$ red and $5$ blue marbles. Two marbles are drawn without replacement. Find the probability that both are red, and the probability that the second is red given the first was red.
Solution
Without replacement means the second draw depends on the first, so the denominator drops by one.
First draw red: $P = \tfrac{3}{8}$. After removing one red, $2$ red remain out of $7$, so second red: $P = \tfrac{2}{7}$.
Both red: multiply along the branch, $\tfrac{3}{8} \times \tfrac{2}{7} = \tfrac{6}{56} = \tfrac{3}{28}$.
The conditional probability $P(\text{2nd red} \mid \text{1st red})$ is exactly that second-branch value, $\tfrac{2}{7}$. Once the first red is fixed, only $7$ marbles and $2$ reds are left.
Drawing without replacement is the heart of this question — once a marble is taken out, it changes what's left for the next draw, so the two events aren't independent.
On the first draw there are $3$ reds out of $8$ marbles, so $P(\text{1st red}) = \tfrac{3}{8}$.
If that first marble was red, the bag now holds $2$ reds among $7$ marbles, so $P(\text{2nd red} \mid \text{1st red}) = \tfrac{2}{7}$.
To get both red we follow that path through the tree and multiply: $\tfrac{3}{8} \times \tfrac{2}{7} = \tfrac{6}{56}$, which simplifies to $\tfrac{3}{28}$.
And the conditional probability they ask for is simply that second branch, $\tfrac{2}{7}$ — it's the chance of the second event given the first has already happened, which is why we used the reduced totals.
Without replacement → totals reduce.
- $P(\text{1st red}) = \tfrac{3}{8}$
- $P(\text{2nd red} \mid \text{1st red}) = \tfrac{2}{7}$
- $P(\text{both red}) = \tfrac{3}{8} \times \tfrac{2}{7} = \tfrac{6}{56} = \tfrac{3}{28}$
Conditional $= \tfrac{2}{7}$.
Where the marks go
- 1 mark: Correct first-draw probability $\tfrac{3}{8}$
- 1 mark: Reduces totals correctly for the second draw $\tfrac{2}{7}$
- 1 mark: Multiplies along the branch for $P(\text{both red}) = \tfrac{3}{28}$
- 1 mark: States the conditional probability $\tfrac{2}{7}$
Key idea
Without replacement, reduce both the favourable count and the total for the second draw; multiply along a branch for 'and', and the conditional probability is just that second branch.
Frequently asked questions
Step-by-step solutions to Statistics & Probability questions in Year 10 Maths, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Year 10 Maths, using the methods and notation expected in exams and assessments.