Probability — Worked Solutions (Preliminary Maths Advanced)
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Worked examples for Preliminary Maths Advanced 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 Preliminary Maths Advanced — 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 — Complementary and addition rules
Question
A standard six-sided die is rolled once. Find the probability of rolling a number that is even or greater than $4$.
Solution
Use the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Even numbers: $\{2, 4, 6\}$, so $P(\text{even}) = \frac{3}{6}$. Greater than $4$: $\{5, 6\}$, so $P(>4) = \frac{2}{6}$.
Overlap (even and $>4$): $\{6\}$, so $P(A \cap B) = \frac{1}{6}$.
$P = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.
Subtract the overlap — counting $6$ twice is the classic mistake here.
"Or" signals the addition rule, but with a catch: if an outcome satisfies both conditions we'd count it twice, so we subtract the overlap. The rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
List the outcomes. Even numbers are $\{2, 4, 6\}$, giving $P(A) = \frac{3}{6}$. Numbers greater than $4$ are $\{5, 6\}$, giving $P(B) = \frac{2}{6}$. The number $6$ is in both lists, so $P(A \cap B) = \frac{1}{6}$.
Now combine: $\frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.
So the probability is $\frac{2}{3}$. You can sanity-check by listing the favourable set directly — $\{2, 4, 5, 6\}$, which is $4$ of the $6$ outcomes.
Addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
- Even $\{2,4,6\}$: $P(A) = \frac{3}{6}$
- $>4$ $\{5,6\}$: $P(B) = \frac{2}{6}$
- Both $\{6\}$: $P(A \cap B) = \frac{1}{6}$
- $\frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
$P = \frac{2}{3}$.
Where the marks go
- 1 mark: Correct individual probabilities $\frac{3}{6}$ and $\frac{2}{6}$
- 1 mark: Identifies the overlap $P(A \cap B) = \frac{1}{6}$
- 1 mark: Correct answer $\frac{2}{3}$
Key idea
For "or" use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, subtracting the overlap so shared outcomes are not double-counted.
Example 2 — Conditional probability
Question
In a class of $30$ students, $18$ study Music and $12$ of those who study Music also study Drama. A student who studies Music is chosen at random. Find the probability that they also study Drama.
Solution
This is conditional probability: $P(D \mid M) = \dfrac{P(D \cap M)}{P(M)}$, but here the condition is given directly — we are already restricted to Music students.
Music students: $18$. Of those, studying Drama: $12$.
So $P(D \mid M) = \dfrac{12}{18} = \dfrac{2}{3}$.
Once you're told the student studies Music, your denominator is $18$, not $30$.
The phrase "a student who studies Music is chosen" tells us we're working inside a smaller world — only the Music students count now. That's the heart of conditional probability: we shrink the sample space to the condition.
There are $18$ Music students in total, and $12$ of them also do Drama. So out of the $18$ we're choosing from, $12$ are favourable.
That gives $P(\text{Drama} \mid \text{Music}) = \dfrac{12}{18} = \dfrac{2}{3}$.
Notice we never use the $30$ — once the condition "studies Music" is fixed, the class total is irrelevant. The denominator is always the size of the group you've conditioned on.
Condition on Music — restrict the sample space.
- Music students: $18$
- Music and Drama: $12$
- $P(D \mid M) = \dfrac{12}{18} = \dfrac{2}{3}$
$P = \frac{2}{3}$.
Where the marks go
- 1 mark: Recognises conditional probability $P(D \mid M)$
- 1 mark: Uses the restricted denominator of $18$ Music students
- 1 mark: Correct answer $\frac{2}{3}$
Key idea
Conditional probability restricts the sample space: $P(D \mid M) = \frac{P(D \cap M)}{P(M)}$, so the denominator becomes the size of the conditioning group.
Frequently asked questions
Step-by-step solutions to Probability questions in Preliminary Maths Advanced, 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 Preliminary Maths Advanced, using the methods and notation expected in exams and assessments.