Worked Solutions
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.
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.