Counting principles (multiplication principle, permutations and combinations), set notation, simple probability, conditional probability and the addition / multiplication rules

What this dot point is asking

VCAA wants you to apply counting principles (multiplication, permutations, combinations) to count outcomes in probability problems, use set notation to describe events, and compute simple probabilities, conditional probabilities and combined probabilities via the addition and multiplication rules.

Counting principles

The multiplication principle

If an event can occur in $m$ ways followed by another in $n$ ways, the combined event can occur in $m \times n$ ways.

Example. A menu has 4 mains and 3 desserts. The number of main-plus-dessert combinations is $4 \times 3 = 12$.

Permutations

A permutation is an arrangement of items in order. The number of ways to arrange $n$ distinct items in order is:

The number of ways to choose and arrange $k$ items from $n$ (order matters) is:

Example. Number of ways to award gold, silver, bronze from 8 finalists: $P(8, 3) = 8! / 5! = 8 \cdot 7 \cdot 6 = 336$.

Combinations

A combination is a selection without regard to order. The number of ways to choose $k$ items from $n$ (order does not matter) is:

Example. Number of ways to choose 5 students from 30: $\binom{30}{5} = 142,506$.

Set notation for events

In probability, an event is a set of outcomes.

• Sample space $S$: the set of all possible outcomes.
• Event $A$: a subset of $S$.
• Union $A \cup B$: outcomes in $A$ or $B$ (or both).
• Intersection $A \cap B$: outcomes in both $A$ and $B$.
• Complement $A'$ or $\bar{A}$: outcomes not in $A$.

Simple probability

For a sample space with equally likely outcomes:

Properties:

• $0 \leq P(A) \leq 1$.
• $P(S) = 1, P(\emptyset) = 0$.
• $P(A') = 1 - P(A)$.

The addition rule

For mutually exclusive events (no overlap), $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.

Conditional probability

The probability of $A$ given that $B$ has occurred:

(provided $P(B) > 0$).

Interpretation: conditional probability restricts attention to outcomes where $B$ has occurred.

The multiplication rule

For independent events, $P(A | B) = P(A)$, so $P(A \cap B) = P(A) \cdot P(B)$.

Independence test

Events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$. Equivalently, $P(A | B) = P(A)$.

Worked example: conditional probability

In a class of 30 students, 18 study Maths, 12 study Physics, and 8 study both.

$P(\text{Maths}) = 18/30 = 0.6$.

$P(\text{Physics}) = 12/30 = 0.4$.

$P(\text{Both}) = 8/30 \approx 0.267$.

$P(\text{Physics} | \text{Maths}) = P(\text{Both}) / P(\text{Maths}) = (8/30) / (18/30) = 8/18 \approx 0.444$.

Among Maths students, the conditional probability of also doing Physics is about 0.444.

Worked example: tree diagrams

A box contains 5 red and 3 blue balls. Two are drawn without replacement.

$P(\text{both red}) = P(\text{first red}) \cdot P(\text{second red} | \text{first red}) = (5/8) \cdot (4/7) = 20/56 = 5/14$.

$P(\text{first red, second blue}) = (5/8) \cdot (3/7) = 15/56$.

Tree diagrams help visualise: at each node, the conditional probability depends on what has been drawn already.

Examples in context

Example 1. Arranging a roster. A cafe must roster $3$ of its $7$ baristas to the three distinct shifts (morning, midday, evening) on a given day. Since each shift is a different role, order matters, so the count is a permutation: $P(7, 3) = \frac{7!}{4!} = 7 \times 6 \times 5 = 210$ ways. If instead the manager just picks a team of $3$ with no shift assignment, order does not matter, giving $\binom{7}{3} = 35$ teams.

Example 2. Conditional probability in screening. Of $200$ surveyed shoppers, $80$ own a loyalty card and $50$ of those used a discount; $30$ non-card-holders used a discount. The probability a shopper used a discount given they hold a card is $P(\text{discount}\mid\text{card}) = \frac{50}{80} = 0.625$. Compare $P(\text{discount}) = \frac{80}{200} = 0.40$: since $0.625 \neq 0.40$, using a discount is not independent of holding a card.

Try this

Q1. A PIN uses $4$ digits from $0$ to $9$ with no repeats. How many are possible? [2 marks]

• Cue. Permutation $P(10, 4) = 10 \times 9 \times 8 \times 7 = 5040$.

Q2. Events $A$ and $B$ have $P(A) = 0.5$, $P(B) = 0.4$, $P(A \cap B) = 0.2$. (a) Find $P(A \cup B)$. (b) Are $A$ and $B$ independent? [2+2 marks]

• Cue. (a) $0.5 + 0.4 - 0.2 = 0.7$. (b) $P(A)P(B) = 0.5 \times 0.4 = 0.2 = P(A \cap B)$, so yes independent.

Q3. A bag has $4$ red and $6$ green discs. Two are drawn without replacement. Find the probability both are green. [3 marks]

• Cue. $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$.