Apply the rules of probability (addition, multiplication, conditional), the counting principles (permutations and combinations), and use these to find probabilities in compound experiments

What this dot point is asking

VCAA wants you to apply the rules of probability and the counting principles to compute probabilities of compound events, distinguishing situations where order matters (permutations) from those where it does not (combinations).

Probability rules

Probability of an event $A$

$P(A) =$ favourable outcomes / total outcomes (for equally likely outcomes).

Complement

$P(A^c) = 1 - P(A)$.

Addition rule (union of events)

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$. For mutually exclusive events, $P(A \cap B) = 0$.

Multiplication rule (intersection)

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

Conditional probability

$P(B|A) = P(A \cap B)/P(A)$, the probability of $B$ given $A$ has occurred.

Counting

Multiplication principle. If task $1$ has $n_1$ ways and task $2$ has $n_2$ ways, then the combined task has $n_1 \cdot n_2$ ways.

Permutations (order matters). Arrangements of $r$ items from $n$:

Combinations (order doesn't matter). Selections of $r$ from $n$:

When to use which

Independence versus mutual exclusivity

Two terms are routinely confused. Events are mutually exclusive when they cannot both occur, so $P(A \cap B) = 0$ and the addition rule simplifies to $P(A \cup B) = P(A) + P(B)$. Events are independent when one occurring does not change the probability of the other, so $P(A \cap B) = P(A)P(B)$, equivalently $P(B \mid A) = P(B)$. These are different (indeed almost opposite) ideas: two events with non-zero probabilities cannot be both mutually exclusive and independent, because mutual exclusivity forces $P(A \cap B) = 0$ while independence forces $P(A \cap B) = P(A)P(B) > 0$. VCAA frequently asks you to test independence by comparing $P(A \cap B)$ with $P(A)P(B)$.

Tree diagrams and Karnaugh tables

For multi-stage experiments, a tree diagram multiplies along branches (the multiplication rule) and adds across distinct paths (the addition rule). A two-by-two probability table (Karnaugh table) is the fastest tool for problems framed with "and", "or" and "given that", because the margins give $P(A)$ and $P(B)$ while the interior cells give the intersections. Filling the table from the information supplied, then reading off the required cell or margin, avoids most conditional-probability slips.

In one sentence

Probability is the ratio of favourable to total outcomes, governed by the addition rule ($P(A \cup B) = P(A) + P(B) - P(A \cap B)$), the multiplication rule ($P(A \cap B) = P(A) P(B|A)$) and conditional probability ($P(B|A) = P(A \cap B)/P(A)$); counting uses the multiplication principle, permutations ($^nP_r$ when order matters) and combinations ($^nC_r$ when it does not).

Examples in context

Example 1. Selecting a sports squad. A coach picks a netball lineup of $7$ from a squad of $10$. Order on the team sheet does not matter, so the number of possible lineups is $^{10}C_7 = \dfrac{10!}{7!\,3!} = 120$. If exactly $3$ specific players must be included, the remaining $4$ are chosen from the other $7$: $^7C_4 = 35$ lineups.

Example 2. Lottery-style probability. A simple draw selects $3$ numbered balls from $20$ without replacement. The probability of matching a chosen set of $3$ (order irrelevant) is $\dfrac{1}{^{20}C_3} = \dfrac{1}{1140} \approx 0.00088$. Using the multiplication rule directly: $\dfrac{3}{20} \times \dfrac{2}{19} \times \dfrac{1}{18} = \dfrac{6}{6840} = \dfrac{1}{1140}$, confirming the same result.

Try this

Q1. How many ways can $5$ books be arranged on a shelf? [1 mark]

• Cue. $5! = 120$.

Q2. A committee of $3$ is chosen from $5$ teachers and $4$ students. Find the probability it contains exactly $1$ teacher. [3 marks]

• Cue. $\dfrac{^5C_1 \cdot {}^4C_2}{^9C_3} = \dfrac{5 \times 6}{84} = \dfrac{30}{84} = \dfrac{5}{14}$.

Q3. From $8$ red and $2$ white balls, $2$ are drawn without replacement. Find $P(\text{at least one white})$. [3 marks]

• Cue. $1 - P(\text{no white}) = 1 - \dfrac{8}{10}\cdot\dfrac{7}{9} = 1 - \dfrac{56}{90} = \dfrac{34}{90} = \dfrac{17}{45}$.