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

Worked example

A bag contains $4$ red and $6$ blue marbles. Two are drawn without replacement.

$P(\text{both red}) = \dfrac{4}{10} \cdot \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$.

Alternatively using combinations: $\dfrac{^4C_2}{^{10}C_2} = \dfrac{6}{45} = \dfrac{2}{15}$.

Common traps

Using $^nP_r$ instead of $^nC_r$. For unordered selections, use combinations.

Forgetting "without replacement" reduces the population. After one draw, the second draw is from $9$, not $10$.

Treating dependent events as independent. Conditional probability is needed when events influence one another.

Forgetting the intersection in the addition rule. $P(A) + P(B)$ counts $P(A \cap B)$ twice.

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).