Apply the rules of probability (addition, multiplication, conditional), permutations and combinations to calculate probabilities of compound events

What this dot point is asking

QCAA wants you to apply the rules of probability and counting techniques to find probabilities of compound events.

Probability rules

For equally likely outcomes, the probability of an event is the proportion of the sample space it covers, $P(A) = \dfrac{\text{favourable}}{\text{total}}$, and every probability lies between $0$ and $1$. The complement rule $P(A^c) = 1 - P(A)$ is often the quickest route, especially for "at least one" questions where the complement "none" is a single easy calculation.

Addition rule

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$, subtracting the overlap so it is not counted twice. For mutually exclusive events $P(A \cap B) = 0$, so the rule reduces to $P(A) + P(B)$.

Multiplication rule

$P(A \cap B) = P(A) \cdot P(B \mid A)$, the chance of the first times the chance of the second given the first. For independent events $P(B \mid A) = P(B)$, so $P(A \cap B) = P(A)P(B)$.

Conditional probability

$P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$, the probability of $B$ restricted to the world in which $A$ has happened. A tree diagram organises multi-stage experiments, with branch probabilities multiplied along a path and added across paths.

Counting

Multiplication principle

$n$ ways for task $1$ and $m$ ways for task $2$ gives $nm$ ways combined.

Permutations (order matters)

$^nP_r = \dfrac{n!}{(n-r)!}$.

Combinations (order does not matter)

$^nC_r = \dfrac{n!}{r! (n-r)!}$.

With repetition

$n^r$ ways to choose $r$ from $n$ with replacement, since each of the $r$ choices independently has $n$ options.

The bridge from counting to probability is that, for equally likely outcomes, a probability is a count of favourable selections divided by a count of total selections. Permutations and combinations supply those counts, which is why fluent counting is a prerequisite for compound-event probability.

Choosing the right tool

Two routes to the same probability

Many selection probabilities can be computed two equivalent ways: by sequential conditional probabilities (multiplying the chance at each draw) or by counting combinations (favourable selections over total selections). The conditional route emphasises that without-replacement draws are dependent, while the combination route is often faster for "exactly $k$ of a type" questions. Knowing both, and that they agree, is a useful self-check.

In one sentence

Probability rules (addition $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, multiplication $P(A \cap B) = P(A) P(B|A)$, conditional $P(B|A) = P(A \cap B)/P(A)$) combine with counting principles (multiplication, permutations $^nP_r = n!/(n-r)!$ for ordered selections, combinations $^nC_r$ for unordered) to compute probabilities of compound events.