Counting techniques (multiplication principle, permutations and combinations), simple probability, conditional probability and the addition and multiplication rules

What this dot point is asking

QCAA wants Year 11 students to apply counting techniques (multiplication, permutations, combinations) to compute probabilities, use set notation, and apply conditional probability and independence.

Counting principles

Multiplication principle. Two events: first in $m$ ways, second in $n$ ways. Combined: $m \times n$ ways.

Permutations. Arrangements where order matters. Choose $k$ from $n$ in order: $P(n, k) = n!/(n-k)!$.

Combinations. Selections where order does not matter. Choose $k$ from $n$: $\binom{n}{k} = n!/(k!(n-k)!)$.

Set notation

• Sample space $S$: all outcomes.
• Event $A$: subset of $S$.
• Union $A \cup B$: in $A$ or $B$.
• Intersection $A \cap B$: in both.
• Complement $A'$: not in $A$.

Simple probability

For equally likely outcomes: $P(A) = |A| / |S|$.

Properties: $0 \leq P(A) \leq 1$. $P(A') = 1 - P(A)$.

Addition rule

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

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

Conditional probability

$P(A | B) = P(A \cap B) / P(B)$ for $P(B) > 0$.

Multiplication rule

$P(A \cap B) = P(A | B) P(B)$.

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

Common errors

Permutation vs combination confusion. Order matters: permutation. Order does not: combination.

Forgetting overlap. Use addition rule with subtraction.

Conditional probability inversion. $P(A | B) \neq P(B | A)$ in general.

Treating dependent as independent. Without-replacement sampling is dependent.

In one sentence

Unit 1 introduces counting principles (multiplication, $P(n,k) = n!/(n-k)!$ for permutations, $\binom{n}{k} = n!/(k!(n-k)!)$ for combinations) and probability rules (simple $P = |A|/|S|$, addition $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, conditional $P(A|B) = P(A \cap B)/P(B)$, multiplication $P(A \cap B) = P(A|B) P(B)$, independence $P(A|B) = P(A)$).