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:

• IMATH_30 .
• IMATH_31 .
• IMATH_32 .

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.

Common errors

Confusing permutations and combinations. Permutations: order matters. Combinations: order does not.

Forgetting the overlap. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Forgetting the subtraction double-counts.

Conditional probability backwards. $P(A | B)$ is generally not equal to $P(B | A)$. Bayes's theorem relates the two.

Treating dependent events as independent. Drawing without replacement: the second draw depends on the first. Always check whether sampling is with or without replacement.

Misreading the question. "At least one" usually means "1 - none". "Exactly one" is different from "at least one".

In one sentence

Unit 1 probability combines counting principles (multiplication rule, permutations, combinations), set notation for events, simple probability ($P = |A|/|S|$ for equally likely outcomes), the addition rule ($P(A \cup B) = P(A) + P(B) - P(A \cap B)$), conditional probability ($P(A|B) = P(A \cap B)/P(B)$), and the multiplication rule, with independence ($P(A \cap B) = P(A) P(B)$) as a special case that simplifies many calculations.