Probability, conditional probability and independence: VCE Math Methods Unit 3

What this dot point is asking

VCAA wants you to compute probabilities of events using the addition rule, conditional probability and the multiplication rule, and to test whether two events are independent. These are the building blocks for the binomial distribution (also in Unit 3) and the normal distribution (Unit 4).

Sample spaces and events

A random experiment has a set of possible outcomes. The sample space $S$ is the set of all outcomes; an event $A$ is a subset of $S$.

When all outcomes in $S$ are equally likely:

That is, the number of outcomes in $A$ divided by the total number of outcomes.

Set operations and probabilities

For events $A$ and $B$:

• Union $A \cup B$: "$A$ or $B$ (or both)" occurs.
• Intersection $A \cap B$: "$A$ and $B$" both occur.
• Complement $A'$: "not $A$".

Basic identities:

• $P(A') = 1 - P(A)$.
• $0 \leq P(A) \leq 1$.
• $P(S) = 1$.

The addition rule

For any two events:

The subtraction avoids double-counting the overlap.

If $A$ and $B$ are mutually exclusive (cannot both happen), then $P(A \cap B) = 0$ and the rule reduces to $P(A \cup B) = P(A) + P(B)$.

Conditional probability

The probability of $A$ given that $B$ has occurred is:

provided $P(B) > 0$. Conditional probability restricts the sample space to outcomes where $B$ has happened, then asks how often $A$ also occurs.

Worked example

In a class of 30 students, 18 study Methods, 12 study Specialist, and 8 study both. A student is chosen at random.

$P(\text{Methods}) = 18/30 = 3/5$.

$P(\text{Specialist}) = 12/30 = 2/5$.

$P(\text{Methods} \cap \text{Specialist}) = 8/30 = 4/15$.

$P(\text{Specialist} | \text{Methods}) = \frac{P(\text{Spec} \cap \text{Meth})}{P(\text{Meth})} = \frac{8/30}{18/30} = \frac{8}{18} = \frac{4}{9}$.

So among students taking Methods, $4/9$ also take Specialist.

The multiplication rule

Rearranging the conditional definition:

This is the key rule for "and" events that are dependent. For drawing without replacement, the second probability is conditional on the outcome of the first.

Independence

Events $A$ and $B$ are independent if knowing one occurred does not change the probability of the other. Equivalent conditions:

• $P(A | B) = P(A)$
• $P(B | A) = P(B)$
• $P(A \cap B) = P(A) \cdot P(B)$

Any one of these tests independence; if it holds, the others hold too.

Caution: independence is not the same as mutual exclusivity. If $A$ and $B$ are mutually exclusive with positive probabilities, they cannot be independent (since $P(A \cap B) = 0 \neq P(A) P(B)$).

Worked example

Roll two fair dice. Let $A$ = "first die is 6" and $B$ = "sum is 7".

$P(A) = 1/6$. $P(B) = 6/36 = 1/6$ (the pairs $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$).

$A \cap B$ = "first die 6 and sum 7" = $\{(6, 1)\}$, so $P(A \cap B) = 1/36$.

Check: $P(A) P(B) = 1/6 \times 1/6 = 1/36 = P(A \cap B)$. Independent.

Tree diagrams and tables

For multi-stage experiments, probability tree diagrams are the standard tool. Each branch shows a conditional probability; multiply along a path to get the probability of that path.

For two events $A$ and $B$ with given probabilities, two-way tables organise the four intersections $A \cap B$, $A \cap B'$, $A' \cap B$, $A' \cap B'$. Use the row and column totals to fill in gaps.

Examples in context

Example 1. Medical screening interpretation. A clinic finds that $30\%$ of patients have a condition ($P(C) = 0.3$), and a test is positive for $90\%$ of those who have it. The probability a patient both has the condition and tests positive is $P(C \cap +) = P(+ \mid C)P(C) = 0.9 \times 0.3 = 0.27$. This multiplication-rule step is the foundation for reasoning about screening accuracy.

Example 2. Drawing components without replacement. A drawer holds $4$ working and $2$ faulty fuses. Two are taken without replacement. The probability both work is $P(W_1 \cap W_2) = \frac{4}{6} \times \frac{3}{5} = \frac{12}{30} = \frac{2}{5}$. The second factor is conditional ($\frac{3}{5}$, not $\frac{4}{6}$) because the first draw is not replaced, so the events are dependent.

Try this

Q1. Events $A, B$ have $P(A) = 0.6$, $P(B) = 0.3$, $P(A \cap B) = 0.18$. Find $P(A \cup B)$ and decide if they are independent. [3 marks]

• Cue. $P(A \cup B) = 0.6 + 0.3 - 0.18 = 0.72$; $P(A)P(B) = 0.18 = P(A \cap B)$, so independent.

Q2. Given $P(A \cap B) = 0.15$ and $P(B) = 0.5$, find $P(A \mid B)$. [2 marks]

• Cue. $P(A \mid B) = \frac{0.15}{0.5} = 0.3$.

Q3. A box has $5$ green and $3$ yellow balls. Two drawn without replacement. Find $P(\text{both green})$. [3 marks]

• Cue. $\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.