Random experiments, sample spaces, events and probabilities, including the addition rule, conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$, the multiplication rule, and the concept of independence

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:

• IMATH_19 .
• IMATH_20 .
• IMATH_21 .

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:

• IMATH_38
• IMATH_39
• IMATH_40

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.

Common Paper 1 traps

Forgetting to subtract the overlap in the addition rule. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Skipping the subtraction overcounts.

Confusing $P(A | B)$ with $P(A \cap B)$. The conditional probability divides by $P(B)$; the intersection probability does not.

Confusing independence with mutual exclusivity. These are different concepts and usually contradict each other (for events with positive probabilities).

Assuming independence without justification. Two events are not independent just because they "feel unrelated". Always test using $P(A \cap B) = P(A) P(B)$.

Treating "without replacement" as independent. Drawing without replacement creates dependence: the second probability is conditional on the first.

In one sentence

Probability of an event is computed from counts when outcomes are equally likely; the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ handles unions, the multiplication rule $P(A \cap B) = P(A | B) P(B)$ handles intersections via conditional probability, and independence holds when $P(A \cap B) = P(A) P(B)$, which is different from mutual exclusivity.