Mutually Exclusive Events Revision Notes for HSC SSCE Mathematics Advanced

Mutually Exclusive Events

What are mutually exclusive events?

Mutually exclusive events are two events that cannot happen at the same time in the same chance experiment. When one event occurs, the other cannot possibly occur in that same trial.

For example, when a fair coin is tossed twice, the events 'getting two heads (HH)' and 'getting two tails (TT)' cannot occur at the same time. These events are mutually exclusive.

When two events $A$ and $B$ are mutually exclusive, they have no outcomes in common. Mathematically, we write this as:

$A \cap B = \emptyset$

This means the intersection of events $A$ and $B$ is the empty set (contains no elements).

infoNote

The symbol $\emptyset$ represents the empty set, which contains no elements at all. When $A \cap B = \emptyset$ , it means there are absolutely no outcomes that belong to both events simultaneously.

Venn diagram representation

In a Venn diagram, mutually exclusive events are shown as separate circles that do not overlap within the sample space $S$ .

This Venn diagram illustrates mutually exclusive events $A$ and $B$ . Notice that:

The two circles do not touch or overlap
The intersection $A \cap B = \emptyset$ (empty set)
The union $A \cup B$ includes all outcomes from both events
Each circle sits within the universal set $S$ (the sample space)

chatImportant

If $A \cap B \neq \emptyset$ , then the events are not mutually exclusive because they share at least one common outcome. In this case, the circles in a Venn diagram would overlap.

Understanding through examples

Card drawing example

Consider drawing a card from a standard deck of 52 cards. Let event $A$ be drawing a heart and event $B$ be drawing a spade.

Are these events mutually exclusive? Yes, because:

A card cannot be both a heart and a spade simultaneously
There are no cards that belong to both suits
Therefore $A \cap B = \emptyset$

This is an example of mutually exclusive events. If we instead defined event $B$ as drawing a face card, then $A$ and $B$ would not be mutually exclusive, since cards like the King of Hearts belong to both events.

Worked example 1: Identifying mutually exclusive events

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Worked Example: Identifying Mutually Exclusive Events with a Die

A fair six-sided die is rolled, with sample space $S = \{1, 2, 3, 4, 5, 6\}$ .

Part a) Define event $A$ as rolling a number less than 3.

To find event $A$ , we identify all outcomes in $S$ that are less than 3.

$A = \{1, 2\}$

Part b) Define event $B$ as rolling a number greater than 4. Show that $A$ and $B$ are mutually exclusive.

First, find event $B$ by identifying outcomes greater than 4:

$B = \{5, 6\}$

Now check if $A \cap B = \emptyset$ :

$A \cap B = \{1, 2\} \cap \{5, 6\} = \emptyset$

Since the intersection is empty, events $A$ and $B$ have no common outcomes.

Therefore, $A$ and $B$ are mutually exclusive.

Part c) Find $P(A \cup B)$ .

For mutually exclusive events, we can use the simplified addition rule:

$P(A \cup B) = P(A) + P(B)$

First, calculate $P(A)$ :

$|A| = 2$ and $|S| = 6$

$P(A) = \frac{|A|}{|S|} = \frac{2}{6} = \frac{1}{3}$

Next, calculate $P(B)$ :

$|B| = 2$ and $|S| = 6$

$P(B) = \frac{|B|}{|S|} = \frac{2}{6} = \frac{1}{3}$

Now find $P(A \cup B)$ :

$P(A \cup B) = P(A) + P(B) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Probability rules for events

Two important probability rules help us calculate the likelihood of events.

The complement rule

The complement rule states that the probability of an event not occurring is:

$P(\overline{A}) = 1 - P(A)$

where:

$P(\overline{A})$ is the probability that event $A$ does not occur
$P(A)$ is the probability that event $A$ occurs

For example, if $P(A) = \frac{1}{3}$ , then:

$P(\overline{A}) = 1 - \frac{1}{3} = \frac{2}{3}$

infoNote

The complement of an event $A$ includes all outcomes in the sample space that are not in $A$ . Since probabilities must sum to 1, we can find $P(\overline{A})$ by subtracting $P(A)$ from 1.

The addition rule

The addition rule for the union of two events is:

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

where:

$P(A \cup B)$ is the probability that $A$ or $B$ occurs
$P(A \cap B)$ is subtracted to avoid double-counting outcomes that appear in both events

Simplified addition rule for mutually exclusive events

For mutually exclusive events, $P(A \cap B) = 0$ because they share no common outcomes.

This simplifies the addition rule to:

$P(A \cup B) = P(A) + P(B)$

chatImportant

This is a key advantage when working with mutually exclusive events - we simply add the individual probabilities together without needing to adjust for overlap.

These rules can be visualised using Venn diagrams, where $P(A \cup B)$ corresponds to the combined area of circles $A$ and $B$ , adjusted for any overlap.

Worked example 2: Applying probability rules

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Worked Example: Applying Complement and Addition Rules

A fair six-sided die is rolled, with sample space $S = \{1, 2, 3, 4, 5, 6\}$ .

Part a) For event $A$ (even numbers), find $P(A)$ .

Event $A$ consists of the even numbers on the die:

$A = \{2, 4, 6\}$

Calculate the probability:

$|A| = 3$ and $|S| = 6$

$P(A) = \frac{|A|}{|S|} = \frac{3}{6} = \frac{1}{2}$

Part b) Find $P(\overline{A})$ .

Use the complement rule $P(\overline{A}) = 1 - P(A)$ , where $P(A) = \frac{1}{2}$ from part (a).

$P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$

Part c) Let $B$ be the numbers greater than 4. Find $P(B)$ and $P(A \cap B)$ .

Event $B$ consists of numbers greater than 4:

$B = \{5, 6\}$

Calculate $P(B)$ :

$|B| = 2$ and $|S| = 6$

$P(B) = \frac{|B|}{|S|} = \frac{2}{6} = \frac{1}{3}$

Now find $A \cap B$ (outcomes in both $A$ and $B$ ):

$A \cap B = \{2, 4, 6\} \cap \{5, 6\} = \{6\}$

So $|A \cap B| = 1$ , and we can calculate:

$P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{1}{6}$

Therefore, $P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{6}$ .

Part d) Find $P(A \cup B)$ .

Use the addition formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6}$

Convert to a common denominator of 6:

$P(A \cup B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

bookmarkSummary

Key Points to Remember:

Mutually exclusive events cannot occur simultaneously in the same experiment, so $A \cap B = \emptyset$
In Venn diagrams, mutually exclusive events are represented by non-overlapping circles within the sample space
The complement rule is $P(\overline{A}) = 1 - P(A)$ , giving the probability that event $A$ does not occur
The addition rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ for any two events
For mutually exclusive events, this simplifies to $P(A \cup B) = P(A) + P(B)$ because $P(A \cap B) = 0$

Mutually Exclusive Events (HSC SSCE Mathematics Advanced): Revision Notes