This question aims to find statements representing mutually exclusive **events** when events $A$ and $B$ are **mutually exclusive.**

Two separate events are called **mutually exclusive** if they do not occur at the same time or simultaneously. For example, when we **toss** one **coin,** there are two possibilities whether the **head** will be displayed or the **tail** will be displayed on its return. It means both heads and tails **cannot occur** at the **same time.** It is a **mutually exclusive** event, and the **probability** of these events occurring at the same time becomes **zero.**

There is another name for mutually exclusive events, and that is **disjoint event.**

**Mutually Exclusive Events** can be represented as:

\[P (A \cap B) = 0\]

## Expert Answer

The addition rule for **disjoint events** is only valid when the sum of two events occurring gives the **probability** of either event occurring. If we consider **two events** $A$ or $B$, then their **probability** of occurrence is given by:

\[P (A \cup B) = P (A) + P (B)\]

When two events, $A$ and $B$, are not **mutually exclusive** events, then the formula changes to:

\[ P (A \cup B) = P (A) + P (B) – P (A \cap B)\]

If we consider that $A$ and $B$ are **mutually exclusive** events which means the **probability** of their occurrence at the same time becomes **zero**, it can be shown as:

\[P (A \cap B) = 0 \hspace {0.4 in} Eq.1\]

From **addition rule** of **probability:**

\[ P (A \cup B) = P (A) + P (B) – P (A \cap B) \hspace {0.4 in} Eq.2\]

By putting $Eq.1$ into $Eq.2$, we get:

\[ P (A \cup B) = P (A) + P (B) – 0\]

## Numerical Solution

We get the following statement:

\[P (A \cup B) = P (A) + P (B)\]

This statement shows that the **two events** $A$ and $B$ are **mutually exclusive.**

## Example

When we **roll** a **die,** the **probability** of **occurrence** of both $3$ and $5$ **simultaneously** is **zero.** In this case, either $5$ will occur or $3$ will occur.

Similarly, the **probability** of a **die** to show a **number** $3$ or $5$ is:

Let $P(3)$ become the **probability** of getting $3$, while $P(5)$ is the **probability** of getting $5$, then:

\[ P (3) = \frac {1} {6} , P (5) = \frac {1} {6}\]

From the formula:

\[P (A \cup B) = P (A) + P (B)\]

\[P (3 \cup 5) = P (3) + P (5)\]

\[P (3 \cup 5) = (\frac {1} {6}) + (\frac {1} {6})\]

\[P (3 \cup 5) = (\frac {2} {6})\]

\[P (3 \cup 5) = \frac {1} {3}\]

**The probability of the die showing $3$ or $5$ is $\frac {1} {3}$.**