 # Events A and B are Mutually Exclusive. Which of the following statements is also true?

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:

$P (A \cap B) = 0$

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:

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

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

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 $\frac {1} {3}$.