**P(E) = 0.38**

**P(F) = 0.57**

The of this question is to find the **probability** of **two mutually exclusive events E** and **F** when either of them can occur.

The question is based on the concept of **probability** of **mutually exclusive events. Two events** are **mutually exclusive events** when both of these events **do not occur** at the **same time,** for example when a **die** is **rolled** or when we **toss** a **coin. **The **probability** that it will come **head** or **tail** is completely separate from each other. These **two events** can not occur at the same time, it will be either **head** or **tail.** These types of events are called **mutually exclusive events.**

## Expert Answer

The **probability** that either **E or F** will occur can be calculated by adding the **probabilities** of both of the **events.** The **probabilities** of the **separate** events is given as:

\[ P (E) = 0.38 \]

\[P (F) = 0.57 \]

The **probability** of **two mutually exclusive events** occurring at the **same time** is given by:

\[ P( E\ and\ F) = 0 \]

As these **two events** are **mutually exclusive,** their **probability** of **occurring** at the same time is always **zero.**

The **probability** that either of these **mutually exclusive events** will occur is given by:

\[ P ( E\ or\ F ) = P (E) + P (F) \]

\[ P ( E\ or\ F ) = 0.38 + 0.57 \]

\[ P ( E\ or\ F ) = 0.95 \]

The **probability** that **either** **E** **or F** will occur is **0.95 or 95%.**

## Numerical Result

The **probability** that either **two mutually exclusive events** **E and F** will **occur** is calculated to be:

\[ P ( E\ or\ F ) = 0.95 \]

## Example

Find the **probability P ( G or H )**, if **G and H** are **two mutually exclusive** events. The **probabilities** of the **separate** events are given below:

\[ P (G) = 0.43 \]

\[ P (H) = 0.41 \]

The **probability** that either **G or H** will occur can be calculated by **adding** the **probabilities** of both of the **events.**

The **probability** that either of these **mutually exclusive events** will occur is given by:

\[ P ( G\ or\ H ) = P (E) + P (F) \]

\[ P ( G\ or\ H ) = 0.43 + 0.41 \]

\[ P ( G\ or\ H ) = 0.84 \]

The **probability** of **G and H**, two **mutually exclusive** events, when either of these events can occur is calculated to be **0.84 or 84%.**