**An experiment yields four results, each with $ P ( E_1 ) = 0.2 $, $ P ( E_2 ) = 0.3 $ and $ P ( E_3 ) = 0.4 $. What is the probability of $E_4 $?****An experiment yields four results, each with $ P ( E_1 ) = 0.2 $, $ P ( E_2 ) = 0.2 $ and $ P ( E_3 ) = 0.4 $. What is the probability of $E_4 $?**

The main objective of this question is to find the **probability of an outcome** when two events are **mutually exclusive**.

This question uses the concept of **mutually exclusive events**. When **two occurrences** do not occur **simultaneously**, such as when a die is thrown or when we flip a coin, they are **mutually exclusive**. The likelihood that it will land on its head or its tail is **entirely independent** of one another. These two things **cannot** happen at the **s****ame time**; either the **head or the tail** will come first. Events of this nature are referred to be **mutually exclusive events**.

## Expert Answer

**1)** In this question, we have to find the **probability** of an event when the two events are **mutually exclusive**.

We know that when **events** are** mutually exclusive**:

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

**And**:

\[= \space P ( A u B) = \space P ( A ) \space + \space P (B )- P ( A n B ) \]

By **putting values**, we get:

\[= \space 0.3 \space + \space 0.5 \space – \space 0 \space = \space 0.8\]

**2) **In this** question**, we have to find the **probability** of an event which is $ E_4 $.

So:

We know that **sum of probability** is equal to $ 1 $.

\[P (E4) \space = \space 1 \space – \space 0.2 \space – \space 0.3 \space – \space 0.4 \space = \space 0.1\]

**3) **In this question, we have to find the** probability** of an **event** which is E_4.

**So**:

We know that **sum of probability** is equal to $ 1 $.

\[P (E4) \space = \space 1 \space – \space 0.2 \space – \space 0.2 \space – \space 0.4 \space = \space 0.2\]

## Numerical Answer

- The
**probability**of $ a \cap b $ is $ 0.8 $. - The
**probability of event**which is $ E_4 $ is $ 0.1 $. - The
**probability of event**which is $ E_4 is $ 0.2 $.

## Example

An experiment yields four results, each with $ P ( E_1 ) = 0.2 $, $ P ( E_2 ) = 0.2 $ and $ P ( E_3 ) = 0.2 $. What is the probability of $E_4 $? Another experiment also yields four results, each with $ P ( E_1 ) = 0.1 $, $ P ( E_2 ) = 0.1 $ and $ P ( E_3 ) = 0.1 $. What is the probability of $E_4 $?

In this question, we have to **find the probability** of an event which is $ E_4 $.

So:

We know that** sum of probability** is equal to $1 $.

\[P (E4) \space = \space 1 \space – \space 0.2 \space – \space 0.2 \space – \space 0.2 \space = \space 0.4\]

Now for the **second experiment** we have to find the** probability** of an **event** which is $E_4 $.

**So**:

We know that **sum of probability** is equal to $1$.

\[P (E4) \space = \space 1 \space – \space 0.1 \space – \space 0.1 \space – \space 0.1 \space = \space 0.7\]