This problem aims to familiarize us with **random events** and their **predictable outcomes.** The concepts required to solve this problem are mostly related to **probability,** and **probability distribution.**

So **probability** is a method to predict the **occurrence** of a **random event,** and its value can be between **zero** and **one.** It measures the likelihood of an **event,** events that are difficult to predict an **outcome.** Its formal definition is that a **possibility** of an event occurring is equal to the **ratio** of favorable outcomes and the total **number** of **tries.**

**Given as:**

\[\text{Pobability of event to occur} = \dfrac{\text{Number of favourable events}}{\text{Total number of events}}\]

## Expert Answer

So as per the **statement,** a total of **two dices** are rolled and we are to find the **probability** that the **sum** of **numbers** on those two dices is an even number.

If we look at a **single dice,** we find that there are a total of $6$ **outcomes,** of which only $3$ **outcomes** are even, the rest are subsequently **odd numbers.** Lets create a sample space for **one dice**:

\[ S_{\text{one dice}} = {1, 2, 3, 4, 5, 6} \]

Out of which the **even numbers** are:

\[ S_{even} = {2, 4, 6} \]

So the **probability** of getting an **even number** with a **single dice** is:

\[ P_1(E) = \dfrac{\text{Even numbers}}{\text{Total numbers}} \]

\[ P_1(E) = \dfrac{3}{6} \]

\[ P_1(E) = \dfrac{1}{2} \]

So the **probability** that the number would be an **even number** is $\dfrac{1}{2}$.

Similarly, we will create a **sample space** for the outcome of **two dies:**

\[ S_2 = \begin{matrix} (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),\\ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),\\ (3,1), (3,2), (3,3), (3,4), (3,5), (3,6),\\ (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), \\ (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), \\ (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \end{matrix}\]

Out of which the **even numbers** are:

\[S_{even}=\begin{matrix} (1,1), (1,3), (1,5),\\ (2,2), (2,4), (2,6),\\ (3,1), (3,3), (3,5),\\ (4,2), (4,4), (4,6),\\(5,1), (5,3), (5,5),\\(6,2), (6,4), (6,6)\end{matrix}\]

So there are $18$ **possibilities** to get an **even number.** Thus, the **probability** becomes:

\[ P_2(E) = \dfrac{\text{Even numbers}}{\text{Total numbers}}\]

\[ P_2(E)=\dfrac{18}{36}\]

\[ P_2(E)=\dfrac{1}{2}\]

Hence, the **probability** that the **sum** would be an even **number** is $\dfrac{1}{2}$.

## Numerical Result

The **probability** that the sum of outcomes of **two dies** would be an **even number** is $\dfrac{1}{2}$.

## Example

**Two dice** are rolled such that the event $A = 5$ is the **sum** of the **numbers** revealed on the **two dice,** and $B = 3$ is the event of at least **one** of the dice showing the **number.** Find whether the **two events** are mutually **exclusive,** or **exhaustive?**

The total number of **outcomes** of **two dice** is $n(S)=(6\times 6)=36$.

Now the **sample space** for $A$ is:

$A={(1,4),(2,3),(3,2),(4,1)}$

And $B$ is:

$A={(3,1),(3,2),(3,3),(3,4),(3,5),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)}$

Let’s check if $A$ and $B$ are **mutually exclusive:**

\[ A \cap B = {(2,3), (3,2)} \neq 0\]

Hence, $A$ and $B$ are not **mutually exclusive.**

Now for an **exhaustive** event:

\[ A\cup B \neq S\]

Thus $A$ and $B$ are not **exhaustive events** as well.