This problem aims to find the probability of the occurrence of a **random event** and its **predictable outcomes.** The concepts required for this problem are mainly related to **probability** and the **product rule.**

Let’s first look at a **fair die,** whose each face has the **identical probability** of coming **faced up.**

The **product rule** is stated as the probability of two **autonomous events** $(m,n)$ happening together can be estimated by **multiplying** the **respective probabilities** of each event **arising independently** $(m\times n)$.

So **probability** is a procedure to predict the **happening **of a **random event,** and its value is mostly between **zero** and **one.** It calculates the possibility of an **event,** events that are a bit tricky to anticipate an **outcome.**

**Given as:**

\[\text{Probability of event to occur} = \dfrac{\text{Number of ways an event can occur}}{\text{Total number of outcomes of that event}}\]

## Expert Answer

So as per the **statement,** a **dice** is rolled $6$ times and we are to find the **probability** that the **outcome** of these events is not an **even number,** or in other words, the **outcome** of these events is an **odd number.**

If we look **at dice,** we find a total of $6$ **faces,** of which only $3$ **faces **are odd, the rest are subsequently **even numbers.** Let’s create a **sample space** for a dice that is rolled only once:

\[S_{\text{first role}}={1, 2, 3, 4, 5, 6} \]

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

\[S_{odd}={1, 3, 5 }\]

So the **probability** of getting an **odd number** with a **single role **is:

\[P_{1 role}(O)=\dfrac{\text{Odd faces}}{\text{Total faces}} \]

\[P_{1 role}(O)=\dfrac{3}{6}\]

\[P_{1 role}(O)=\dfrac{1}{2}\]

So the **probability** that the number would be **odd** after the **first** role is $0.5$.

Similarly, in every role there are a total of $6$ outcomes:

\[S_{2^{nd} … 6^{th}} = {1, 2, 3, 4, 5, 6}\]

Here we are going to use the **property** of the **product rule** to calculate the **total number** of **outcomes** after six roles:

\[\text{Total outcomes}=6\times 6\times 6\times 6\times 6\times 6\]

\[\text{Total outcomes}=6^6 = 46656\]

Since there are only $3$ **odd numbers** in a **die,** the total number of **outcomes** becomes:

\[\text{Odd outcomes} = 3\times 3\times 3\times 3\times 3\times 3\]

\[\text{Odd outcomes} = 3^6 = 729\]

So $729$ of the $46656$ outcomes **results** in an **odd** number.

Now the **probability** becomes:

\[P_{6\space roles}(O)=\dfrac{729}{46656}\]

\[P_{6\space roles}(O)=0.0156\]

## Numerical Result

The **probability** that the outcome of a **fair die** rolled **six times** would not be an **even number** is $0.0156$.

## Example

A **dice** is rolled **six times,** find the **probability** of getting the **number six.**

Let’s assume $P$ is the **probability** of getting a $6$:

\[P=\dfrac{1}{6}\]

Similarly, the **probability** of getting any **number other than** $6$ is:

\[P’= 1-P=\dfrac{5}{6}\]

Now we are going to use the **property** of the **product rule** to calculate the **total number** of outcomes after **six** roles:

\[\text{P(Not getting a 6 for n times)} = \text{P’ to the n_{th} power} \]

So it **becomes:**

\[(\dfrac{5}{6})^6 = \dfrac{15,625}{46,656} \approx 0.334 \]

Hence, the **probability** of getting a **six** at **least once** is $1-0.334=0.666$.