Question

As in Problem $11$, show that the expected number of$5\text{'}s$ in n tosses of a die is $\frac{n}{6}$.

Short answer

$E\left(x_1\right),E\left(x_2\right),E\left(x_2\right),\cdots ,E\left(x_n\right)$all are equal.

The statement has been proven.

Step-by-step solution

Given Information

A coin is tossed twice.

Definition of the cumulative distribution function.

The likelihood that a comparable continuous random variable has a value less than or equal to the function's argument is the value of the function.

Prove the statement

A dice is tossed, the outcomes is$S=\{1,2,3,4,5,6\}$ .

$\begin{array}{c}E\left(x\right)=\sum _{\left(1\right)}{x}_i{p}_{i}n\\ =\frac{\left(1\right)}{6}\end{array}$

If the dice are tossed n times, then the expectation is given below.

$\begin{array}{c}E\left(x_1+x_2+\cdots +x_n\right)=E\left(x_1\right)+E\left(x_2\right)+E\left(x_2\right)+\cdots +E\left(x_n\right)\\ \frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\cdots \\ =\frac{n}{6}\end{array}$

Hence, $E\left(x_1\right),E\left(x_2\right),E\left(x_2\right),\cdots ,E\left(x_n\right)$ all are equal.

Hence, the statement has been proven.