Question

Two dice are throw simultaneously. If $X$ denotes the number of sixes, find the expectation of $X.$

Answer 1

Given:
Two dice are thrown simultaneously then,
sample space

$S=\left\{\begin{array}{cccccc}(1,1),& (1,2),& (1,3),& (1,4),& (1,5)& (1,6),\\ (2,1),& (2,2),& (2,3),& (2,4),& (2,5)& (2,6),\\ (3,1),& (3,2),& (3,3),& (3,4),& (3,5),& (3,6),\\ (4,1),& (4,2),& (4,3),& (4,4),& (4,5),& (4,6),\\ (5,1),& (5,2),& (5,3),& (5,4),& (5,5),& (5,6),\\ (6,1),& (6,2),& (6,3),& (6,4),& (6,5)& (6,6),\end{array}\right\}$

Total number of possible outcomes=$36$

Let$X:$ be the number of sixes occur
So, the value of $X$ can be $0,1\text{or}2.$

Hence, random variables and its
probabilities are:
$\begin{array}{ccc}X& \text{Number of outcomes}& P\left(X\right)\text{otherwise}\\ 0& 25& \frac{25}{36}\\ 1& 10& \frac{10}{36}\\ 2& 1& \frac{1}{36}\end{array}$

Now, probability distribution table is
$\begin{array}{cccc}X& 0& 1& 2\\ P\left(X\right)& \frac{25}{36}& \frac{10}{36}& \frac{1}{36}\end{array}$

We know that,
The expectation of $X$ is given by

$E\left(X\right)=\mu ={\sum }_{i=1}^n{x}_i{p}_i$

$\Rightarrow E\left(X\right)=0\times \frac{25}{36}+1\times \frac{10}{36}+2\times \frac{1}{36}$

$\Rightarrow E\left(X\right)=\frac{10}{36}+\frac{2}{36}$

$E\left(X\right)=\frac{12}{36}$
$\Rightarrow E\left(X\right)=\frac{1}{3}$

$\therefore$ The expectation of $X$ is $E\left(X\right)=\frac{1}{3}$