Question

Let a pair of dice be thrown and the random variable $X$ be the sum of the numbers that appear on the two dice. Then the mean or expectation of $X$ is:

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

The sample space of the experiment consists of $36$ elementary events in the form of ordered pairs $(x_i,y_i),\text{where}{x}_i=1,2,3,4,5,6\text{and}{y}_i=1,2,3,4,5,6.$
The random variable $X$ i.e. the sum of the numbers on the two dice takes the value $2,3,4,5,6,7,8,9,10,11,12.$
Now,
The probability distribution of $X$ is
$\begin{array}{cccccccccccc}\text{X or}{x}_i& \text{2}& \text{3}& \text{4}& \text{5}& \text{6}& \text{7}& \text{8}& \text{9}& \text{10}& \text{11}& \text{12}\\ \text{P(X) or}{p}_i& \frac{1}{36}& \frac{2}{36}& \frac{3}{36}& \frac{4}{36}& \frac{5}{36}& \frac{6}{36}& \frac{5}{36}& \frac{4}{36}& \frac{3}{36}& \frac{2}{36}& \frac{1}{36}\end{array}$
Therefore,
$\mu =E\left(X\right)={\sum }_{i=1}^n{x}_i{p}_i=2\times \frac{1}{36}+3\times \frac{2}{36}+4\times \frac{3}{36}+5\times \frac{4}{36}+6\times \frac{5}{36}+7\times \frac{6}{36}+8\times \frac{5}{36}+9\times \frac{4}{36}+10\times \frac{3}{36}+11\times \frac{2}{36}+12\times \frac{1}{36}=\frac{2+6+12+20+30+42+40+36+30+22+12}{36}=7$
Thus, the mean of the sum of the numbers that appear on throwing two fair die is $7.$