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. Find the mean of $X$ .

Answer 1

$P(X=2)=P(1,1)=\frac{1}{36}$
$P(X=3)=P\left[\right(1,2),(2,1\left)\right]=\frac{2}{36}$
$P(X=4)=P\left[\right)1,3),(2,2),(3,1\left)\right]=\frac{3}{36}$
$P(X=5)=P\left[\right(1,4),(2,3),(3,2),(4,1\left)\right]=\frac{4}{36}$
$P(X=6)=P\left[\right(1,5),(2,4),(3,3),(4,2),(5,1\left)\right]=\frac{5}{36}$
$P(X=7)=P\left[\right(1,6),(2,5),(3,4),(4,3),(5,2),(6,1\left)\right]=\frac{6}{36}$
$P(X=8)=P\left[\right(2,6,(3,5),(4,4),(5,3),(6,2)]=\frac{5}{36}$
$P(X=9)=P\left[\right(3,6),(4,5),(5,4),(6,3\left)\right]=\frac{4}{36}$
$P(X=10)=P\left[\right(4,6),(5,5),(6,4\left)\right]=\frac{3}{36}$
$P(X=11)=P\left[\right(5,6),(6,5\left)\right]=\frac{2}{36}$
$P(X=12)=P\left[\right(6,6\left)\right]=\frac{1}{36}$

$X$	$P=P\left(X\right)$	$p_i{x}_i$
2	$\frac{1}{36}$	$\frac{2}{36}$
3	$\frac{2}{36}$	$\frac{6}{36}$
4	$\frac{3}{36}$	$\frac{12}{36}$
5	$\frac{4}{36}$	$\frac{20}{36}$
6	$\frac{5}{36}$	$\frac{30}{36}$
7	$\frac{6}{36}$	$\frac{42}{36}$
8	$\frac{5}{36}$	$\frac{40}{36}$
9	$\frac{4}{36}$	$\frac{36}{36}$
10	$\frac{3}{36}$	$\frac{30}{36}$
11	$\frac{2}{36}$	$\frac{22}{36}$
12	$\frac{1}{36}$	$\frac{12}{36}$

Total $1$ $\frac{252}{36}$
Mean of $X==\frac{\sum p_i{x}_i}{\sum p_i}=\frac{252}{36\times 1}=7$
Mean $=7$