Question

Two dice are thrown together and the number appearing on them noted. X denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely, what is the probability distribution of X?

Answer 1

Let X denote the sum of the numbers on two die. Then, X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Sample space : {(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)}

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

Thus, the probability distribution of X is given by

X	P(X)
2	$\frac{1}{36}$
3	$\frac{2}{36}$
4	$\frac{3}{36}$
5	$\frac{4}{36}$
6	$\frac{5}{36}$
7	$\frac{6}{36}$
8	$\frac{5}{36}$
9	$\frac{4}{36}$
10	$\frac{3}{36}$
11	$\frac{2}{36}$
12	$\frac{1}{36}$