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

\[P\left( X = 2 \right) = \frac{1}{36}\]

\[P\left( X = 3 \right) = \frac{2}{36}\]

\[P\left( X = 4 \right) = \frac{3}{36}\]

\[P\left( X = 5 \right) = \frac{4}{36}\]

\[P\left( X = 6 \right) = \frac{5}{36}\]

\[P\left( X = 7 \right) = \frac{6}{36}\]

\[P\left( X = 8 \right) = \frac{5}{36}\]

\[P\left( X = 9 \right) = \frac{4}{36}\]

\[P\left( X = 10 \right) = \frac{3}{36}\]

\[P\left( X = 11 \right) = \frac{2}{36}\]

\[P\left( X = 12 \right) = \frac{1}{36}\]

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	\[\frac{1}{36}\]