Finding the distribution law Throwing a cube twice,
let $X$ be the sum of the two throws

Find the distribution law

My attempt
$$
\begin{array}{c|lcr}
 &2&3&4&5&6&7&8&9&10&11&12 \\
\hline
\text{P}_\text{X}(x) & 1/12 & 1/12 & 1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12 \\
\end{array}
$$
But shouldn't the sum be $1$? Currently the sum is $11/12$

What you've tried is wrong:
You can get $2$ just getting a one in each die (1+1), so the probability is $\frac{1}{6} \frac{1}{6} = \frac{1}{36}$. Instead, you have more options to get $5$ (1+4,2+3,3+2,4+1), giving you a total probability of $\frac{2}{18}$
Doing this process for all the numbers you get the correct result.
