<s>[INST] 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 $11/12$
 [/INST] The probabilities are not equal. Have a look at the following table.
$$
\left.\begin{array}{c|c|c|c|c} & 1 & 2 & 3 & 4 & 5& 6\\\hline 1 & 2 & 3 & 4 & 5& 6& 7 \\\hline 2 & 3 & 4 & 5 & 6 & 7& 8\\\hline 3 & 4 & 5 & 6 & 7 & 8& 9\\\hline 4 & 5 & 6 & 7 & 8 & 9& 10\\\hline 5 & 6 & 7 & 8 & 9 & 10& 11\\\hline 6 & 7 & 8 & 9 & 10 & 11& 12\end{array}\right.
$$
It is easy to tell various probabilities of the outcomes of this experiment using the table above. For example, the probability that we get $7$ is $6/36=1/6$.
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