Let's go over the possible results and see how likely each is to happen.
We have two six sided dice, so we have a total of 36 possbile combinations.
Out of those combinations, if X is the sum of the numbers, then
P(X=1)=0, because the result will always be greater than 1.
P(X=2)=1/36, because there is only one possible combination that gives us a result of 2, (1,1).
P(X=3)=2/36 since there are two combinations that give this result, (2,1), (2,1).
P(X=4)=3/36 as there are three combination that results in this (1,3), (3,1), (3,1)
P(X=5)=4/36, these are the four combinations that produce it (1,4), (2,3), (2,3), (4,1).
P(X=6)=5/36, the combinations are (1,5), (2,4), (2,4), (3,3), (3,3).
P(X=7)=6/36, the combinations are (1,6), (2,5), (2,5), (3,4), (3,4), (4,3)
P(X=8)=5/36, the combinations are (2,6), (2,6), (3,5), (3,5), (4,4)
P(X=9)=4/36, the combinations are (1,8), (3,6), (3,6), (4,5)
P(X=10)=3/36, the combinations are (2,8), (2,8), (4,6)
P(X=11)=2/36, the combinations are (3,8), (3,8)
P(X=12)=1/36, the only combination is (4,8)
We can quickly verify our results by summing all the probabilities and checking that the sum equals 1, which is indeed the case.
In any other case (that is X equals any other number n other than he ones above) P(X=n)=0 as it is impossible to get that result.
In summary, the probabilty distribution of X is