When two fair dice are thrown there are total 36 possible outcomes.

∵ X denotes the sum of 2 numbers appearing on dice.

∴ X can take values 2,3,4,5,6,7,8,9,10,11 and 12

As appearance of a number on a fair die is equally likely

i.e. P(appearing of 1) = P(appearing of 2) = P(appearing of 3) = P(appearing of 4) = P(appearing of 5) = P(appearing of 6) = 1/6

And also the appearance of numbers on two different dice is an independent event. So two find conditions like P(1 in the first dice and 2 in the second dice) can be given using multiplication rule of probability.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

P(X = 2) = {∵ (1,1) is the only combination resulting sum = 2}

P(X = 3) =

{∵ (1,2) and (2,1) are the combinations resulting in sum = 3}

P(X = 4) =

{∵ (1,3), (3,1) and (2,2) are the combinations resulting in sum = 4}

P(X = 5) =

{∵ (3,2) (2,3) (1,4) and (4,1) are the combinations resulting in sum = 5}

P(X = 6) =

{∵ (1,5) (5,1) (2,4) (4,2) (3,3) are the combinations resulting in sum = 6}

P(X = 7) =

{∵ (1,6) (6,1) (2,5) (5,2) (3,4) (4,3) are the combinations resulting in sum = 7}

P(X = 8) =

{∵ (3,5) (5,3) (2,6) (6,2) (4,4) are the combinations resulting in sum = 8}

P(X = 9) =

{∵ (3,6) (6,3) (5,4) and (4,5) are the combinations resulting in sum = 9}

P(X = 10) =

{∵ (6,4), (4,6) and (5,5) are the combinations resulting in sum = 10}

P(X = 11) =

{∵ (5,6) and (6,5) are the combinations resulting in sum = 11}

P(X = 12) = {∵ (6,6) is the only combination resulting sum = 2}

Now we have p_i and x_i.

∴ Required probability distribution is:-