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

In our question, we are throwing two dice 2 times

∵ Y denotes the number of times, a sum of two numbers appearing on dice is equal to 9.

∴ Y can take values 0,1 or 2

Means In both the throw sum of 9 is not obtained, in one of throw of two dice 9 is obtained and in both the throws of two dice a sum of 9 is obtained.

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

Let A denotes the event of getting a sum of 9 in a throw of 2 dice

∴ P(A) = probability of getting the sum of 9 in a throw of 2 dice is

And P(A’) = probability of not getting the sum of 9 in the throw of 2 dice =

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

P(Y = 0) = P(A’)×P(A’) =

P(Y = 1) = P(A)×P(A’)+ P(A’)×P(A)+ =

P(Y = 2) = P(A)×P(A) =

Now we have p_i and x_i.

∴ The required probability distribution is:-