Using the sample space for rolling two dice, illustrates a probability distribution for random variable X respending the sum of the number that appear

we define the following as the outcome

probability that the sum is 2 is as below

p(2)= (1/6) * (1/6) = 1/36  (1,1)

probability that the sum is 3

p(3) = (1/6)*(1/6)*(2 )= 1/18 (1,2) or (2,1)

Probability that the sum is 4

p(4) = ( (1/6) * (1/6) * ( 2 ) ) + ( (1/6) * (1/6) ) = 1/12 (1,3) or (3,1) or (2,2)

probability that the sum is 5

p(5) = ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6)*( 2 ) ) = 1/9 (1,4) or (4,1) or (2,3) or (3,2)

probability that the sum is 6

p(6) = ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6) ) = 5/36 (1,5) or (5,1) or (2,4) or (4,2) or (3,3)

probability that the sum is 7

p(7) = ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6) * (2) ) =1/6 (1,6) or (6,1) or (2,5) or (5,2) or (3,4) or (4,3)

probability that the sum is 8

p(8) = ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6) ) = 5/36 (2,6) or (6,2) or (5,3) or (3,5) or (4,4)

probability that the sum is 9

p(9) = ( (1/6) * (1/6)*( 2 ) ) + ( (1/6) * (1/6)*( 2 ) ) = 1/9 (3,6) or (6,3) or (5,4) or (4,5)

probability that the sum is 10

p(10) = ( (1/6) * (1/6) * ( 2 ) ) + ( (1/6) * (1/6) ) = 1/12 (4,6) or (6,4) or (5,5)

probability that the sum is 11

p(11) = ( (1/6) * (1/6) * ( 2 ) ) = 1/18 (5,6) or (6,5)

probability that the sum is 12

p(12) = ( (1/6) * (1/6) ) = 1/36 (6,6)

Hence we may represent the probability distribution as below

$\begin{matrix} X&2&3&4&5&6&7&8&9&{10}&{11}&{12}\\ p&{\frac{1}{{36}}}&{\frac{1}{{18}}}&{\frac{1}{{12}}}&{\frac{1}{9}}&{\frac{5}{{36}}}&{\frac{1}{6}}&{\frac{5}{{36}}}&{\frac{1}{9}}&{\frac{1}{{12}}}&{\frac{1}{{18}}}&{\frac{1}{{36}}} \end{matrix}$