Two dice are thrown. Find the probability of getting :

1) The sum of the numbers on their upper faces is at least 9.

2) The sum of the numbers on their upper faces is 15.

3) The number on the upper face of the second die is greater than the number on the upper face of the first die.

#### Solution

S= {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2, 2) (2,3) (2, 4)(2,5)(2,6)

(3,1)(3, 2) (3,3) (3, 4)(3,5) (3,6)

(4,1) (4, 2) (4,3) (4, 4)(4,5)(4,6)

(5,1)(5, 2)(5,3)(5, 4)(5,5) (5,6)

(6,1) (6, 2)(6,3) (6, 4) (6,5) (6,6)}

n(S) = 36

Let A = sum of the numbers in their upper faces is at least 9.

A = {(3,6) (4,5) (4,6) (5, 4) (5,5) (5,6) (6,3) (6, 4) (6,5) (6,6)}

n(A) = 10

`P(A) = (n(A))/(n(S)) = 10/36 = 5/18`

Let B = sum of the number on their upper faces is 15.

B = {} (Null Set)

n(B) = 0

`P(B) = (n(B))/(n(S))= 0/36 = 0`

Let C = number on the upper face of second die is greater than the number on the upper face of first die.

C = {(1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 3) (2, 4) (2, 5) (2, 6) (3, 4) (3, 5) (3, 6) (4,5) (4, 6) (5, 6)}

n(C) = 15

`P(C) = (n(C))/(n(S)) = 15/36 = 5/12`