In a single of two dice the probability of getting more then 7?

There are 36 possible outcomes of the sum of the rollings two dice

(1,1) 1 way to get two

(2,1) (1,2) 2 ways to get 3

(1,3) , (2,2) , (3,1) 3 ways to get 4

there a pattern

4 ways to get 5

5 ways (1,5), (2,4) , (3,3), (4,2) , (5,1) ways to get 6

6 ways to get 7

5 (2,6), (3,5), (4,4), (5,3) , (2,6) ways to get 8

4 (3,6) , (4,5) , (6,3) (5,4) ways to get 9

3 (6,4) , (5,5) , (4,6) ways to get 10

2 (5,6) , (6,5) ways to get 11

1 (6,6) way to get 12

so more than 7

5 (2,6), (3,5), (4,4), (5,3) , (2,6) ways to get 8

4 (3,6) , (4,5) , (6,3) (5,4) ways to get 9

3 (6,4) , (5,5) , (4,6) ways to get 10

2 (5,6) , (6,5) ways to get 11

1 (6,6) way to get 12

so 5+4+3+2 +1 = 9+6 = 15

so 15/36 = (3*5)/ (3* 12)= (3/3)*(5/12) = 5/12

P( X> 7) = 5/12