Apair of dice are tossed. what is the probability that doubles are rolled, given that the sum on the two dice is less than 8?

(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)

the sum is less than 8, we will remove  s which have sum of 8 or more
we have 21

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

the doubles  (1.1)   (2.2)  (3.3) 
So 3 out of the 21 are doubles

3/21 = 1/7 

Add the cents that add the left overs to get the first number