Two dice are thrown. find the probability of getting
a) a pair of 3's
b) any pair
c) at least one 4
d) a total of 9
e) a total greater than 8

There are 36 possible ways of throwing two dices. The following lists are shown in ordered pair:
(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)

a) a pair of 3's
The probability is 1/36 because there is only one pair which is (3,3).
b) any pair
The probability is 6/36 or 1/6 because the number of pairs are (1,1) (2,2) (3,3) (4,4) (5,5) and (6,6).
c) at least one 4
The probability is 10/36 or 5/18 because there would be (4,1) (4,2) (4,3) (4,5) (4,6) (1,4) (2,4) (3,4) (5,4) (6,4) for any combination.
d) a total of 9
The probability is 4/26 or 2/13 because (4,5) (5,4) (3,6) (6,3) are the totals of 9.
5) a total greater than 8
The probability is 10/36 or 5/18. Just find the ordered pair having a sum greater than 8.

Hope this helps. 

Understanding what? lol