Question 1078225
Total Universe=6^2=36 cases-

First Condition only (First is multiple of 3 and the sum isn't 8) :
(3,1),(3,2),(3,3),(3,4),(3,6),(6,1),(6,3),(6,4),(6,5),(6,6).
 Favorable cases= 10.

Second Condition only (First is not a multiple of 3 and the sum is 8) :

(2,6),(4,4),(5,3)
Favorable cases=3.

First and Second condition at the same time:
(3,5),(6,2)
favorable cases=2.

Total favorable cases=10+3+2=15.

So P()= 15/36 =5/12

#natolino