Edward C. answered 04/01/15
Tutor
5.0
(438)
Caltech Grad for math tutoring: Algebra through Calculus
For a single die roll there are 6 possible outcomes {1,2,3,4,5,6}. When you roll two dice there are 36 possible outcomes, 6 on the 1st die and 6 on the 2nd, which can be listed as
(1,1), (2,1), (3,1), (4,1), (5,1), (6,1),
(1,2), (2,2), (3,2), (4,2), (5,2), (6,2),
(1,3), (2,3), (3,3), (4,3), (5,3), (6,3),
(1,4), (2,4), (3,4), (4,4), (5,4), (6,4),
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5),
(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)
There are 2 outcomes with a sum of 3, namely (1,2) and (2,1)
There are 5 outcomes with a sum of 6 {(1,5),(2,4),(3,3),(4,2),(5,1)}
There are 4 outcomes with a sum of 9 {(3,6),(4,5),(5,4),(6,3)}
There is 1 outcome with a sum of 12, namely (6,6)
These are all the outcomes with the sum being a multiple of 3 or equal to 9, so the probability that this occurs is
(2 + 5 + 4 + 1) / 36 = 12/36 = 1/3