Edward C. answered 10/06/15
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Caltech Grad for math tutoring: Algebra through Calculus
When you roll a single die there are 6 equally likely possible outcomes: 1,2,3,4,5 or 6
When you roll 2 dice there are 36 possible outcomes that can be represented by ordered pairs (X,Y) where each of X and Y are 1 of the 6 possible outcomes listed above. These outcomes are
(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)
There are 6 outcomes that result in a sum of 7: they are
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
Of these 6 outcomes, only 1 [the (3,4) outcome] has the first number equal to 3. So if the sum is 7, the probability that the first number is a 3 is 1/6
There are 5 outcomes that result in a sum of 8: they are
(2,6) (3,5) (4,4) (5,3) (6,2)
Of these 5 outcomes, only 1 [the (3,5) outcome] has the first number equal to 3. So if the sum is 8, the probability that the first number is a 3 is 1/5