Edward C. answered • 03/05/15

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Caltech Grad for math tutoring: Algebra through Calculus

There are 36 possible outcomes when you roll two dice (6 on each die)

There are 4 ways to get a sum of 9: (3,6), (4,5), (5,4), or (6,3) so P(9) = 4/36

There are 2 ways to get a sum of 3: (1,2) or (2,1) so P(3) = 2/36

There are 5 ways to get a sum of 6: (1,5), (2,4), (3,3), (4,2) or (5,1) so P(6) = 5/36

So P(sum = 9, 3, or 6) = 4/36 + 2/36 + 5/36 = 11/36

To find the probability that the sum is at least 7 you need to figure out P(7), P(8), P(9), P(10), P(11) and P(12). I won't enumerate them (although you should for practice) but the values are

P(7) = 6/36

P(8) = 5/36

P(9) = 4/36

P(10) = 3/36

P(11) = 2/36

P(12) = 1/36

Do you see the pattern? So

P(sum >=7) = 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36 = 21/36 = 7/12