Question 1127577
There are 11 possible outcomes for the sum of two dice (2-12, inclusive):
<pre>
Sum      Number of ways to get this sum
2                   1
3                   2
4                   3
5                   4
6                   5
7                   6
8                   5
9                   4
10                  3
11                  2
12                  1<br>

a)   Pr(sum is 9) = 4/36 = 1/9
b)   Pr(sum > 10) = (2+1)/36 = 3/36 = 1/12
c)   Pr(sum < 6) =  (1+2+3+4)/36 = 10/36 = 5/18