Answer
a. $\frac{1}{9}$
b. $\frac{2}{9}$
c. $\frac{1}{6}$
Work Step by Step
a. A sum of 9
There will be 5 favorable outcomes (3+6, 4+5, 5+4, 6+3) while the total outcomes are 36 ($6\times6$).
So the probability of get a sum of 9 is $p=\frac{4}{36}=\frac{1}{9}$
b. A sum of 7 or 11
For the sum of 7, there are 6 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) favorable outcomes.
For the sum of 11, there are 2 (5+6, 6+5) favorable outcomes..
So the probability of getting a sum of 7 or 11 is: $p=\frac{6+2}{36}=\frac{2}{9}$
c. Doubles
There are 6 (11,22,33,44,55,66) favorable outcomes.
So the probability of getting a double is: $p=\frac{6}{36}=\frac{1}{6}$
