Question 137573
Event A : The sum of two dice
First look at all of the possible outcomes.
First number rolled, second number rolled, then the sum.
1 1=2
1 2=3
1 3=4
1 4=5
1 5=6
1 6=7
2 1=3
2 2=4
2 3=5
.
.
.
6 1=7
6 2=8
6 3=9
6 4=10
6 5=11
6 6=12
All sums from 2 to 12 are possible out of 36 possible outcomes, with the following probabilities.
2=1/36
3=2/36
4=3/36
5=4/36
6=5/36
7=6/36
8=5/36
9=4/36
10=3/36
11=2/36
12=1/36
For the probability of the sum being greater than 7, you would add the probabilities of getting 8, 9, 10, 11, and 12.
P(A)=(5+4+3+2+1)/36=15/36 or 5/18.
Event B : The sum is an odd number. 
Add up all of the probabilities for the sum to be 3, 5, 7, 9, and 11.
P(B)=(2+4+6+4+2)/36=18/36 or 1/2. 
Since half of the sums are even and half of the sums are odd, you could have come to the same conclusion.