The propability is not my strength but, I remember that we have 1/36 to have 2 and 1/36 to have 12 as overall outcome.
so for not 2 could be $\displaystyle 1 - \frac{1}{36}$
for 4 $\displaystyle \frac{3}{36}$
and for 6 $\displaystyle \frac{5}{36}$
=> think to 2 you have 1/6 for the first dice and 1/6 for the second so 1/36
Hope that it is useful
Hello, ndcruz!
It's just matter of counting, isn't it?
There are 36 possible outcomes.
There are: .$\displaystyle 1 + 3 + 5 \:=\:9$ ways to get a sum of 2, 4 or 6.
So there are: .$\displaystyle 36 - 9 \:=\:27$ ways for some other sum.
Therefore: .$\displaystyle P(\text{sum not 2, 4, or 6}) \;=\;\frac{27}{36} \;=\;\frac{3}{4} $