I have this problem from a text book, and the answer it gives for that probability is $\frac{25}{108}$.
I assumed that they are refering to the probability of obtaining two even and an odd. I created the sample space and counted all the points with two even and an odd using R since I also don't know how to use combinatorics for this case. I have obtained 81 so I suppose that the probability is $\frac{81}{216}=\frac{3}{8}$.
What I'm doing wrong?
On
That looks right: you have 3 possibilities for each even number, and 3 for odd, and finally 3 possible dice to be the odd one. So that's 81. You are correct!
Also, notice that likewise you have 81 possibilities to get 1 even and 2 odd.
And you have 27 options for 3 even numbers, and also 27 for 3 odd ones.
Total: 27+81+81+27=216. Which is the whole space. So that's further confirmation you're doing this all ok!
The book is wrong and you are right! To confirm:...
If you roll one fair dice you will obtain an even result with a probability of $1/2$.
If you roll three fair die, you will obtain two consecutive even results, then an odd result with a probability of $1/2^3$.
There are three such arrangements which yeild a result of two even and one odd, and they are all equally probable. Thus the probability for obtaining two even and one odd result is $3/8$, which is also $81/216$.
However, there is an additional arrangement which is a result of three even numbers, which has the same probability as the other three arrangements. So the probability for at least two even results is $4/8$, $108/216$, or $1/2$.
To confirm the result, hindsight shows that there is a equal probability for obtaining at least two even results as there is for obtaining at least two odd results, and that these events are exclusive and exhaustive.
Neither answer is $25/108$, nor should they be.