How do I compute the probability in this exercise? I need to define the probability space $\Omega$  and it's probability function $\Bbb{P}$ for the following exercise:

For a fair die which we toss two times, compute the probability that the parity of the two numbers that show up matches.

My Idea was the following. We define $\Omega=\{1,...,6\}^2$, then $|\Omega|=36$. We have the following partition into even and odd numbers:$$\Omega=\{1,3,5\}~\dot\cup~\{2,4,6\}=:A~\dot \cup ~B$$Now let us define $\Lambda\subset \Omega$ such that it contains all pairs $(u,v)$ where $u,v\in A$ or $u,v\in B$.

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*Case $1$: $u,v\in A$. Then we have $3^2=9$ possibilities.

*Case $2$: $u,v\in B$. Then we have $3^2=9$ possibilities.

Thus we have $18$ possibilities and $$\Bbb{P}(\Lambda)=\frac{|\Lambda|}{|\Omega|}=\frac{18}{36}=\frac{1}{2}$$ Is this correct so?

Alternative approach:
When you define the sample space, it is very helpful if each element in the space is equally likely.  Within this parameter, you could simplify the problem by defining the sample space to have only $4$ elements, instead of $36$ elements.
Let $E$ be the event that the die roll is even. 
Let $O$ be the event that the die roll is odd.
Then, you could define the sample space to be
$\{E,O\}^2~$ (four elements)
Here, the favorable outcomes would be the union of $\{E\}^2$ (one element) with $\{O\}^2$ (one element).
So, the probability would be computed as
$$\frac{1 + 1}{4}.$$
The underlying idea behind this alternative approach is that the problem asks you to focus (only) on the die's odd/even parity on each of the  two rolls.  This can be (validly) done without focusing on the (exact) number of spots face-up on the die rolls.
