You have 4 balls in a bag, 2 yellow, 1 black, 1 red. If you drawn 2 balls I tell you that at least 1 is yellow, what is the probability that you've drawn 2 yellow balls?
There are two ways I see that this can be solved: using Bayes formula and looking at the probability space.
Using Bayes formula we get $P(2\ yellow | al1\ yellow) = \frac{P(al1\ yellow | 2\ yellow) \cdot P(2\ yellow)}{P(al1\ yellow)}$, here "$al1$" us at lease one.
Hence we get $$P(2\ yellow | al1 \, blue) = \frac{1 \cdot \left(\frac{2}{4} \cdot \frac{1}{3}\right)}{\left(1-all\ non\ yellow\right)}=\frac{1 \cdot \left(\frac{2}{4} \cdot \frac{1}{3} \right)}{\left(1-\frac{2}{4} \cdot \frac{1}{3}\right)}=\frac{1}{5}.$$
Using probability space approach we get these possible combinations: YY,YB, YR, RY, BY hence the probability of 2 yellows is 1/5
Now the above solutions only work in case we draw balls in order, if we draw them simultaneously the solution is different and from probability space we get the answer of 1/3
Please explain what solution is correct and why
In the first approach, your mistake is in the calculation of $P(Y=2)$ and $P(Y \geq 1)$.
$ \displaystyle P(Y = 2) = \frac 2 4 \cdot \frac 1 3 = \frac 16$
$ \displaystyle P(Y \geq 1) = 1 - P(Y = 0) = 1 - \frac 24 \cdot \frac 1 3 = \frac 56$
So, $ \displaystyle ~P(Y = 2 \mid Y \geq 1) = \frac 1 5$
If we were drawing simultaneously,
$ \displaystyle P(Y = 2) = {2 \choose 2} / {4 \choose 2} = \frac 16$
$ \displaystyle P(Y \geq 1) = 1 - P(Y = 0) = 1 - {2 \choose 2} / {4 \choose 2} = \frac 56$
That gives the same answer.
While your second approach gives correct answer, you should be careful in how you write the reduced sample space. You need to consider each ball different to make sure the outcomes in the sample space are equally probable. So it is in fact,
$\{Y ~y\}, \{y ~Y\}, \{Y ~B\}, \{y ~B\}, \{B ~Y\}, \{B ~y\}, \{Y ~R\}, \{y ~R\}, \{R ~Y\}, \{R ~y\}$
and that gives probability of $\dfrac 2{10}$.