Simple joint probability question of two dice throw

Question

I have a simple joint probability question I cannot understand. If I throw two dice and define the following events:

$X$ – number of times 4 was obtained.

$Y$ – number of even results obtained.

Now I need to find the joint probability function, so my problem is with $P(x=0, y=1) = P(\text{obtaining zero times four and one time an even result})$, so it should be equal to probability of obtaining uneven results, and probability of obtaining an even result times $2/3$ (since 4 was not received). From this reasoning, I think the result should be: $\frac{3}{6} \cdot (\frac{3}{6} \cdot \frac{2}{6})$, i.em $1/2$ chance to obtain uneven number times the probability to obtain an even number that is not ($3/6 \cdot 2/6$).

However, the correct result should be: $\frac{2 \cdot 2 \cdot3}{6 \cdot 6}$, and I don't understand why. I would appreciate an explanation.

EDIT: similarly, why in $P(x=1, y=2) = P(\text{two even results, of which only one is four})$ = $\frac{2 \cdot (1 \cdot 2)}{6 \cdot 6}$ and not: even ($3/6$) * even that is not four ($2/6$)?

Answer 1

Give the dice a number.

Let $D_1$ denote the result of the first and let $D_2$ denote the result of the second die.

Then:

$$P(X=0,Y=1)=P(D_1\in\{2,6\}, D_2\in\{1,3,5\})+P(D_1\in\{1,3,5\}, D_2\in\{2,6\})=$$$$2P(D_1\in\{2,6\})P(D_2\in\{1,3,5\})=2\frac26\frac36$$

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Edit (after edit of question):

$$P(X=1,Y=2)=P(D_1=4, D_2\in\{2,6\})+P(D_1\in\{2,6\}, D_2=4)=$$$$2P(D_1=4)P(D_2\in\{2,6\})=2\frac16\frac26$$