We have 6 plates on the table (2 red, 2 white and 2 blue). We also have 6 cups in the cupboard (2 red, 2 white and 2 blue). We randomly take cups out of the cupboard and place them on the plates. What is the probability that the cups and the plates, on which they are placed, match colors?
I don't know if it's correct but how I understood the question was that first we take, let's say, one of the red cups, which has 2 options, and put it on one of the 6 plates. Then the second red cup is placed on the remaining 5, and so on with the other colored cups.
My solution is $$\frac26 \cdot \frac15 \cdot \frac24 \cdot \frac13 \cdot \frac22 \cdot \frac11 = \frac4{720}$$
Is this solution correct or maybe I should've also focused on the color of the plates?
There is a computational mistake in your working, it comes out to $\frac{8}{720} = \frac1{90}$. which is the correct answer.
Another way to confirm the answer is to realize that had the plates and cups been numbered, there would be $6!=720$ ways the plates could have been arranged, and only one matching arrangements of cups, but as pairs of plates are identical, there are $2^3 = 8 $ matching arrangements of cups, thus $Pr = \frac1{90}$