Putting 12 balls of different colors in 4 urns, 3 balls in each. 5 red, 4 green and 3 blue balls are put in 4 boxes so that every box has exactly 3 balls.
Find the probability that:

*

*At least 1 box has a ball of every color;

*Only 1 box has a ball of every color

I've tried to compute the probability of at least 1 box having an RGB as follows:
$$\mathbb{P}(\geq 1 RGB)=\frac{\binom{5}{1}\cdot\binom{4}{1}\cdot\binom{3}{1}\cdot\frac{9!}{(3!)^3}}{\frac{12!}{(3!)^4}}=\frac{3}{11}$$
since we choose an RGB and then put the remaining 9 balls into boxes at random.
The problem is, I don't know whether I need to account for the allocation of boxes. If, say, we choose one box to put the RGB into and the total number of ways to allocate boxes is $4!$, then the $\mathbb{P}(\geq 1 RGB)$ will multiply by $\frac{4}{4!}$ and be equal to $\frac{2}{11}$.
No progress with the other probability so far.
From the examples found online, this problem may be solved with the inclusion-exclusion formula, but I struggle to figure out what the events $A_1,\dots,A_n$ are in this model.
How do I apply the inclusion-exclusion formula here or is there any other way to obtain the probabilities in question?

Line up the urns from left to right.  There are
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
to distribute three balls to each of the four urns.
An urn with a ball of each color:  There are four ways to select the urn which will receive a ball of each color, five ways to select a red ball, four ways to select a green ball, and three ways to select a blue ball for that urn.  There are $\binom{9}{3}\binom{6}{3}\binom{3}{3}$ ways to distribute the remaining nine balls to the remaining three urns so that they each receive three balls.  Hence, there are
$$\binom{4}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
such distributions.
Two urns with a ball of each color:  There are $\binom{4}{2}$ ways to select the two urns which will receive a ball of each color.  There are five ways to select a red ball, four ways to select a green ball, and three ways to select a blue ball for the leftmost of those two urns.  There are four ways to select a red ball, three ways to select a green ball, and two ways to select a blue ball for the rightmost of those two urns.  There are $\binom{6}{3}\binom{3}{3}$ ways to distribute the remaining balls to the remaining two urns so that they each receive three balls.  There are
$$\binom{4}{2}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{6}{3}\binom{3}{3}$$
such distributions.
Three urns with a ball of each color:  There are $\binom{4}{3}$ ways to select which three urns will receive a ball of each color.  There are five ways to select a red ball, four ways to select a green ball, and three ways to select a blue ball for the leftmost of those three urns.  There are four ways to select a red ball, three ways to select a green ball, and two ways to select a blue ball for the rightmost of those three urns.  There are three ways to select a red ball, two ways to select a green ball, and one way to select a blue ball for the middle of those three urns.  All three of the remaining balls must be placed in the other urn.  Hence, there are
$$\binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1}\binom{3}{3}$$
such distributions.
It is not possible for all four urns to each receive a ball of each color since there are only three blue balls.  By the Inclusion-Exclusion Principle, the probability that there is at least one ball of each color is
$$\frac{\binom{4}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{9}{3}\binom{6}{3}\binom{3}{3} - \binom{4}{2}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{6}{3}\binom{3}{3} + \binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1}\binom{3}{3}}{\dbinom{12}{3}\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}$$
