Distribution of $10$ different marbles to $6$ different color boxes The question is:

There are $6$ boxes, with each box a different color, and $10$ different marbles. The marbles are scattered randomly in the boxes.

*

*What is the probability that all marbles are in the same box?

*What is the probability that all marbles are in $2$ boxes exactly?

*What is the probability that in the yellow box, red box, green box and blue box will be equal number of marbles in each box?


To solve Q1, I treat that the sample space will be $\binom{10}{6}$, meaning that there are $10$ marbles scattered in $6$ boxes.
I have trouble describing the event space. Should it be $\binom{10}{1}$, picking all $10$ marbles in the same one box? Or $\binom{6}{1}$, picking $1$ box for all the marbles?
About Q2 I have the same situation, about describing the event space. $\binom{10}{2}$, picking marbles, or $\binom{6}{2}$, picking boxes?
 A: We are placing balls in boxes.  Therefore, we have to choose which box receives which ball.
The number of elements in our sample space is $6^{10}$ since there are six choices for each of the ten balls.

There are six boxes, with each box a different color, and $10$ different marbles.  The marbles are scattered randomly in the boxes.  What is the probability that all marbles are in the same box.

There are six favorable cases, depending on which of the six boxes receives all the balls.  Hence,
$$\Pr(\text{all balls are placed in the same box}) = \frac{\binom{6}{1}}{6^{10}} = \frac{6}{6^{10}} = \frac{1}{6^9}$$

There are six boxes, with each box a different color, and $10$ different marbles.  The marbles are scattered randomly in the boxes.  What is the probability that all marbles are in two boxes exactly.

We have to choose which two boxes will receive the balls, then distribute the balls to the selected boxes so that neither of the selected boxes is empty.  There are $\binom{6}{2}$ ways to select the two boxes which will receive the balls.  One the boxes have been selected, there are two choices for each ball, so there are $2^{10}$ ways to distribute the balls to those two boxes.  However, two of those $2^{10}$ distributions leave one of those two boxes empty.  Thus, there are $2^{10} - 2$ ways to distribute the balls so that both of those boxes receive at least one ball.  Thus,
$$\Pr(\text{all balls are placed in exactly two boxes}) = \frac{\binom{6}{2}(2^{10} - 2)}{6^{10}}$$
