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?
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 favorable cases, depending on which of the six boxes receives all the balls. Hence, $$\Pr(\text{all balls are placed in the same box}) = \frac{\binom{6}{1}}{6^{10}} = \frac{6}{6^{10}} = \frac{1}{6^9}$$
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, $$\Pr(\text{all balls are placed in exactly two boxes}) = \frac{\binom{6}{2}(2^{10} - 2)}{6^{10}}$$