Probability of exact event occurring? Going through a discrete math textbook on my own time with no answer key and just want to check my work here and see if my thinking is correct. Lets say we have 20 boxes and 3 marbles. We choose 1 box at a time, uniformly at random, to drop a single marble into. What is the probability that exactly 2 boxes will have marbles? In other words, what is the probability that, out of 3 marbles, exactly 2 land in the same bucket?
So here is my thinking:
1.) prob of choosing a specific box $\frac{1}{20}$
2.) prob of second marble going in that same box $(\frac{1}{20})^2$
3.) prob of third marble going to any other open box $\frac{19}{20}$
4.) result: $\left(\frac{1}{20}\right)^2 \cdot\frac{19}{20}$ right?
 A: Notice that the box which contains $2$ marbles can be in any box, so the probability of the first marble would actually be $1$ instead of $\frac{1}{20}$.
Since we are dealing with probability, we can order the marbles. Call the marbles 1,2, and 3. Since each marble can go in $20$ different boxes, the total amount of ways, or the denominator, would be $20^3=8000$.
We can have $1,2$ in the same box, $1,3$ in the same box or $2,3$ in the same box. For $1,2$ in the same box, there are $20\cdot19=380$ ways to do that since $2$ boxes are occupied and they are not identical (one with 1 marble and the other with 2 marbles). Since the marbles are ordered, the other two cases are identical and would also have $380 $ ways.
Therefore, our answer would be $\frac{380\cdot3}{8000}=\frac{57}{400}$.
A: Probability all three same box $=(\frac{1}{20})^2=\frac{1}{400}$
Probability all three in different boxes $=\frac{19\times 18}{20^2}=\frac{342}{400}$
Probability of exactly two in one box $=(1-$ above summed) $=\frac{57}{400}$.
