The continuous case is easier, but the same logic works here.
Then it gets messy.
Here you have to think about all the possible strings of length n of numbers 1 through k.
All sequences are equally likely.
To have exactly one 2 (and no 1's) that's
To have exactly two 2's (and no 1's) that's
Then exactly three 2's....
The sum of all of these will give you P(Y=2).
The last is easy.