You can compute E[X] and E[Y] I guess ^^ (same binomial distributions)
For the distribution , note that it's equivalent to saying "what is the probability to get l 2's, while there already had k 1's ?
You can also see that l can get values only between 0 and n-k.
And that it follows a binomial distribution (n-k,1/5) ~ Z
So we actually have :
But looks like a lot , doesn't it ?
Can you try to do this ?