Hello,

Why not ?

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 ?