a) Perhaps it's my fault but I don't understand your notation. If p (X 1) represents the probability that X is 1, then what is the X Y doing there?

b) Y = 2, so the situation is the same as choosing 4 balls out of three white and six red. C(n, k) means n choose k.

E[X|Y=2] = P(X=1|Y=2)*1 + P(X=2|Y=2)*2 + P(X=3|Y=2)*3

= (1/C(9,4))*(C(3,1)*C(6,3)*1 + C(3,2)*C(6,2)*2 + C(3,3)*C(6,1)*3)

c) No. As a simple counterexample, P(X=2) > 0, but P(X=2|Y=5) = 0.