An urn contains three white, six red and five black balls. Six balls are selected randomly. Let

X and Y, respectively, be the numbers of white and black balls selected.

a) Find p (X 1) X Y

b) Find E[X | Y = 2]

c) Are X and Y independent?

Results 1 to 2 of 2

- April 20th 2010, 08:10 AM #1

- Joined
- Apr 2010
- Posts
- 9

- April 20th 2010, 11:53 AM #2
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.