a non urn question

[audia4]

Question: We have a distribution in which P{X=0} = 0.4 and P{X=1} = 0.6. From this distribution we draw two samples X_1 and X_2, and we let their average be Y. What is Var(Y)

Well, I thought this was an easy question for I did the following.

E(X) = 0(0.4) + 1(0.6) = 0.6
E(X^2) = 0^2(0.4) = 1^2(0.6) = 0.6

Therefore, Var(Y) = Var((X1 + X2)/2) = 1/4Var(X1 + X2 + 2cov(x1,x2))
= .25[var(x1) + var(x2) + 2cov(x1,x2]

but cov(x1,x2) = 0, since x1 and x2 are drawn from two different samples
However, doing the above calcululation yields 0.25(0.24 + 0.24) = 0.12

but, this answer does not match one of the following except (e)

a 0.6 b. 0.174 c. 0.16 d. 0.36 e. none of the above. AS I am reluctant to choose answer e, I have made the assumption that the answer to this question has to be part of a, b,c or d.

Thank you.