Question 1029572
Since E(X) = 2, we must have P(X = 1) = P(X = 3) = p and P(X = 2) = 1-2p for some p.


Since *[tex \large E(X^2) = 5], we have *[tex \large 1p + 4(1-2p) + 9p = 5] or *[tex \large 2p + 4 = 5]. So p = 1/2, so 


P(X = 1) = 1/2
P(X = 2) = 0
P(X = 3) = 1/2