Statistics & DataEasyRandom Variables ShareRandom Variables · Easy problemA random variable XXX has P(X=0)=14P(X = 0) = \dfrac14P(X=0)=41, P(X=1)=12P(X = 1) = \dfrac12P(X=1)=21 and P(X=2)=14P(X = 2) = \dfrac14P(X=2)=41. Find Var(X)\mathrm{Var}(X)Var(X). Express your answer as a common fraction.Check Show answer Walk through it step by step Practice more like this Review this concept