The random variable X can take only the values 0, 1, 2. Given that P (X = 0) = P (X = 1) = p and that E(X2) = E[X], find the value of p.


Given that-


X=0,1,2 and P(X) at X=0 and 1,


Let at X=2, P(X) is x.


p+p+x=1


x=1-2p


We get the following distribution



E(X)= XP(X)


= 0×P + 1×P+ 2(1-2P)


= P+2-4P = 2-3P


And E(X)2= X2 P(X)


= 0×P+ 1× P+ 4×(1-2P)


= P+4-8P=4-7P


Also, given that E(X2) =E(X)


4-7p= 2-3p


4p=2



36
1