The random variable X can take only the values `0; 1;2`. Giventhat `P(X=0) = P(X=1) = p` and that `E(X^2) = E(X)`; find the value of `p`.

Explanation

It is Given that ,
X=0,1,2 and P(X=0)=P(X=1)=p

Let P(X) at X=2 is x


`Rightarrow p+p+x=1`
`Rightarrow x=1-2 p`


Now, we have the following distribution:


` {X} 0 1 2
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