The random variable X can take only the values 0,1,2. If
$P(X = 0) = P(X = 1) = p$ and $E\left( {{X^2}} \right) = E[X]$
then find the value of $p$.

Since, $X = 0,1,2$ and $P(X)$ at $X = 0$ and 1 is $p$, let at $X = 2,P(X)$ is $x$.

$\Rightarrow$ $p + p + x = 1$

$\Rightarrow$ $x = 1 - 2p$

We get,

the following distribution.

$\therefore$ $E[X] = \Sigma XP(X)$

$= 0 \cdot p + 1 \cdot p + 2(1 - 2p)$
$= p + 2 - 4p = 2 - 3p$

and $E\left( {{X^2}} \right) = \Sigma {X^2}P(X)$
$= 0 \cdot p + 1 \cdot p + 4 \cdot (1 - 2p)$
$= p + 4 - 8p = 4 - 7p$

Also, given that $E\left( {{X^2}} \right) = E[X]$
$\Rightarrow$ $4 - 7p = 2 - 3p$

$\Rightarrow$ $4p = 2 \Rightarrow p = \frac{1}{2}$