Question

The random variable $X$ can take only the values $0,1,2,3$. Given that $P(X=0)=P(X=1)=p$ and $P(X=2)=P(X=3)$ such that $\sum p_i{x}_i^2=2\sum p_i{x}_i$ find the value of $p$.

Answer 1

Let $P(X=2)=P(X=3)=a$

$\sum p_i=1⟹a=\frac{1}{2}-p$

$\sum p_i{x}_i^2=2\sum p_i{x}_i$

$0\left(p\right)+1\left(p\right)+4\left(a\right)+9\left(a\right)=2\left(0\right(p)+1(p)+2(a)+3(a\left)\right)$

$p+13a=2p+10a$

$p=3(\frac{1}{2}-p)⟹p=\frac{3}{8}$