JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
Let a random variable $X$ take values $0,1,2,3$ with $P(X=0)=P(X=1)=p$, $P(X=2)=P(X=3)$ and $E(X^{2})=2E(X)$. Then the value of $8p-1$ is :
Statistics
Solution
$2p+2q=1$. $E(X)=p+2q+3q=p+5q$, $E(X^2)=p+4q+9q=p+13q$. $p+13q=2p+10q\Rightarrow p=3q$. Then $8q=1$, $q=\frac18$, $p=\frac38$. $8p-1=3-1=2$. Option B.