Practicing Success
Practicing Success
CUET Preparation Today
The random variable x can take values 0, 1, 2, 3, Given that P(x=0)=P(x-1)=P and P(x-2)=P(x-3) such that $E(x^2)=2E(x)$, then value of 'P' is : |
$\frac{1}{4}$ $\frac{3}{8}$ $\frac{2}{3}$ $\frac{5}{8}$ |
$\frac{3}{8}$ |
