If $Y=\frac{ln{U}_1}{ln{U}_1+ln(1-U_2)}$

Where $U_1$ and $U_2$

are independent

$U(0,1)$ random

variables, then

variance of $Y$ equals

$($A$)\frac{1}{12}$

$($B$)\frac{1}{3}$

$($C$)$

D$)\frac{1}{6}$