Random variables $X, Y$ are independent with densities $f_{X}$ and $f_{Y}$ where $f_{X}(x) = \begin{cases} 4x^{2}e^{-2x} & \text{if $x>0$}\\ 0 & \text{otherwise} \end{cases}$
$f_{Y}(x) = \begin{cases} \frac{8}{3}x^{3}e^{-2x} & \text{if $x>0$}\\ 0 & \text{otherwise} \end{cases}$
Let $V=\frac{X}{X+Y}$. Find $Var(V)$.
My problem is that this might be done right from the definition of $Var$ (and the answer is $\frac{3}{98}$) but it takes a lot of time. I am looking for some smarter solution.
So to answer your question...
With $\text{G}(\alpha,p)$ denoting a Gamma distribution having pdf $\displaystyle f(u)=\frac{\alpha^p e^{-\alpha u}u^{p-1}}{\Gamma(p)}\mathbf1_{u>0}$ where $\alpha$ and $p$ are positive, you have $X\sim\text{G}(2,3)$ and $Y\sim\text{G}(2,4)$ independent of each other.
So, $\displaystyle V=\frac{X}{X+Y}\sim\text{Beta}(3,4)$, the beta distribution of the first kind.
Hence $\displaystyle\mathrm{Var}(V)=\frac{3\times4}{(3+4)^2(3+4+1)}$