Probability of $(X \leq 2Y)$ $X$ and $Y$ are independent and their probability density functions are
$$f_X(t)=f_Y(t)=\left\{\begin{array}{l} e^{-t},\:\text{if $t \geq 0$;} \\ 0,\:\text{otherwise.}\end{array}\right.$$
$P(X \leq 2Y)$=? (The probability of $X \leq 2Y$)

Given that $X$ and $Y$ are independent, the joint density is $f_{X,Y}(x,y)=e^{-x}e^{-y}$.
To find the probability that $X\leq 2Y$ you need to evaluate
$$\text{P}(X\leq 2Y)=\text{P}\left(Y\geq \frac{X}{2}\right)=\int_{x=0}^{x=\infty}\int_{y=\frac{x}{2}}^{y=\infty}e^{-x}e^{-y}dxdy\\=\int_{x=0}^{x=\infty}\left[\int_{y=\frac{x}{2}}^{y=\infty}e^{-y}dy\right]e^{-x}dx\\=\int_{x=0}^{x=\infty}\left[-e^{-y}\right]_{y=\frac{x}{2}}^{\infty}e^{-x}dx\\=\int_{x=0}^{x=\infty}e^{-\frac{3x}{2}}dx=-\frac{2}{3}\left[e^{-\frac{3x}{2}}\right]_0^\infty=\frac{2}{3}$$ 
