Determining how to find $P(X \leq Y )$

Question

Let X and Y denote two independent continuous random variables, where $X \sim \text{Exponential(1)}$ and $Y \sim \text{Exponential(2)}$.

Calculate $P(X \leq Y)$

I know that i have: $$ F_X(x) = 1 - e^{-1x}, \; \; \; x > 0 $$ and $$ F_Y(y) = 1 - e^{-2x}, \; \; \; x > 0 $$ By using the law of total probability i think that i can get this: $$ P(X \leq Y) = \sum_{y = 1}^{x} P(X \leq Y | Y = y)P(Y = y) $$ But if i'm going to be honest then i don't know if i'm using the law of total probability correct or even if i'm supposed to use it.

Update: Thanks for the help i'm pretty sure i understand it now and know how to solve it.

Answer 1

By independence, the joint distribution of $(X, Y)$ is given by $$ f_{X, Y}(x, y) = f_X(x)f_Y(y) = 2e^{-x - 2y}1_{x > 0}1_{y > 0} $$

Then, the probability to be calculated is $$ \mathbb{P}(X \le Y) = \iint_{0 < x \le y} 2e^{-x - 2y}dxdy $$

Answer 2

Since $X$ and $Y$ are independent, their joint density is: $$f(x, y)=f_X(x)\cdot f_Y(y) = 2e^{-(x+2y)},$$ which is defined on non negative real numbers. We want the probability of $y$ taking any value in $[0,\infty)$ and $x$ taking any value in $[0,y]$, so $$P(X\leq Y)=\int_{0}^\infty\int_0^y 2e^{-(x+2y)}dxdy=\frac{1}{3}.$$