Given that $X$~$Exp(1)$ and $Y$~$Exp(1)$, what is the likelihood of $x+y \gt 2$

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So, to sum it up swiftly, $X$ symbolizes the likelihood of $X$'s age and if $X$ breaks, $y$ will hopefully live longer. They both have their probability density functions as $e^{-x}$ and $e^{-y}$.

I'm trying to calculate the probability of them exceeding at least the 2-year-mark, but I'm having difficulties.

What I've done thus far is that I've divided the problem into two events:

  1. X exceeds the 2-year-mark alone, the probability of this happening is $e^{-2}=0.13533...$
  2. $X$ breaks between 0 and 2 years and $Y$ has to reach at least $2-x$ years of age

For #2 I'm having trouble at deciding what I should use as $f(x)$ when I'm trying to calculate $\iint_{x,y}f(x)dydx$ and I'm not sure if my boundaries are correct.

The boundaries I've brainstormed are as follows:

$0 \lt x \lt 2$

$2-x \lt y \lt \infty$

$x \lt y \lt \infty$

$0 \lt y \lt 2-x$

(The last boundary would be inverse case for $y$ for when $x+y \lt 2$

Can someone help me by telling me how I'm approaching this problem wrong?

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"1. X exceeds the 2-year-mark alone, the probability of this happening is $e^{-2}=0.13533\ldots$" i.e.

$$\int_{x=2}^\infty e^{-x} \, dx$$

and similarly

"2. $X$ breaks between $0$ and $2$ years and $Y$ has to reach at least $2-x$ years of age" so

$$\int_{x=0}^2 \int_{y=2-x}^\infty e^{-x}e^{-y}\, dy \, dx$$