How to find $P(X+Y \geq c)$ cumulative probability distribution Suppose X and Y are both exponentially distributed with the same $\lambda$, what is the cumulative density distribution for the sum of the two? 
i.e. how do we find $P(X+Y \geq c)$
 A: Assuming $X$ and $Y$ are independent random variables with densities $f_X$ and $f_Y$,
$$f_X(x) = f_Y(x) = \begin{cases} 0 & \mbox{if } x < 0 \\ \lambda e^{-\lambda x} &\mbox{if } x \geq 0. \end{cases}$$
If we let $Z = X+Y$ and denote its density as $f_Z$, for $z \geq 0$ we have
\begin{align*}
f_Z(z) &= \int_{-\infty}^\infty f_X(z-x)\,f_Y(x) \; \mathrm{d}x & \text{convolution formula} \\
&= \int_0^\infty f_X(z-x)\,f_Y(x) \; \mathrm{d}x & f_Y(x) = 0 \text{ for }x < 0 \\
&= \int_0^z f_X(z-x)\,f_Y(x) \; \mathrm{d}x & f_X(z-t) = 0 \text{ for }t > z \\
&= \int_0^z \lambda e^{-\lambda(z-x)}\,\lambda e^{-\lambda x} \; \mathrm{d}x & \text{substitution} \\
&= \lambda^2 \int_0^z e^{-\lambda z} \; \mathrm{d}x & \text{simplification} \\
&= \lambda^2 z e^{-\lambda z}.  & e^{-\lambda z} \text{ is constant in $x$} \\
\end{align*}
When $z < 0$, we have $f_Z(z) = 0$. It follows that for $c > 0$, 
\begin{align*}
P(X + Y < c) &= \int_0^c \lambda^2 z e^{-\lambda z} \; \mathrm{d}z & \text{definition of CDF} \\
&= \left.\left(-\lambda z e^{-\lambda z}\right)\right|_0^c + \lambda \int_0^c e^{-\lambda z} \; \mathrm{d}z  & \text{integrate by parts} \\
&= -\lambda c e^{-\lambda c} + 1 - e^{-\lambda c}  & \text{integrate and simplify} \\
&= 1-e^{-c \lambda } (1 + c \lambda).  & \text{factor} \\
\end{align*}
It follows that 
$$P(X + Y \geq c) = 1 - P(X + Y < c) = e^{-c \lambda } (1 + c \lambda).$$
Since the sum of two positive random variables is always positive, the final solution is
$$P(X + Y \geq c) =  \begin{cases} 1 & \mbox{if } c \leq 0 \\
e^{-c \lambda } (1 + c \lambda) & \mbox{if } c > 0. \end{cases}$$
A: We calculate $\Pr(X+Y)\le c$. This is obviously $0$ if $c\le 0$. So from now on assume that $c\gt 0$.
Draw  the line $x+y=c$. We want the probability that $(X,Y)$ lands in the triangle $T$ with corners $(0,0)$, $(c,0)$, and $(0,c)$. This is
$$\iint_T \lambda^2 e^{-\lambda x}e^{-\lambda y}\,dx\,dy.$$
Here we assumed that the joint density is the product of the individual densities. That requires assuming that $X$ and $Y$ are independent. That condition was regrettably left out of the problem statement. 
Express this as an iterated integral, integrating first, say, with respect to $y$ (it doesn't really matter). We get
$$\int_{x=0}^c \left(\int_{y=0}^{c-x}\lambda^2 e^{-\lambda x}e^{-\lambda  y}\,dy\right)\,dx.$$
Now calculate.
