How to find $\Pr(z<X<Y)$ if $X$ and $Y$ are independent exponential r.v.'s with parameters $\lambda$ and $\mu$
So $x$ is bounded by $z$ and $y$, and y must be go from $z$, (and not from $0$) to infinity, I tried:
$\displaystyle\int_z^{\infty}\mu e^{-\mu y}\int_z^y\lambda e^{-\lambda x}dxdy=\int_z^{\infty}\mu e^{-\mu y}(e^{-\lambda z}-e^{-\lambda y})dy$
$\displaystyle=e^{-\lambda z}\int_z^{\infty}\mu e^{-\mu y}dy-\int_z^{\infty}\mu e^{-y(\lambda+\mu)}dy=e^{-\lambda z}(-e^{-\mu y}\bigg|_z^{\infty})-(\frac{\mu}{\lambda+\mu}e^{-(\lambda+\mu )y}\bigg|_z^{\infty}$
$\large=e^{-z(\lambda+\mu)}-\frac{\mu}{\mu+\lambda}e^{-z(\lambda+\mu)}=\frac{\lambda}{\lambda+\mu}e^{-z(\lambda+\mu)}$
Is my solution correct ?
Thanks in advance