How to calculate the joint probability: $\Pr \left( \tfrac{g_1}{g_3} \geq \theta_1, \tfrac{g_2}{g_3} \geq \theta_2, g_3 > \theta_3 \right)$? Question: 
How to calculate the following?
$$\Pr \left( \dfrac{g_1}{g_3} \geq \theta_1, \dfrac{g_2}{g_3} \geq \theta_2, g_3 > \theta_3 \right),$$
where $g_i, i \in \{1, 2, 3\}$ is an exponentially distributed random variable with pdf given by 
$$f_{g_i}(x) = \dfrac{1}{\Omega_i} \exp\left( \dfrac{-x}{\Omega_i}\right),$$
and $\theta_i, \Omega_i \in \mathbb R^+ \forall i$. Also, all the $g_i$s are assumed to be mutually independent.
Attempt:
For the case when we have to calculate the following
$$\Pr\left( \dfrac{g_1}{g_3} \geq \theta_1, g_3 > \theta_3 \right) = \dfrac{1}{\Omega_1 \Omega_3} \int_{y = \theta_1 \theta_3}^{y = \infty} \int_{x = \theta_3}^{x = y/\theta_1} f_{g_3}(x) f_{g_1}(y) \ \mathrm dx \ \mathrm dy.$$
However, I am stuck due to the third variable $g_2$. Any leads appreciated. 
 A: I assume those $G_i$ are independent. Then conditional on $G_3$ will simplify the calculations:
$$ \begin{align} & \Pr\left\{\frac {G_1} {G_3} \geq \theta_1, \frac {G_2} {G_3} \geq \theta_2, G_3 > \theta_3\right\} \\
= & \int_{\theta_3}^{+\infty} \Pr\left\{\frac {G_1} {G_3} \geq \theta_1, 
\frac {G_2} {G_3} \geq \theta_2, G_3 > \theta_3 ~\Bigg|~ G_3 = g  \right\} 
\frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\ 
= & \int_{\theta_3}^{+\infty} \Pr\{G_1 \geq g\theta_1, G_2 \geq g\theta_2\} \frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\
= & \int_{\theta_3}^{+\infty} \Pr\{G_1 \geq g\theta_1\}\Pr\{G_2 \geq g\theta_2 \} \frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\
= & \int_{\theta_3}^{+\infty} \exp\left\{- \frac {g\theta_1} {\omega_1} \right\}  \exp\left\{- \frac {g\theta_2} {\omega_2} \right\}  
\frac {1} {\omega_3} \exp\left\{- \frac {g} {\omega_3} \right\} dg \\
= & \frac {1} {\omega_3} \int_{\theta_3}^{+\infty} \exp\left\{- \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2}  + \frac {1} {\omega_3} \right)g\right\} dg \\ 
= & - \frac {1} {\omega_3} \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2}  + \frac {1} {\omega_3} \right)^{-1}  
\left. \exp\left\{- \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2}  + \frac {1} {\omega_3} \right)g\right\} \right|_{\theta_3}^{+\infty} \\
=& \left(\frac {\theta_1\omega_3} {\omega_1} + \frac {\theta_2\omega_3} {\omega_2}  + 1 \right)^{-1} \exp\left\{- \left(\frac {\theta_1} {\omega_1} + \frac {\theta_2} {\omega_2}  + \frac {1} {\omega_3} \right)\theta_3 \right\}
\end{align} $$
