Let's consider two independent variables $X \sim Exp(4)$ and $Y \sim Exp(12)$. I want to calculate $$P(\frac{X}{X-3Y} \le \frac19)$$
My work so far
$$P(\frac{X}{X-3Y} \le \frac19) = P(\frac{9X-X+3Y}{X-3Y} \le 0)= P(\frac{8X+3Y}{X-3Y} \le 0)= $$ $$ = P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) + P(\{8X+3Y \le 0\} \cap \{X-3Y\ge0\})$$
I'm not sure what I should do next... Intuitevly I would just decompose $P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) = P(\{8X+3Y \ge 0\}) \cdot P(\{X-3Y\le0\})$ but thinking more it's not so obvious for me that these events are independent.
Am I going in the right direction ?
Because the support of $X$ and $Y$ is $[0, \infty)$, $P(\{8X+3Y \le 0\}=0$
$$ P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) + P(\{8X+3Y \le 0\} \cap \{X-3Y\ge0\})= P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) = P(\{X-3Y\le0\}),$$
$$P(\{X-3Y\le0\}) = \int_0^{\infty}\int_0^{3y}f_y(y)f_x(x)dxdy= \\\int_0^{\infty}\int_0^{3y}12e^{-12y}4e^{-4x}dxdy = 0.5$$
Interpretation: Exp($\lambda$) represents the waiting times between Poisson($\lambda$)-distributed events. $E_Y$, the Poisson event associated with $Y$, is $\frac{12}{4} = 3$ times more frequent than $E_X$, the Poisson event associated with $X$. The probability that $E_X$ will occur once before $E_Y$ happens three times $\big(P(\{X-3Y\le0\})\big)$ is 50%, because the waiting times between different $E_Y$s are independent.