Finding $\mathbb{P}(Y
Let $X$ and $Y$ be two independent random variables. The pdf of $Y$ is $f_Y(y)$ with $y \geq a$, while the pdf of $X$ is
  \begin{equation}
f_X(x) =  
\begin{cases} 
f_{X,1}(x), \quad b \leq x\leq c \\
f_{X,2}(x), \quad  c < x \leq d 
\end{cases}
\end{equation} 
  where $a$, $b$, $c$ and $d$ are all positive real numbers such that $a\leq b<c<d<\infty$. Find $\mathbb{P}[Y<g(X)]$ with $g(X)=(c_1 X^{-c_2})^{-1/c_3}$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.
Here is my latest attempt: 
$$
\mathbb{P}(Y<g(X)) =\int_a^\infty(1-F_{g(X)}(y)) f_Y(y) \, dy
$$
Letting $Z=g(X)$.
\begin{align*}
F_Z(z)&=\mathbb{P}(Z<z) \\
&=\mathbb{P}(g(X)<z)\\
&=\mathbb{P}(X<g^{-1}(z)) \\
& = F_X(g^{-1}(z)) \\
&=
\begin{cases}
\int_b^{g^{-1}(z)} f_{X,1}(x) \,dx, \quad g(b) \leq z\leq g(c) \\
\int_b^c f_{X,1}(x) \,dx+\int_c^{g^{-1}(z)} f_{X,2}(x) \,dx, \quad  g(c) <z \leq g(d)
\end{cases}
\end{align*}
 Then,
\begin{align*}
\mathbb{P}(Y<g(X))= {} &\int_a^\infty (1-F_{g(X)}(y)) f_Y(y) \, dy
= {}  \int_a^\infty(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy
\end{align*}
Now $g(.)$ is a monotonically increasing function. Hence, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the required probability is zero. Otherwise, we have three cases: 
(1) If $a<g(b)$
\begin{align*}
\mathbb{P}(Y<g(X)) = {} & \int_a^{g(b)}(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy \\
+ {} &\int_{g(b)}^{g(c)}(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy \\
+ {} &\int_{g(c)}^{g(d)}(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy
\end{align*} 
(2) If $g(b)<a<g(c)$
\begin{align*}
\mathbb{P}(Y<g(X)) = {} \int_{a}^{g(c)}(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy 
+ \int_{g(c)}^{g(d)}(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy
\end{align*} 
(3) If $g(c)<a<g(d)$
\begin{align*}
\mathbb{P}(Y<g(X)) = {}  \int_{a}^{g(d)}(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy
\end{align*} 
I am unsure how to proceed from here (or even if my attempt is correct to begin with). Any help with that?
 A: I'm posting this just in case someone else is tackling a problem similar to mine. I've come up with this solution after several trials and have verified it through numerical evaluation. I may be missing something though, given that I am a neophyte to probability.
\begin{align*}
\mathbb{P}(Y<g(X))= {} &\int_a^\infty (1-F_{g(X)}(y)) f_Y(y) \, dy
= {}  \int_a^\infty(1-F_{X}(g^{-1}(y)) f_Y(y) \,dy
\end{align*}
with
\begin{align*}
F_X(g^{-1}(z))
=
\begin{cases}
\int_b^{g^{-1}(z)} f_{X,1}(x) \,dx, \quad \quad \quad \quad \quad \quad \quad \, \, \, \, g(b) \leq z\leq g(c) \\
\int_b^c f_{X,1}(x) \,dx+\int_c^{g^{-1}(z)} f_{X,2}(x) \,dx, \quad  g(c) <z \leq g(d)
\end{cases}
\end{align*}
Since $g(.)$ is a monotonically increasing function, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the desired probability is zero. Otherwise, we have three cases: 
(1) If $a<g(b)$
\begin{align*}
\mathbb{P}(Y<g(X)) = {} & \int_a^{g(b)}f_Y(y) \,dy \\
+ {} &\int_{g(b)}^{g(c)}\left(1-\int_b^{g^{-1}(y)} f_{X,1}(x) \,dx\right) f_Y(y) \,dy \\
+ {} &\int_{g(c)}^{g(d)}\left(1- \left[ \int_b^c f_{X,1}(x) \,dx+\int_c^{g^{-1}(z)} f_{X,2}(x) \,dx\right]\right) f_Y(y) \,dy
\end{align*} 
(2) If $g(b)<a<g(c)$
\begin{align*}
\mathbb{P}(Y<g(X)) = {} &\int_{a}^{g(c)}\left(1-\int_b^{g^{-1}(y)} f_{X,1}(x) \,dx\right) f_Y(y) \,dy \\
+ {} &\int_{g(c)}^{g(d)}\left(1- \left[ \int_b^c f_{X,1}(x) \,dx+\int_c^{g^{-1}(z)} f_{X,2}(x) \,dx\right]\right) f_Y(y) \,dy
\end{align*} 
(3) If $g(c)<a<g(d)$
\begin{align*}
\mathbb{P}(Y<g(X)) = {}  \int_{a}^{g(d)}\left(1- \left[ \int_b^c f_{X,1}(x) \,dx+\int_c^{g^{-1}(z)} f_{X,2}(x) \,dx\right]\right) f_Y(y) \,dy
\end{align*} 
