$P[XLet $X$ and $Y$ be two independent random variables. The range of $X$ is infinite, say $[a,\infty)$ while the support of the $Y$ is finite, say $[b,c]$, where $a$, $b$, and $c$ are all positive real numbers such that $a\leq b<c<\infty$. I would like to find $\mathbb{P}[X<Y]$. Here is my trial:
\begin{align*}
\mathbb{P}(X<Y)&=\mathbb{P}(X<Y|X\in[a,b])\mathbb{P}(X \in [a,b]) +\mathbb{P}(X<Y|X\in[b,c])\mathbb{P}(X \in [b,c]) +\mathbb{P}(X<Y|X\in[c,\infty])\mathbb{P}(X \in [c,\infty])\\
&=\mathbb{P}(X∈[a,b))+\mathbb{P}(X<Y|X∈[b,c])P(X∈[b,c])\\
&= \int_{a}^{b}\, f_X(x)\,\mathrm{d}x + \int_{b}^{c}F_X(y)\, f_Y(y) \,\mathrm{d}y 
\end{align*}
Is this correct?
 A: Your law ot total probability is wrong - you should have:
$$\mathbb{P}(X<Y)=\mathbb{P}(X<Y|X\in[a,b])\mathbb{P}(X \in [a,b])+\mathbb{P}(X<Y|X\in[b,c])\mathbb{P}(X \in [b,c])+\mathbb{P}(X<Y|X\in[c,\infty])\mathbb{P}(X \in [c,\infty])$$
A: In the first place it looks as if you accidently went for $\mathbb P(X>Y)$ instead of $\mathbb P(X<Y)$.
A good start is:
$$\begin{aligned}\mathbb{P}\left(X<Y\right) & =\mathbb{P}\left(X<Y\mid X<b\right)\mathbb{P}\left(a\leq X<b\right)+\mathbb{P}\left(X<Y\mid b\leq X\leq c\right)\mathbb{P}\left(b\leq X\leq c\right)+\mathbb{P}\left(X<Y\mid X>c\right)\mathbb{P}\left(X>c\right)\\
 & =\mathbb{P}\left(a\leq X<b\right)+\mathbb{P}\left(X<Y\mid b\leq X\leq c\right)\mathbb{P}\left(b\leq X\leq c\right)
\end{aligned}
$$
Secondly we have:$$\mathbb{P}\left(X<Y\mid b\leq X\leq c\right)\mathbb{P}\left(b\leq X\leq c\right)=\mathbb P(X<Y\wedge b\leq X\leq c)=\mathbb E\mathbf1_{X<Y}\mathbf 1_{b\leq X\leq c}$$$$=\int^c_b\mathbb P(x<Y)f_X(x)dx$$In your answer you seem to think however that the LHS of this equality should be $\mathbb{P}\left(X<Y\mid b\leq X\leq c\right)$ which is wrong.  
So our final result is:$$\mathbb P(X<Y)=\mathbb P(a\leq X<b)+\int^c_b\mathbb P(x<Y)f_X(x)dx=F_X(b)+\int_b^c(1-F_Y(x))f_X(x)dx$$
