Answer
The probability that $X \le Y$ is $P\left( {X \le Y} \right) = \frac{4}{7}$.
Work Step by Step
In this case, time is always non-negative and the time until next claim can be infinitely long. So, the ranges of $x$ and $y$ are
$0 \le x \le \infty $, ${\ \ \ \ \ }$ $0 \le y \le \infty $
To find the probability that $X \le Y$, we must define a domain ${\cal D}$ such that $x \le y$. Thus, the description:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le \infty ,0 \le x \le y} \right\}$
Evaluate the probability that $X \le Y$:
$P\left( {X \le Y} \right) = \mathop \smallint \limits_{y = 0}^\infty \mathop \smallint \limits_{x = 0}^y p\left( {x,y} \right){\rm{d}}x{\rm{d}}y$
$ = 12\mathop \smallint \limits_{y = 0}^\infty \mathop \smallint \limits_{x = 0}^y {{\rm{e}}^{ - 4x - 3y}}{\rm{d}}x{\rm{d}}y$
$ = - 3\mathop \smallint \limits_{y = 0}^\infty \left( {\left( {{{\rm{e}}^{ - 4x - 3y}}} \right)|_0^y} \right){\rm{d}}y$
$ = - 3\mathop \smallint \limits_{y = 0}^\infty \left( {{{\rm{e}}^{ - 7y}} - {{\rm{e}}^{ - 3y}}} \right){\rm{d}}y$
$ = - 3\left( {\left( { - \frac{1}{7}{{\rm{e}}^{ - 7y}} + \frac{1}{3}{{\rm{e}}^{ - 3y}}} \right)|_0^\infty } \right)$
$ = - 3\left( {\frac{1}{7} - \frac{1}{3}} \right) = \frac{4}{7}$
So, the probability that $X \le Y$ is $P\left( {X \le Y} \right) = \frac{4}{7}$.
