Answer
$P\left( {X + Y \le 2} \right) \simeq 0.074$
Work Step by Step
As in Exercise 49, we have the joint probability density function:
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{{72}}\left( {2xy + 2x + y} \right)}&{{\rm{if}{\ }}0 \le x \le 4{\ }{\rm{and}{\ }}0 \le y \le 2}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
The condition $X + Y \le 2$ corresponds to the domain ${\cal D}$ given by:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 2,0 \le y \le 2 - x} \right\}$
By definition, the probability that $X$ and $Y$ satisfy the condition: $X + Y \le 2$ is given by
$P\left( {X + Y \le 2} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} p\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^{2 - x} \left( {2xy + 2x + y} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {\left( {x{y^2} + 2xy + \frac{1}{2}{y^2}} \right)|_0^{2 - x}} \right){\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {x{{\left( {2 - x} \right)}^2} + 2x\left( {2 - x} \right) + \frac{1}{2}{{\left( {2 - x} \right)}^2}} \right){\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {2 - x} \right)\left( {x\left( {2 - x} \right) + 2x + \frac{1}{2}\left( {2 - x} \right)} \right){\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {2 - x} \right)\left( {2x - {x^2} + 2x + 1 - \frac{1}{2}x} \right){\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {2 - x} \right)\left( { - {x^2} + \frac{7}{2}x + 1} \right){\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {{x^3} - \frac{{11}}{2}{x^2} + 6x + 2} \right){\rm{d}}x$
$ = \frac{1}{{72}}\left( {\left( {\frac{1}{4}{x^4} - \frac{{11}}{6}{x^3} + 3{x^2} + 2x} \right)|_0^2} \right)$
$ = \frac{1}{{72}}\left( {4 - \frac{{44}}{3} + 12 + 4} \right) = \frac{2}{{27}} \simeq 0.074$
So, $P\left( {X + Y \le 2} \right) \simeq 0.074$.
