Answer
The probability:
$P\left( {0 \le X \le 2;1 \le Y \le 2} \right) \simeq 0.18$
Work Step by Step
We have
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{{72}}\left( {2xy + 2x + y} \right)}&{{\rm{if}{\ }}0 \le x \le 4{\ }{\rm{and}{\ }}0 \le y \le 2}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
By definition:
$P\left( {0 \le X \le 2;1 \le Y \le 2} \right) = \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 1}^2 p\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 1}^2 \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)|_1^2} \right){\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {4x + 4x + 2 - x - 2x - \frac{1}{2}} \right){\rm{d}}x$
$ = \frac{1}{{72}}\mathop \smallint \limits_{x = 0}^2 \left( {5x + \frac{3}{2}} \right){\rm{d}}x$
$ = \frac{1}{{72}}\left( {\left( {\frac{5}{2}{x^2} + \frac{3}{2}x} \right)|_0^2} \right) = \frac{1}{{72}}\left( {10 + 3} \right) = \frac{{13}}{{72}} \simeq 0.18$
So, the probability: $P\left( {0 \le X \le 2;1 \le Y \le 2} \right) \simeq 0.18$
