Answer
$C = \frac{1}{{33}}$
$P\left( {X \le 1;Y \le 2} \right) = \frac{8}{{33}}$
Work Step by Step
We have a probability distribution:
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{C\left( {4x - y + 3} \right)}&{{\rm{if}}{\ }0 \le x \le 2{\ }{\rm{and}}{\ }0 \le y \le 3}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
Recall the conditions that a joint probability density function must satisfy:
1. First condition: $p\left( {x,y} \right) \ge 0$ for all $x$ and $y$, since probabilities cannot be negative.
2. Second condition: the normalization condition must hold, namely Eq. (5) in Section 16.5 must be satisfied:
$\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = 1$
From the definition of $p\left( {x,y} \right)$ we obtain the domain ${\cal D}$, where $p$ is nonzero. So, the description of ${\cal D}$ is
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 2,0 \le y \le 3} \right\}$
Then, using Eq. (5) in Section 16.5 we evaluate:
$\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^3 C\left( {4x - y + 3} \right){\rm{d}}y{\rm{d}}x = 1$
$C\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^3 \left( {4x - y + 3} \right){\rm{d}}y{\rm{d}}x = 1$
$C\mathop \smallint \limits_{x = 0}^2 \left( {\left( {4xy - \frac{1}{2}{y^2} + 3y} \right)|_0^3} \right){\rm{d}}x = 1$
$C\mathop \smallint \limits_{x = 0}^2 \left( {12x - \frac{9}{2} + 9} \right){\rm{d}}x = 1$
$C\mathop \smallint \limits_{x = 0}^2 \left( {12x + \frac{9}{2}} \right){\rm{d}}x = 1$
$C\left( {\left( {6{x^2} + \frac{9}{2}x} \right)|_0^2} \right) = 1$
$C\left( {24 + 9} \right) = 1$
So, $C = \frac{1}{{33}}$ such that $p\left( {x,y} \right)$ is a joint probability distribution.
Thus,
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{{33}}\left( {4x - y + 3} \right)}&{{\rm{if}}{\ }0 \le x \le 2{\ }{\rm{and}}{\ }0 \le y \le 3}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
Evaluate
$P\left( {X \le 1;Y \le 2} \right) = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^2 \frac{1}{{33}}\left( {4x - y + 3} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{{33}}\mathop \smallint \limits_{x = 0}^1 \left( {\left( {4xy - \frac{1}{2}{y^2} + 3y} \right)|_0^2} \right){\rm{d}}x$
$ = \frac{1}{{33}}\mathop \smallint \limits_{x = 0}^1 \left( {8x + 4} \right){\rm{d}}x$
$ = \frac{1}{{33}}\left( {4{x^2} + 4x} \right)|_0^1 = \frac{8}{{33}}$
So, $P\left( {X \le 1;Y \le 2} \right) = \frac{8}{{33}}$.
