Answer
$P\left( {3X + 2Y \ge 6} \right) = \frac{{19}}{{33}}$
Work Step by Step
Recall the probability density in Exercise 52:
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{C\left( {4x - y + 3} \right)}&{{\rm{if}}{\ }0 \le x \le 2{\ }{\rm{and}}{\ }0 \le y \le 3}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
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
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 2,0 \le y \le 3} \right\}$
Since the normalization condition must hold, namely Eq. (5) in Section 16.5 must be satisfied:
$\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = 1$
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,
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{{33}}\left( {4x - y + 3} \right)}&{{\rm{if}}{\ }0 \le x \le 2{\ }{\rm{and}}{\ }0 \le y \le 3}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
To evaluate $P\left( {3X + 2Y \ge 6} \right)$ we must define a domain ${{\cal D}_1}$ such that $3x + 2y \ge 6$ and satisfies the domain of $p\left( {x,y} \right)$ where $p\left( {x,y} \right)$ is nonzero, that is, $0 \le x \le 2$, $0 \le y \le 3$.
We sketch the domain and from the figure attached we can describe ${{\cal D}_1}$ as a horizontally simple region bounded left by the line $3x + 2y = 6$ and bounded right by the line $x=2$.
Using equation $3x + 2y = 6$, we obtain $x = \frac{1}{3}\left( {6 - 2y} \right)$.
Thus, the description of ${{\cal D}_1}$:
${{\cal D}_1} = \left\{ {\left( {x,y} \right)|0 \le y \le 3,\frac{1}{3}\left( {6 - 2y} \right) \le x \le 2} \right\}$
Evaluate $P\left( {3X + 2Y \ge 6} \right)$ on ${{\cal D}_1}$:
$P\left( {3X + 2Y \ge 6} \right) = \mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{x = \left( {6 - 2y} \right)/3}^2 \frac{1}{{33}}\left( {4x - y + 3} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{{33}}\mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{x = \left( {6 - 2y} \right)/3}^2 \left( {4x - y + 3} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{{33}}\mathop \smallint \limits_{y = 0}^3 \left( {\left( {2{x^2} - xy + 3x} \right)|_{\left( {6 - 2y} \right)/3}^2} \right){\rm{d}}y$
$ = \frac{1}{{33}}\mathop \smallint \limits_{y = 0}^3 \left( {8 - 2y + 6 - \frac{2}{9}{{\left( {6 - 2y} \right)}^2} + \frac{1}{3}\left( {6 - 2y} \right)y - \left( {6 - 2y} \right)} \right){\rm{d}}y$
$ = \frac{1}{{33}}\mathop \smallint \limits_{y = 0}^3 \left( { - \frac{{14}}{9}{y^2} + \frac{{22}}{3}y} \right){\rm{d}}y$
$ = \frac{1}{{33}}\left( {\left( { - \frac{{14}}{{27}}{y^3} + \frac{{22}}{6}{y^2}} \right)|_0^3} \right)$
$ = \frac{1}{{33}}\left( { - 14 + \frac{{66}}{2}} \right) = \frac{{19}}{{33}}$
So, $P\left( {3X + 2Y \ge 6} \right) = \frac{{19}}{{33}}$.
