Answer
$C=24$
(a) $P\left( {X \le \frac{1}{2};Y \le \frac{1}{4}} \right) \simeq 0.094$
(b) $P\left( {X \ge Y} \right) = \frac{1}{2}$
Work Step by Step
We have
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{Cxy}&{{\rm{if}{\ }}0 \le x{\ }{\rm{and}}{\ }0 \le y \le 1 - x}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
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.
Since $x \ge 0$ and $y \ge 0$, we require that $C \ge 0$.
2. Second condition: the normalization condition must hold, namely Eq. (5) must be satisfied:
(5) ${\ \ \ \ }$ $\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = 1$
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
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 1 - x} \right\}$
Then, using Eq. (5) we evaluate:
$\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} Cxy{\rm{d}}y{\rm{d}}x = 1$
$\frac{C}{2}\mathop \smallint \limits_{x = 0}^1 x\left( {{y^2}|_0^{1 - x}} \right){\rm{d}}x = 1$
$\frac{C}{2}\mathop \smallint \limits_{x = 0}^1 x\left( {1 - 2x + {x^2}} \right){\rm{d}}x = 1$
$\frac{C}{2}\mathop \smallint \limits_{x = 0}^1 \left( {x - 2{x^2} + {x^3}} \right){\rm{d}}x = 1$
$\frac{C}{2}\left( {\left( {\frac{1}{2}{x^2} - \frac{2}{3}{x^3} + \frac{1}{4}{x^4}} \right)|_0^1} \right) = 1$
$\frac{C}{2}\left( {\frac{1}{2} - \frac{2}{3} + \frac{1}{4}} \right) = 1$
$\frac{C}{{24}} = 1$
So, $C=24$ such that $p\left( {x,y} \right)$ is a joint probability density function.
Thus,
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{24xy}&{{\rm{if}{\ }}0 \le x{\ }{\rm{and}}{\ }0 \le y \le 1 - x}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
(a) Since $x \ge 0$ and $y \ge 0$,
$P\left( {X \le \frac{1}{2};Y \le \frac{1}{4}} \right) = \mathop \smallint \limits_{x = 0}^{1/2} \mathop \smallint \limits_{y = 0}^{1/4} 24xy{\rm{d}}y{\rm{d}}x$
$ = 12\mathop \smallint \limits_{x = 0}^{1/2} x\left( {{y^2}|_0^{1/4}} \right){\rm{d}}x$
$ = \frac{3}{4}\mathop \smallint \limits_{x = 0}^{1/2} x{\rm{d}}x$
$ = \frac{3}{8}\left( {{x^2}|_0^{1/2}} \right) = \frac{3}{{32}} \simeq 0.094$
So, $P\left( {X \le \frac{1}{2};Y \le \frac{1}{4}} \right) \simeq 0.094$.
(b) To evaluate $P\left( {X \ge Y} \right)$ we must define the domain ${{\cal D}_1}$ such that $x \ge y$ and satisfies the domain of $p\left( {x,y} \right)$ where $p$ is nonzero, that is, $0 \le x \le 1$, $0 \le y \le 1 - x$.
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 $x=y$ and bounded right by the line $y=1-x$.
Notice that the upper boundary is the intersection of the two lines: $x=y$ and $y=1-x$:
$y=1-y$
$2y=1$, ${\ \ \ \ }$ $y = \frac{1}{2}$
So, the range of $y$ is $0 \le y \le \frac{1}{2}$.
Thus, the description of ${{\cal D}_1}$:
${{\cal D}_1} = \left\{ {\left( {x,y} \right)|0 \le y \le \frac{1}{2},y \le x \le 1 - y} \right\}$
Evaluate
$P\left( {X \ge Y} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} p\left( {x,y} \right){\rm{d}}x{\rm{d}}y = 24\mathop \smallint \limits_{y = 0}^{1/2} \mathop \smallint \limits_{x = y}^{1 - y} xy{\rm{d}}x{\rm{d}}y$
$ = 12\mathop \smallint \limits_{y = 0}^{1/2} \left( {{x^2}|_y^{1 - y}} \right)y{\rm{d}}y$
$ = 12\mathop \smallint \limits_{y = 0}^{1/2} \left( {1 - 2y} \right)y{\rm{d}}y$
$ = 12\mathop \smallint \limits_{y = 0}^{1/2} \left( {y - 2{y^2}} \right){\rm{d}}y$
$ = 12\left( {\left( {\frac{1}{2}{y^2} - \frac{2}{3}{y^3}} \right)|_0^{1/2}} \right)$
$ = 12\left( {\frac{1}{8} - \frac{1}{{12}}} \right) = \frac{1}{2}$
So, $P\left( {X \ge Y} \right) = \frac{1}{2}$.
