Joint density of random variables

Question

$f_{X,Y}(x,y) = \frac{(xy-2x-2y+4)}{32}$; $2\le x \le y \le 6$ for random vars $X$ and $Y$; find $P(X \gt 3 \mid Y = 5)$

Answer 1

Notice $xy-2x-2y+4 = (x-2)(y-2)$

Then $$\begin{align} f(x,y)&=\tfrac{1}{32}(x-2)(y-2)\mathbf 1_{2\leq x\leq y\leq 6} \\[1ex]f(x,5)&= \tfrac{3}{32}(x-2)\mathbf 1_{2\leq x\leq 5} \\[3ex]\mathsf P(X>3\mid Y=5) &= \dfrac{\int_3^5 (x-2)~\mathrm d x}{\int_2^5(x-2)~\mathrm d x}\\[1ex]&~~\vdots\end{align}$$

Answer 2

This is what I came up with so far, where

$f_{X,Y}(X = x/Y = y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$

$\Rightarrow f_Y(y) = \int_{2}^{s} \frac{1}{32}(xy-2x-s+4)dx$

$\Rightarrow f_Y(y) = \frac{1}{32} \int_{2}^{s}(xy-2x-s+4)dx$

$\Rightarrow f_Y(y) = \frac{1}{32}[\frac{x^{2}y}{2} - x^2 - xy + 4x]_{r=2}^{r=s}$

$\Rightarrow f_Y(y) = \frac{1}{64} (y^3-4y^2+8y-8)$

$\Rightarrow f_{X,Y}(X = x \mid Y = y) = \frac{\frac{1}{32}(xy - 2x - y + 4)} {\frac{1}{64}(y^3-4y^2+8y-8)}$

then provided $y=5$

$\Rightarrow f_{X,Y}(X = x \mid Y = y) \frac{2}{57}(3r-1); 2 \le x \le 5$

Now going back to $P(X \gt 3 \mid Y = 5)$

$\Rightarrow P(X \gt 3 \mid Y = 5) = \int_{3}^{5} f_{X,Y}(X = x \mid Y = 5)dx$

$\Rightarrow P(X \gt 3 \mid Y = 5) = \int_{3}^{5} \frac{2}{57}(3x-1)dx$

$\Rightarrow P(X \gt 3 \mid Y = 5) = \frac{2}{57} \int_{3}^{5} (3x-1)dx$

$\Rightarrow P(X \gt 3 \mid Y = 5) = \frac{2}{57} [\frac{3x^2}{2}-x]_{r=3}^{r=5}$

$\implies P(X \gt 3 \mid Y = 5) = \frac{46}{57} = .807$