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**lllll** $\displaystyle f(x,y) = \left\{ \begin{array}{rcl}

6(1-y) & \mbox{for} & 0 \leq x \leq y \leq 1 \\

0 & \mbox{for} & \mbox{other}

\end{array}\right. $

so far I have:

for the marginal density functions:

$\displaystyle \begin{array}{l} f_1(x) = \int_{x}^{1} 6-6y \ dy \ 0 \leq x \leq 1 \\ \\

= 6y - 3y^2 \bigg{|}^{1}_{x} \\ \\

= 3 - (6x- 3x^2) \\ \\

= 3(x-1)^2 \\

\end{array} $

$\displaystyle f_1(x) = 3(x-1)^2 \ \ 0 \leq x \leq 1 \\ $

$\displaystyle \begin{array}{l} f_2(y) = \int_{1}^{0} 6-6y \ dx \ x \leq y \leq 1 \\ \\

= 6x - 6xy \bigg{|}^{1}_{0} \\ \\

= 6(1-y) \\ \\

f_2(y) = 6(1-y) \ \ x \leq y \leq 1 \\

\end{array} $

**so my marginals are:**

$\displaystyle f_1(x) = \left\{ \begin{array}{rcl}

3(x-1)^2 & \mbox{for} & 0 \leq x \leq 1 \\

0 & \mbox{for} & \mbox{other}

\end{array}\right. $

$\displaystyle f_2(y) = \left\{ \begin{array}{rcl}

6(1-y) & \mbox{for} & x \leq y \leq 1 \\

0 & \mbox{for} & \mbox{other}

\end{array}\right. $

Find $\displaystyle P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) $

for $\displaystyle X \leq \frac{3}{4} $:

$\displaystyle \int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = - \frac{1}{64}$ **which is clearly wrong.**

and $\displaystyle Y \leq \frac{1}{2} $:

$\displaystyle \int^{1/2}_{x} 6(1-y) \ dy = 6y-3y^2 \bigg{|}^{1/2}_{x} = \frac{9}{4} - 6x-3x^2 $ **which doesn't make any sense**.

once you get the appropriate values do you just divide Y by X?

$\displaystyle P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{32}{63}$ according to the solution in the back of the book.

I have no clue what I'm doing wrong and any help clarifying this would be greatly appreciated.