
Originally Posted by
mr fantastic
$\displaystyle \Pr \left( X = \frac{1}{4}, 0 < Y < 3 \right) = \Pr \left( 0 < Y < 3 \bigg{|} X = \frac{1}{4} \right) \Pr \left( X = \frac{1}{4} \right)$.
But since you have a continuous distribution, $\displaystyle \Pr \left( X = \frac{1}{4} \right) = 0$ ......
So, do you mean find $\displaystyle \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) \,$ ?
If so:
First draw the region of integration. Then:
$\displaystyle \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) = \int_{0}^{1/4} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....$