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:

$\displaystyle \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) = \int_{0}^{1/4} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....$