$P(Y<.5 | X >.5), P(Y>2X), P(.5Let X and Y have the joint probability density function  
$f(x,y) = \frac{3}{2}(x^2 +y^2)$, $0<x<1,0<y<1$  
a.) Find $P(Y<.5|X>.5)$. My answer is $\frac{5}{11}$.  
b.) Find $P(Y>2X)$. I got $\frac{1}{4}$ but I'm not sure if I'm supposed to integrate the joint pdf or the marginal function $f_Y$.  
c.) Find $P(.5<X+Y<1.5)$. I set up the integral as $\int_{.5+y}^{1.5-y}$ but again, not sure. This is the one I'd prefer to be answered most. My book has one example but it's only for $X+Y<a$, and the integrand is $f_Y(y)dyf_X(x)dx$.
 A: For part b, I got .203125.
I set it up as follows:
$$ \frac{3}{2} \int_0^\frac{1}{2} \int_{2x}^1 x^2 + y^2 dy dx $$
My explanation is, set y = 2x.  Find the area bound by the space of (0,1) for x and y, that is above y = 2x.  It needs to be a double integral (thus the joint probability) to get an actual probability.  (The marginal would result in a function, not a probability value.)
For part c, I am pretty sure this is a convolution which requires the convolution formula, which is a lot of tedious integration with the joint pdf you have.
You'd have to break it up into the intervals $.5<z<1$ and $1<z<1.5$
I am learning this too, so I post my answer with limited confidence, arithmetic errors or otherwise.  I will watch to see what the more experienced have to say.
A: This looks a bit like homework, but I thought it fun to check some of your solutions using the mathStatica add-on to Mathematica. Presumably, you will need to show workings anyway :).
Given: random variables X and Y have joint pdf f(x,y):
f = (3/2)*(x^2 + y^2);     domain[f] = {{x, 0, 1}, {y, 0, 1}}; 

Then, the solution to part b is:
Prob[y > 2 x, f]


13/64

and the solution to part c is:
Prob[1/2 < x + y < 3/2, f]


23/32

Hope this helps
