# Find P(X+Y<1) in a different way

1. Oct 17, 2012

### Scootertaj

1. Let the pdf of X,Y be $$f(x,y) = x^2 + \frac{xy}{3}, 0<x<1, 0<y<2$$
Find P(X+Y<1) two ways:

a) P(X+Y<1) = P(X<1-Y)
b) Let U = X + Y, V=X, and finding the joint distribution of (U,V), then the marginal distribution of U.

3. The attempt at a solution
a) P(X<1-Y) = ?
$$P(x<1-y) = \int_0^1 \int_0^{1-y} (x^2 + \frac{xy}{3})dx = 7/72$$
b) I have no idea how to start.

Last edited: Oct 17, 2012
2. Oct 17, 2012

Resolved.