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

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Scootertaj
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1. Let the pdf of X,Y be [tex]f(x,y) = x^2 + \frac{xy}{3}, 0<x<1, 0<y<2[/tex]
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.

The Attempt at a Solution


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