(adsbygoogle = window.adsbygoogle || []).push({}); 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.

3. 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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# Homework Help: Find P(X+Y<1) in a different way

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