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

  1. Oct 17, 2012 #1
    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.
    Last edited: Oct 17, 2012
  2. jcsd
  3. Oct 17, 2012 #2
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