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Probability problem

  1. Mar 4, 2005 #1
    I'm trying to learn probability on my own and I'm stuck.

    My multiple-variable-calculus is not so strong so the following problem got me stuck.

    I have density function

    f(x,y) = x^2 + xy/3 for 0<x<1; 0<y<2 otherwise 0

    And I need to calculate Prob(X > Y). X and Y are random variables.

    I know how to do Prob(X <= 0.5) etc.

    Also would be nice if someone could explain Prob(Y < 1/2 and X < 1/2) and
    Prob(X+Y < 1)

    Thanks
    /Niels
     
  2. jcsd
  3. Mar 4, 2005 #2
    Just to make sure, this is just integrating over all values of Y, such that X <= 0.5, which is a rectangle.

    Similarly, this is an integration of the density function over the region where Y < 1/2 and X < 1/2. This is a square of sidelength 1/2 with one corner at the origin. It is a double integral that can be written as:
    [tex]\int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} f(x,y) dy dx[/tex]
    Do you know how to find P(X<1/2 OR Y<1/2) ?
    If you cannot picture the region, rewrite it in a friendlier form; ie., Y < -X + 1. You're then integrating the density function for all points below the line y = -x + 1. This integral can be written:
    [tex]\int_0^1 \int_0^{-x+1} f(x,y) dy dx[/tex]
    So your original problem, P(Y<X) is just the set of points below the line y=x.
     
    Last edited: Mar 4, 2005
  4. Mar 4, 2005 #3
    thank you!
     
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