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Math Help - another one of Joint Probability density of X and Y

  1. #1
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    another one of Joint Probability density of X and Y

    If the independent random variables X and Y have the mariginal densities

    f(x)= 0.5 for 0<x<2 and zero elsewhere

    g(y)=1/3 for 0<y<3 and zero elsewhere

    find the value of P(X^2+Y^2 > 1)

    I have the joint prob. density h(x,y)=1/6 for 0<x<2, 0<y<3 and zero elsewhere, it's because they are independent, so their joint p density is the product of those two. and don't know what to do next...

    And I thank you all in advance, "Thank you"
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  2. #2
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    Quote Originally Posted by Judi View Post
    If the independent random variables X and Y have the mariginal densities

    f(x)= 0.5 for 0<x<2 and zero elsewhere

    g(y)=1/3 for 0<y<3 and zero elsewhere

    find the value of P(X^2+Y^2 > 1)

    I have the joint prob. density h(x,y)=1/6 for 0<x<2, 0<y<3 and zero elsewhere, it's because they are independent, so their joint p density is the product of those two. and don't know what to do next...

    And I thank you all in advance, "Thank you"
    If you draw the picture, you'll see it is easiest to evaluate P(X^2 + Y^2 < 1) = 1 - P(X^2 + Y^2 > 1). This is just the quarter circle with radius 1 with probability pi/24.

    If you have to use an integral to evaluate this, write the inequality x^2 + y^2 < 1 as y < sqrt(1 - x^2) with 0 < x < 1. The iterated integral is

    int (0,1) int (0,sqrt(1 - x^2)) 1/6 dy dx = int (0,1) sqrt(1 - x^2)/6 dx.

    You can use The Integrator to find the antiderivative of sqrt(1 - x^2).
    Last edited by JakeD; September 24th 2006 at 01:03 AM.
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  3. #3
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    If you don't mind, could you replain how you got pi/24

    thankyou
    Last edited by Judi; September 24th 2006 at 03:44 PM.
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  4. #4
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    Quote Originally Posted by Judi View Post
    If you don't mind, could you replain how you got pi/24

    thankyou
    1/6 of area of quarter circle. For more on converting area to probability, see my answer to your other post
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