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Math Help - Density function problem?

  1. #1
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    Density function problem?

    I'm not even sure if that's the proper thing to call it, but here's the question.

    A point is uniformly distributed within the disk of radius 1. That is, its density is
    f(x,y)=C, 0≤x^2+y^2≤1
    Find the probability that its distance from the origin is less than x, 0≤x≤1.



    I'm not sure of what I'm supposed to do here. What are my steps?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Density function problem?

    Quote Originally Posted by downthesun01 View Post
    I'm not even sure if that's the proper thing to call it, but here's the question.

    A point is uniformly distributed within the disk of radius 1. That is, its density is
    f(x,y)=C, 0≤x^2+y^2≤1
    Find the probability that its distance from the origin is less than x, 0≤x≤1.

    I'm not sure of what I'm supposed to do here. What are my steps?
    The requested probability is the ratio between the area of a circle af radious \rho and the circle of radious 1...

    P \{d < \rho\} = \frac{\pi\ \rho^{2}}{\pi}= \rho^{2}

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Density function problem?

    I still don't understand the question. I'm trying to picture it right now.

    There's a point that's uniformly distributed within a circle that has a radius of 1, and I'm supposed to find the probability that the point's distance from the origin of the circle is less than x.

    I don't understand where this second circle with a radius of rho comes from.

    I can solve for C. That seems rather straightforward.

    1=\int \int_R Cdxdy=c\pi r^2=c\pi
    c=\frac{1}{\pi}

    But from there I'm pretty confused. If someone could offer some insight, I'd appreciate it. Thanks

    What does c=\frac{1}{\pi} mean? Is it the probability that the point occupies a certain spot in the circle?

    And what is x supposed to be?
    Last edited by downthesun01; July 20th 2011 at 11:16 PM.
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