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Probability that distance from the origin of a uniformly distributed point < x

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A point is uniformly distributed within the disk of radius 1.
    That is, its density is f(x,y) = C

    For 0 ≤ x2 + y2 ≤ 1

    Find the probability that its distance from the origin is less than x, 0 ≤ x ≤ 1.

    [Note] My book says that the answer is supposed to be x2.

    2. The attempt at a solution
    Let D be the distance of the point from the origin.
    D = x2 + y2, where the x and y values correspond to the point's x and y components respectively.

    My interpretation is that the unit circle is centered at origin 0,0.

    I can see that D can be GREATER than the point's x component (e.g. as we move away from the x axis and towards the y-axis). However, how can D be less than x? Isn't this x the same as the circle's x-component? Or is this x independent of the point? For example, we can assume that x is not the same as the circle's x-component (let's denote that as ω). Hence, 0 ≤ x ≤ 1 and 0 ≤ ω ≤ 1, where x may or may not be equal to ω.

    If this is the case, do we have to find the probability that ω2 + y2 < x?

    Any explanations or guidance on how to proceed with this problem will be immensely appreciated!
     
    Last edited: Sep 28, 2012
  2. jcsd
  3. Sep 28, 2012 #2

    Ray Vickson

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    What is the area of a circle of radius x?

    RGV
     
  4. Sep 28, 2012 #3
    [itex]\pi(r^{2})[/itex]. I'm not sure how this relates to the question, since I don't understand how D can ever be less than x for a given value of x (unless I'm understanding this incorrectly).
     
  5. Sep 28, 2012 #4

    Ray Vickson

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    The problem is one of bad notation: the person who wrote the question is using the same symbol (x) to stand for two different things in the same problem. The question should have asked for the probability that the distance is <= d, where 0 < d <= 1.

    RGV
     
  6. Sep 28, 2012 #5

    HallsofIvy

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    As you stated initially, a "uniform" probability density, means that the density function is a constant. Since the integral over the circle is 1, that density function must be one over the area of the disk.
     
  7. Sep 28, 2012 #6
    Thank you for your suggestions Ray and HallsofIvy. From what you're saying, I understand the following:

    1) The probability that the point lies within the circle is 1.
    2) We want to find the probability that the point lies within a distance d of the circle's center, where 0 ≤ d ≤ 1. Hence I divide the area of a circle of radius d by the area of the unit circle.

    Can I jump straight to the solution (I get the correct answer) or do I need to back myself up with more logic and precision?
     
  8. Sep 28, 2012 #7

    Ray Vickson

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    That would be OK with me, but you need to be consistent with the way your instructor/notes/textbook presents the material. Basically, though, a distribution is uniform within a region S if the probability of a subregion of area A is cA, where c is chosen to make P(S) = 1. (The point is that the probability does not vary with the location or shape of the subregion; the only thing that counts is its area.)

    RGV
     
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