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Dartboards and Normal Distribution: Probability of Hitting

  1. Jul 28, 2013 #1
    I was reading this problem on calculating the probability of hitting a certain region on a dartboard. The number of hits the dart thrower will land at a certain radius R on the dartboard is proportional to e^(-R^2). The task is to take a certain portion of a ring (or annulus) on the dartboard and calculate the probability of hitting it. The full problem statement is at this link, if you could read the first two pages to see the parts relevant to my question, that would be nice: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-c-average-value-probability-and-numerical-integration/session-62-integrals-and-probability/MIT18_01SCF10_Ses62d.pdf

    My question is: why is the probability a ratio of volumes and not areas? The pdf only partially addresses it. Given that the height of the graph ce^(r^2) is the number of hits, my intuitive feeling is that the area under the curve is the number of hits for that range of radii. Could someone explain why it's a volume?
     
  2. jcsd
  3. Jul 28, 2013 #2
    By the way, should I be asking this in homework help? I felt my question was mostly related to the concept of integration and not the specific question, so I posted it here.
     
  4. Jul 28, 2013 #3

    pasmith

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    Homework Helper

    What you have is a random variable [itex]X[/itex] (the point where a dart hits) which takes values on the plane [itex]\mathbb{R}^2[/itex], with an associated probability density function [itex]f : \mathbb{R}^2 \to \mathbb{R}[/itex].

    When you calculate the probability that [itex]X \in A \subset \mathbb{R}^2[/itex], what you are by definition calculating is the double integral
    [tex]
    P(X \in A) = \iint_A f(\mathbf{x})\,\mathrm{d}A.
    [/tex]
    The nature of a probability density is that [itex]f(\mathbf{x}) \geq 0[/itex] for all [itex]\mathbf{x} \in \mathbb{R}^2[/itex] and
    [tex]
    P(X \in \mathbb{R}^2) = \iint_{\mathbb{R}^2} f(\mathbf{x})\,\mathrm{d}A = 1.
    [/tex]
    Those integrals look like an area times a "height", making them "volumes".
     
  5. Jul 28, 2013 #4
    A lot of that notation is new to me, but I've done a little bit of Googling to make some sense out of it.

    I'm still confused. What is the probability density function in this case? It wouldn't be c*e^(-r^2) would it? Because to me that seems like a function that takes a real number and puts out a real number, while the definition of the probability function you listed seems to to suggest that the function takes a two dimensional value and puts out a real number.

    What is the reasoning behind the probability of X being within A (which I take it is what you're calling the area of the dartboard?) being defined as a double integral?
     
  6. Jul 28, 2013 #5

    Stephen Tashi

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    Science Advisor

    A function such as f(x,y) = x^2 only employs the variable x to compute a result, but it is still a function of two variables, just by the way it is defined. So the fact that a function only employs R to determine a result doesn't mean it is a function of 1 variable. How many variables are involved depends on how the article defines the function. It may be that the variable R is not involved at all! For example if the density function is defined a function of f(x,y), it may be that you are expected to find R from x and y as an intermediate step.
     
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