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Math Help - three points question

  1. #1
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    three points question

    This is just for fun, I found it on the internet but there was no 'definitive' solution.

    "Three points x_1, x_2 and x_3 lie on an interval [0;1]. What is the probability that x_1>x_2+x_3"

    The suggestions there was to find probability distribution of (x_1-(x_2+x_3)) and P((x_1-(x_2+x_3))>0). In addtion, I want to apply uniform continuous distribution for each x_1,...x_3 - would it be a reasonable assumption (since I am not given the distribution of these points, just the interval)?
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  2. #2
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    Yeah, you'll want to assume a uniform distribution. It seems like the typical thing to do.

    I got 1/3, by integrating over the appropriate region of R^3.
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  3. #3
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    I was hoping I would be able to solve it myself, but I cannot. The only thing I knew how to do was to find distribution function of (x2+x3) (convolution integral), and I got two different intervals for that. How can I move from there to a function of x1-(x2+x3)?
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  4. #4
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    Whoops, did it again and found a small error. I think it is 1/6.

    P(x_1 > x_2 + x_3)<br />
= P(x_1 > x_2 + x_3 \cap 1 > x_2 + x_3)<br />
= P(x_1 > x_2 + x_3 \cap x_2 < 1 - x_3) \displaystyle<br />
= \int_0 ^ 1 \int_0 ^{1 - x_3} \int _{x_2 + x_3} ^ 1 \ dx_1 \ dx_2 \ dx_3

    1/3 is what you get if you incorrectly integrate up from 0 to x_2 + x_3; this should give 1/6.

    If you want to do it the hard way, you should find that x_2 + x_3 has the triangle distribution on (0, 2). Then, x_1 and x_2 + x_3 are independent, and you would integrate over the appropriate area of R^2.
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