# three points question

• Mar 2nd 2011, 03:54 AM
Volga
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)?
• Mar 2nd 2011, 01:47 PM
theodds
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.
• Mar 5th 2011, 02:57 AM
Volga
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)?
• Mar 5th 2011, 06:40 AM
theodds
Whoops, did it again and found a small error. I think it is 1/6.

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

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.