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last Question: (not verbatim) if x, y, z is between [0, 1], what is the probability that $|x-y| >z$ and $|x-z| >y$

In the contest, I had a very hard time trying to visualize and graph the graph of a cube which represents the x, y, z value and the inequalities itself. I'm trying to look for techniques or maybe references so I can easily graph 3d equations and inequalities.

I am also open to another solution besides 3d geometry that can be applied

SuperMage1
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  • First thought: If all the variable are between 0 and 1, are the absolute values necessary? – Matti P. Mar 09 '20 at 10:29
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    What do you mean by "between $[0,1]$"? What are the probability distributions of $x$, $y$ and $z$? Why do you take absolute values if all terms are positive anyway? Also note that the two inequalities together are equivalent to $x>|y-z|$. – Servaes Mar 09 '20 at 10:29
  • sorry, supposed to be subtraction – SuperMage1 Mar 09 '20 at 10:45
  • The conditions translate to either ( x>y+z ) or ( y>x+z and z>x+y). Second condition isn't possible with given constraint [0,1]. So, the only condition that i see here is x-y-z>0. Now we need to find the volume of this region inside the unit cube – Mathsmerizing Mar 09 '20 at 10:59
  • @Mathsmerizing in addition, $x-y-z\leq 1$. The region is formed by $4$ planes: $x-y-z=1$; $x=0$; $y=0$; and $z=0$. A pyramid with volume $\frac{1}{6}$ – acat3 Mar 09 '20 at 12:30

1 Answers1

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From the inequalities, it is obvious that neither $y$ nor $z$ can be the biggest of the three. Thus $x$ is the biggest and the inequalities become $x>y+z$. Now we need to find the probability of $0\leq y< x$ and $0\leq z < x-y$ with probability density function of $x,y,z$ is a constant: $1$.

$$ \begin{aligned} P(x>y+z)&=\int_{0}^{1}{\int_{0}^{x}{\int_{0}^{x-y}{dz}\ dy}\ dx}\\ \\ &= \int_{0}^{1}{\int_{0}^{x}{\left(x-y\right)\ dy}\ dx}\\ \\ &= \int_{0}^{1}{\frac{x^{2}}{2} dx}\\ \\ &=\frac{1}{6} \end{aligned} $$

Not sure if integral is permitted in this kind of competition though.

In $3D$ space, the region is a pyramid with its $4$ points: $(0,0,0);(1,0,0);(1,1,0);(1,0,1)$. This pyramid’s volume is $\frac{1}{6}$.

acat3
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  • Building off your last line, the conditions in the question are given by 4 planes, and we want the volume contained within. Thus, we find the intersection point of each set of 3 planes, which defines the vertices of the pyramid like you have shown. (I suspect that's their approach the problem from an area perspective) – Calvin Lin Mar 09 '20 at 16:18