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I tried the $2D$ case with $x,y \in (0,1)$ and $P(x+y < 1) = \frac{1}{2}$

I got this by sketching the inequalities in the question, namely $ 0\leq x,y\leq 1, $ and $y < 1-x$ and seeing that the region that fits the criteria is only half the region allowed to make a choice from. I'm thinking of extending this to $3$ numbers chosen at random between $0$ and $1$, I'd get the inequalities $ 0 \leq x,y,z \leq 1 $ and $ z < 1-x-y $. The former of the inequalities represents a unit cube and the latter is some plane that chops the cube up, so the probability is probably the volume of the allowed region divided by the volume of the unit cube. The problem arises when I try to calculate the volume of the allowed region, I don't know how to set up the integrals that define that region. For the $2D$ case it was simply $$ A = \int_0^1 1-x\,dx = \frac{1}{2} $$ I can't come up with the integrals for the $3D$ case though. I've been told before the answer has something to do with the volume of a $3$ dimensional simplex and that it'd be $\frac{1}{3!}$ but I can't derive how. I think what I need to understand is how to set up the integrals defining a region like that

In general, if you have $ 0\leq x_0,x_1, \cdots,x_n \leq 1$ and $$x_n < 1 -\sum_{i=0}^{n-1} x_i $$ How do you set up the integrals that calculate the $nD$ volume of the region allowed by these inequalities? Could anyone please explain the thought process behind that I'm only aware of the $2D$ case

StubbornAtom
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Instagram-creative_math_
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    See https://math.stackexchange.com/q/1186863/321264 and in general https://math.stackexchange.com/q/769545/321264. – StubbornAtom Jun 01 '20 at 19:42
  • @StubbornAtom It is indeed the right answer though I don't understand how they set up the integrals. My question is more specifically asking about setting up the integrals that define a certain region and how I do that – Instagram-creative_math_ Jun 02 '20 at 04:36
  • Okay then check out the answers to a similar question for the 3D case: https://math.stackexchange.com/q/3177178/321264. – StubbornAtom Jun 02 '20 at 06:22

1 Answers1

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Well for $x+y+z < 1$ to happen:

  • $x$ can be any value $0$ to $1$, as long as,

  • $y$ is any value $0$ to $1-x$, and,

  • $z$ is any value $0$ to $1-x-y$

And those are your integral limits:

$$\int_0^1 dx \int_0^{1-x} dy \int_0^{1-x-y} dz$$

Can you integrate this to get $\frac16$ or do you need more help?

Incidentally, if you consider the probability space to be the cube $[0,1]^3$, then the region is the pyramid with vertices $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$. This has base area $b = \frac12$ and height $h=1$, so the volume is $\frac13 bh= \frac16$.

antkam
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