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The solution for two uniformly iid variables can be found here:

Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?

However I have generalising to the case where there are three variables:

I try to generalize Rahul's answer:

For the general case, if $X$, $Y$ and $Z$ are three independent, identically distributed variables with probability distribution function $f$ supported in $[0,1]$, then the probability that their sum exceeds $1$ is $$\begin{align} P[X+Y+Z\ge1]=\iint\limits_{\substack{0\le x\le1\\0\le y\le1\\0\le z\le1\\{x+y+z\ge1}}} f(x)f(y)f(z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^1f(x)\left(\int_{1-x}^1f(y)\left(\int_{1-x-y}^1f(z)\,\mathrm dz\right)\,\mathrm dy\right)\,\mathrm dx\\=\int_0^1f(x)\int_{1-x}^1f(y)\big(1-F(1-x-y)dy\big)\,\mathrm dx, \end{align}$$ where $F(t)=\int_0^tf(u)\,\mathrm du$ is the cumulative distribution function of the random variables.

Is this correct so far?

How can I use this expression to calculate the probability that X+Y+Z>1?

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Bazman
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    Easier to compute the probability that they add up to less than $1$, because that's the volume of a relatively simple region, a pyramid with corners $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$. – Thomas Andrews Oct 11 '15 at 20:31
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    The probability that $n$ such numbers adds up to less than $1$ is $\frac{1}{n!}$. – Thomas Andrews Oct 11 '15 at 20:33
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    Here's a general argument for $X_1+X_2+\cdots+X_n\leq 1$ has probability $1/n!$. http://math.stackexchange.com/a/315300/7933 – Thomas Andrews Oct 11 '15 at 20:37
  • Hi Thomas can you show me how you get to the 1/n! expression? – Bazman Oct 11 '15 at 20:37
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    Take a cube of side $1$, and let $P$ be a vertex. Let $A,B,C$ be the adjoining vertices, and draw the plane through these. This cuts off a tetrahedral chunk from the cube. By geometry this chunk has volume $\frac{1}{6}$. – André Nicolas Oct 11 '15 at 20:38
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    My basic argument is that if you pick a sequence of $n$ random numbers, the probability that they come out sorted is "obviously" $1/n!$, and then I prove that the set ${\sum x_i\leq 1}$ is the image of the "sorted" set under a linear transformation that preserves volumes. There's probably a more direct geometric approach, but I like the "obvious" part about sorting probabilities. – Thomas Andrews Oct 11 '15 at 20:40
  • The title says that you're interested in uniformly distributed numbers, but the body doesn't -- please bing them into agreement. – joriki Oct 11 '15 at 22:21

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