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I was trying to solve the following problem

If we pick k numbers from the interval (0, 1), what is the probability the sum of those numbers is <1 when ? (k = 6)

for k = 2, got $\displaystyle\lim_{n \to \infty}\frac{\sum_{i = 1}^{n-1} i}{n^2}=0.5$

for k = 3, I got the expression $\displaystyle \lim_{n \to \infty}\frac{\sum_{i=1}^{n-2} \sum_{j = 1} ^{n-i-1} ( n-(i + j))}{n^3}$

how to compute it and is there a way to generalize it for picking k numbers from the interval (0,1)?

Mehdi K.
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2 Answers2

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Presumably each number is chosen independently and uniformly from $(0,1)$.

A more direct way to compute the probability is is to think of the three-dimensional cube consisting of points $(x,y,z)$ such that $x,y,z \in (0,1)$. Then the probability you seek is the volume of the piece of the cube that lies below the plane $x+y+z=1$. This will be a pyramid, whose volume you can compute easily.

angryavian
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty}{\sum_{i = 1}^{n - 2}\sum_{j = 1}^{n - i - 1} \bracks{n - \pars{i + j}} \over n^{3}}}$

\begin{align} &\lim_{n \to \infty}{\sum_{i = 1}^{n - 2}\sum_{j = 1}^{n - i - 1} \bracks{n - \pars{i + j}} \over n^{3}} \\[5mm] = &\ \lim_{n \to \infty}\braces{{1 \over n^{3}} \sum_{i = 1}^{n - 2}\bracks{\pars{n - i}\pars{n - i - 1} - {\pars{n - i - 1}\pars{n - i} \over 2}}} \\[5mm] = &\ {1 \over 2}\lim_{n \to \infty}\bracks{{1 \over n^{3}} \sum_{i = 1}^{n - 2}\pars{n - i - 1}\pars{n - i}} \\[5mm] = &\ {1 \over 2}\lim_{n \to \infty}\braces{{1 \over n^{3}} \sum_{i = 1}^{n - 2}\bracks{n - \pars{n - 1 - i} - 1} \bracks{n - \pars{n - 1 - i}}} \\[5mm] = &\ {1 \over 2}\lim_{n \to \infty}\bracks{{1 \over n^{3}} \sum_{i = 1}^{n - 2}i\pars{i + 1}} = {1 \over 2}\lim_{n \to \infty}\pars{{1 \over n^{3}} \sum_{i = 1}^{n - 2}\pars{i + 1}^{\underline{2}}} = {1 \over 2}\lim_{n \to \infty}\pars{{1 \over n^{3}} \left.{\pars{i + 1}^{\underline{3}} \over 3}\right\vert_{\ 1}^{\ n - 1}} \\[5mm] = &\ {1 \over 6}\lim_{n \to \infty}\bracks{{1 \over n^{3}}\, \pars{n^{\underline{3}} - 1^{\underline{3}}}} = {1 \over 6}\lim_{n \to \infty}{\pars{n - 1}\pars{n - 2} \over n^{2}} = \bbx{\ds{1 \over 6}} \end{align}

Felix Marin
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