Probability of the sum of three random variables being less than 1
Given three Uniform random variables between 0 and 1, $(x,y,z)$, that are i.i.d., what is the probability $x+y+z < 1$?
Solution 1:
Your probaility is $$\int_0^1\int_0^{1-x}\int_0^{1-x-y}1\,dzdydx=\frac{1}{6}$$
Solution 2:
Does this help?
Histogram[Total@RandomReal[{0, 1}, 3] & /@ Range@1000000, 50]
With[{nn = 6*10^6}, Length@Select[Total@RandomReal[{0, 1}, 3] & /@ Range@nn, #
< 1 &]/nn // N]
appears to tend towards $\dfrac{1}{6}.$
Solution 3:
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
A generalization for $\ds{k}$ variables is straightforward by using the Laplace Transform 'technique'. Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson Bracket.
Namely, \begin{align} &\bbox[10px,#ffe]{\ds{\int_{0}^{1}\cdots\int_{0}^{1} \bracks{x_{1} + \cdots + x_{k} < 1}\dd x_{1}\ldots\dd x_{k}}} \\[5mm] = &\ \int_{0}^{1}\cdots\int_{0}^{1} \bracks{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\exp\pars{\bracks{1 - x_{1} - \cdots - x_{k}}s} \over s}\, {\dd s \over 2\pi\ic}} \dd x_{1}\ldots\dd x_{k} \\[5mm] = &\ \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\expo{s} \over s} \pars{\int_{0}^{1}\expo{-sx}\dd x}^{k}{\dd s \over 2\pi\ic} = \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\expo{s} \over s} \pars{1 - \expo{-s} \over s}^{k}{\dd s \over 2\pi\ic} \\[5mm] = & \sum_{n = 0}^{k}{k \choose n}\pars{-1}^{n}\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\expo{\pars{1 - n}s} \over s^{k + 1}}{\dd s \over 2\pi\ic} = \sum_{n = 0}^{k}{k \choose n}\pars{-1}^{n}\bracks{n < 1} \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{\pars{1 - n}s} \over s^{k + 1}}{\dd s \over 2\pi\ic} \\[5mm] = & \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{s} \over s^{k + 1}}{\dd s \over 2\pi\ic} = \bbx{\ds{1 \over k!}} \end{align}