Uniformly distributed over $( 0,1)$

Question

If $x, y,$ and $z$ are selected independently at random from the interval $[0,1]$, what is the probability that $x<y<z$?

Answer 1

The probability that two of the variables are equal is zero. Thus the event whose probability you seek is the occurrence of one of $3!=6$ orders that are mutually exclusive and all equally probable by symmetry. Thus the probability for this order to occur is $\frac16$.

(I'm assuming that you forgot to state that the variables are identically distributed. If this is not known to be the case, then there's not enough information to answer the question.)

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\bracks{x < y < z}\dd x\,\dd y\,\dd z} = \int_{0}^{1}\int_{0}^{1}\bracks{y < z}\int_{0}^{1}\bracks{x < y}\dd x\,\dd y\,\dd z \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}\bracks{y < z}\int_{0}^{y}\dd x\,\dd y\,\dd z = \int_{0}^{1}\int_{0}^{1}\bracks{y < z}y\,\dd y\,\dd z = \int_{0}^{1}\int_{0}^{z}y\,\dd y\,\dd z \\[5mm] = &\ \int_{0}^{1}{1 \over 2}\,z^{2}\,\dd z = \bbx{1 \over 6} \end{align}

$\ds{\bracks{\cdots}}$ is an Iverson Bracket.