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\section*{Question 2}
\noindent An urn contains six balls, numbered $1$ to $6$. Two balls are drawn at random, without replacement. Let $X$ be the difference between the larger and smaller of the two numbers drawn. What is the probability distribution of $X$?
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\begin{enumerate}[label={},leftmargin=0in]\item
\subsection*{Solution}
By applying the principle of symmetry, easy to define the probability space as
\[
\begin{aligned}
\Omega &= \{1,2,\dots,6\}\\
\mathcal{F} &= \mathcal{P}(\Omega)\\
\bP &:\enspace \bP(\{1,2\}) = \bP(\{1,3\}) = \dots = \bP(\{6,5\}) = \frac{1}{6}\times \frac{1}{5} = \frac{1}{30}
\end{aligned}
\]
Then we have the $pdf$ of $X$
\[
p(x) =
\begin{cases}
\begin{aligned}
\bP(X = 1) &= \bP(\{1,2\},\{2,1\},\dots,\{6,5\}) &= \frac{5}{15} &\approx 0.333,&\quad x = 1\\
\bP(X = 2) &= \bP(\{1,3\},\{2,4\},\dots,\{6,4\}) &= \frac{4}{15} &\approx 0.267,&\quad x = 2\\
\bP(X = 3) &= \bP(\{1,4\},\{2,5\},\dots,\{6,3\}) &= \frac{3}{15} &= 0.2,&\quad x = 3\\
\bP(X = 4) &= \bP(\{1,5\},\{2,6\},\dots,\{6,2\}) &= \frac{2}{15} &\approx 0.133,&\quad x = 4\\
\bP(X = 5) &= \bP(\{1,6\},\{6,1\}) &= \frac{1}{15} &\approx 0.067,&\quad x = 5
\end{aligned}
\end{cases}
\]
Or a general formula without cases
\[p(x) = \frac{6-x}{\dbinom{6}{2}} = \frac{6-x}{15},\quad x\in\{1,2,\dots,5\}\]
\subsection*{Answer}
\[\boxed{p(x) =
\begin{cases}
\begin{aligned}
\frac{5}{15} &\approx 0.333,&\quad x = 1\\
\frac{4}{15} &\approx 0.267,&\quad x = 2\\
\frac{3}{15} &= 0.2,&\quad x = 3\\
\frac{2}{15} &\approx 0.133,&\quad x = 4\\
\frac{1}{15} &\approx 0.067,&\quad x = 5
\end{aligned}
\end{cases}}\]
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