Q3
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Probability
└── Homework
└── W5
└── Q3.tex
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\begin{document}
\section*{Question 3}
\noindent Three fair dice are thrown. Let $X$ denote the number of dice that land with the same number of dots. Describe the probability distribution function of $X$.
\bigskip
\begin{enumerate}[label={},leftmargin=0in]\item
\subsection*{Solution}
By applying the principle of symmetry, easy to define the probability space as
\[
\begin{aligned}
\Omega &= \{1,2,\dots,6\}\\
\mathcal{F} &= \mathcal{P}(\Omega)\\
\bP &:\enspace \bP(\{1\}) = \bP(\{2\}) = \dots = \bP(\{6\}) = \frac{1}{6}
\end{aligned}
\]
Then as given, we have
\[X:\enspace \Omega \rightarrow R,\quad R = \{1,2,3\}\]
Then we have the $pdf$ of $X$
\[
p(x) =
\begin{cases}
\begin{aligned}
\frac{6\times 5\times 4}{6^3} &= \frac{5}{9},&\quad x = 1\\
\frac{6\times 3\times 5}{6^3} &= \frac{5}{12},&\quad x = 2\\
\frac{6}{6^3} &= \frac{1}{36},&\quad x = 3
\end{aligned}
\end{cases}
\]
Or a general formula, with $N_i \in \{1,2,3\},\, i\in \{1,2,\dots,6\}$ denotes the absolute frequency of every face ($i.e.$ $N_1 = 2$ means that face $1$ shows $2$ times)
\[
p(x) = \frac{1}{6^3}\sum_{\sum_{i=1}^6N_i=3}\frac{3!}{N_1!N_2!\dots N_6!},\quad \max(\{N_1,N_2,\dots,N_6\}) = x,\quad x\in R
\]
\subsection*{Answer}
\[\boxed{p(x) =
\begin{cases}
\begin{aligned}
\frac{5}{9},&\quad x = 1\\
\frac{5}{12},&\quad x = 2\\
\frac{1}{36},&\quad x = 3
\end{aligned}
\end{cases}}\]
\end{enumerate}
\end{document}