Q1
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Probability
└── Homework
└── W6
└── Q1.tex
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\fancyhead[c]{Homework \#6}
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\begin{document}
\section*{Question 1}
\noindent Hugo is trying to complete his card collection and needs just one last card. He can obtain cards from biscuit packs, and each pack has a $0.2$ probability of containing the card he needs. Hugo has enough money to buy at most $4$ packs, but he will stop buying as soon as he finds the missing card. Let $X$ be the number of packs Hugo buys.
\bigskip
\begin{enumerate}[start=1,label={\bfseries Part \arabic*:},leftmargin=0in]
\bigskip\item What is the probability distribution of $X$?
\subsection*{Solution}
As given, easy to know that $X$ has a geometric distribution with parameter $p$, that is
\[X\sim\text{Geo}(p),\quad p = 0.2,\quad X \in \{1,2,3,4\}\]
We know that
\[\bP(X = k) = (1-p)^{k-1}p,\quad k\in X\]
Then we have
\[
\bP(X = k) =
\begin{cases}
\begin{aligned}
(1-0.2)^{1-1} 0.2 &= \frac{1}{5} &= 0.2, &\quad k = 1\\
(1-0.2)^{2-1} 0.2 &= \frac{4}{25} &= 0.16, &\quad k = 2\\
(1-0.2)^{3-1} 0.2 &= \frac{16}{125} &= 0.128, &\quad k = 3\\
1 - \sum_{i=1}^3\bP(X = i) &= \frac{64}{125} &= 0.512, &\quad k = 4
\end{aligned}
\end{cases}
\]
\subsection*{Answer}
\[\boxed{\bP(X = k) =
\begin{cases}
\begin{aligned}
\frac{1}{5} &= 0.2, &\quad k = 1\\
\frac{4}{25} &= 0.16, &\quad k = 2\\
\frac{16}{125} &= 0.128, &\quad k = 3\\
\frac{64}{125} &= 0.512, &\quad k = 4
\end{aligned}
\end{cases}}\]
\bigskip\item What is the probability that Hugo successfully completes his collection?
\subsection*{Solution}
Let event $C$ denotes that Hugo completes his collection and let $C_4$ denotes that Hugo completes his collection when opening the fourth pack, easy to know that
\[C = C_4 \cup \bigcup_{k=1}^3(X = k),\quad \bP(C_4) = (1-p)^{4-1}p\]
Then we have
\[
\begin{aligned}
\bP(C) &= \bP(C_4) + \sum_{k=1}^3\bP(X = k)\\
&= (1-0.2)^{3}0.2 + \frac{1}{5} + \frac{4}{25} + \frac{16}{125}\\
&= \frac{369}{625}\\
&= 0.590\,4
\end{aligned}
\]
\subsection*{Answer}
\[\boxed{\bP(C) = \frac{369}{625} = 0.590\,4}\]
\bigskip\item Given that Hugo completes the collection, what is the probability that he only buys one pack?
\subsection*{Solution}
By applying the Bayes’ Theorem, we know that
\[
\begin{aligned}
\bP_C(X = 1) &= \frac{\bP_{X = 1}(C)\bP(X = 1)}{\bP(C)}\\
&= \frac{1\times \frac{1}{5}}{\frac{369}{625}}\\
&= \frac{125}{369}\\
&\approx 0.339
\end{aligned}
\]
\subsection*{Answer}
\[\boxed{\bP_C(X = 1) = \frac{125}{369} \approx 0.339}\]
\end{enumerate}
\end{document}