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└── Homework
└── W3
└── Q3.tex
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\title{Question 3: Conditional Prize Award in a Sequence of Matches}
\author{}
\date{}
\begin{document}
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\section*{Problem Statement}
To obtain a prize, you need to win \emph{at least two consecutive matches} out of the three you play, alternating between matches against your father and your coach. You win 75\% of the matches against your father and 40\% of the matches against your coach.
\bigskip
\section*{Preliminaries}
Let $\mathrm{win}$ and $\mathrm{lose}$ denote the outcomes of a match, and let $F$ and $C$ denote the events of winning against the father and the coach, respectively. Define the probability space as
\[
\begin{aligned}
\Omega &= \{\mathrm{win},\,\mathrm{lose}\},\\[1mm]
\mathcal{F} &= \mathcal{P}(\Omega),\\[1mm]
\mathbb{P} &: \quad \mathbb{P}(F) = 75\%, \quad \mathbb{P}(C) = 40\%.
\end{aligned}
\]
Let $W$ be the event of obtaining a prize (i.e., winning at least two consecutive matches) and denote by $A$, $B$, and $C$ the outcomes of the first, second, and third matches respectively. To be specific, we use subscripts $f$ and $c$ to indicate matches played against the father and the coach. For example, $A_f$ means winning the first match against the father, while $A_c$ means winning the first match against the coach.
Assuming that the match outcomes are stochastically independent, we have:
\[
\begin{alignedat}{4}
\mathbb{P}(A_f) &= \mathbb{P}(B_f) = \mathbb{P}(C_f) &= \mathbb{P}(F) &= 75\%,\\[1mm]
\mathbb{P}(A_c) &= \mathbb{P}(B_c) = \mathbb{P}(C_c) &= \mathbb{P}(C) &= 40\%.
\end{alignedat}
\]
\bigskip
\section*{Part 1: Comparing Two Playing Sequences}
We consider the following two sequences of matches:
\begin{itemize}
\item \textbf{Father-Coach-Father} (denoted by $W_{FCF}$)
\item \textbf{Coach-Father-Coach} (denoted by $W_{CFC}$)
\end{itemize}
\subsection*{Sequence 1: Father-Coach-Father}
In this sequence, a prize is obtained if at least two consecutive wins occur. One can show that the event of winning a prize in the Father-Coach-Father sequence is given by:
\[
\begin{aligned}
\mathbb{P}(W_{FCF}) &= \mathbb{P}\Big((A_f \cap B_c \cap C_f) \cup (A_f^c \cap B_c \cap C_f) \cup (A_f \cap B_c \cap C_f^c)\Big)\\[1mm]
&= 2\,\mathbb{P}(F)\,\mathbb{P}(C) - \mathbb{P}(F)^2\,\mathbb{P}(C).
\end{aligned}
\]
\subsection*{Sequence 2: Coach-Father-Coach}
Similarly, for the Coach-Father-Coach sequence, the probability of obtaining a prize is:
\[
\begin{aligned}
\mathbb{P}(W_{CFC}) &= \mathbb{P}\Big((A_c \cap B_f \cap C_c) \cup (A_c^c \cap B_f \cap C_c) \cup (A_c \cap B_f \cap C_c^c)\Big)\\[1mm]
&= 2\,\mathbb{P}(C)\,\mathbb{P}(F) - \mathbb{P}(C)^2\,\mathbb{P}(F).
\end{aligned}
\]
Since $\mathbb{P}(F)=75\%$ and $\mathbb{P}(C)=40\%$, we have
\[
\mathbb{P}(F) > \mathbb{P}(C),
\]
which implies
\[
\mathbb{P}(W_{FCF}) - \mathbb{P}(W_{CFC}) = \mathbb{P}(F)\,\mathbb{P}(C)\bigl(\mathbb{P}(C) - \mathbb{P}(F)\bigr) < 0.
\]
Thus,
\[
\boxed{\mathbb{P}(W_{CFC}) > \mathbb{P}(W_{FCF})}.
\]
\textbf{Answer for Part 1:} The sequence \emph{Coach-Father-Coach} is preferable.
\bigskip
\section*{Part 2: Including Additional Outcomes}
Let $W'$ be the event of obtaining a prize when additional outcomes are considered.
\subsection*{Sequence 1: Modified Father-Coach-Father ($W'_{FCF}$)}
For the Father-Coach-Father sequence, we now have:
\[
\begin{aligned}
\mathbb{P}(W'_{FCF}) &= \mathbb{P}\Big(W_{FCF} \cup \bigl(A_f \cap B_c^c \cap C_f\bigr)\Big)\\[1mm]
&= 2\,\mathbb{P}(F)\,\mathbb{P}(C) + \mathbb{P}(F)^2\Bigl(1-2\,\mathbb{P}(C)\Bigr)\\[1mm]
&\approx 71.25\%.
\end{aligned}
\]
\subsection*{Sequence 2: Modified Coach-Father-Coach ($W'_{CFC}$)}
Similarly, for the Coach-Father-Coach sequence, we have:
\[
\begin{aligned}
\mathbb{P}(W'_{CFC}) &= \mathbb{P}\Big(W_{CFC} \cup \bigl(A_c \cap B_f^c \cap C_c\bigr)\Big)\\[1mm]
&= 2\,\mathbb{P}(C)\,\mathbb{P}(F) + \mathbb{P}(C)^2\Bigl(1-2\,\mathbb{P}(F)\Bigr)\\[1mm]
&\approx 52\%.
\end{aligned}
\]
Since
\[
\mathbb{P}(W'_{FCF}) \approx 71.25\% > 52\% \approx \mathbb{P}(W'_{CFC}),
\]
\textbf{Answer for Part 2:} The strategy should be changed to \emph{Father-Coach-Father}.
\end{document}