Q2
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Probability
└── Homework
└── W4
└── Q2.tex
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\begin{document}
\section*{Question 2}
\noindent You have three coins in your pocket, two fair ones but the third is biased with probability of heads $p$ and tails $1-p$. One coin selected at random drops on the floor, landing heads up. How likely is it that it is one of the fair coins?
\begin{enumerate}[label={},leftmargin=0in]\item
\subsection*{Solution}
Let $\it{head}$ and $\it{tail}$ denote the results of a drop. Define the probability space as
\[
\begin{aligned}
\Omega &= \{\mathrm{head},\,\mathrm{tail}\}\\
\mathcal{F} &= \mathcal{P}(\Omega)\\
\bP &:\enspace \text{the probability measure on $\mathcal{F}$}
\end{aligned}
\]
Let $F$ and $B$ denote the events of selecting a fair coin and a biased coin, respectively, then we have
\[
\begin{aligned}
\bP_F(\{\mathrm{head}\}) &= \frac{1}{2}\\
\bP_B(\{\mathrm{head}\}) &= p
\end{aligned}
\]
By applying the principle of symmetry, we have
\[
\begin{aligned}
\bP(F) &= \frac{2}{3}\\
\bP(B) &= \frac{1}{3}
\end{aligned}
\]
We know that $\{F,B\}$ is a partition of $\Omega$, then by applying the law of total probability, we have
\[\bP(\{\mathrm{head}\}) = \bP_F(\{\mathrm{head}\})\bP(F) + \bP_B(\{\mathrm{head}\})\bP(B) = \frac{p + 1}{3}\]
Then by applying the Bayes’ Theorem, we have
\[{\bP_{\{\mathrm{head}\}}}(F) = \frac{\bP_F(\{\mathrm{head}\})\bP(F)}{\bP(\{\mathrm{head}\})} = \frac{1}{p+1}\]
\subsection*{Answer}
\[\boxed{{\bP_{\{\mathrm{head}\}}}(F) = \frac{1}{p+1}}\]
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