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\title{Conditional Probability\footnote{ This slide show is an open-source document. See last slide for copyright information.}}
\subtitle{STA 256: Fall 2018}
\date{} % To suppress date

\begin{document}

\begin{frame}
  \titlepage
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\begin{frame}
\frametitle{Conditional Probability: The idea} \pause
%\framesubtitle{} 
\begin{itemize}
    \item If event $B$ has occurred, maybe the probability of $A$ is different from the probability of $A$ overall.  \pause
    \item Maybe the chances of an auto insurance claim are different depending on the type of car.  \pause
    \item We will talk about the \emph{conditional} probability of an insurance claim \emph{given} that the car is a Dodge Charger.
\end{itemize}
\end{frame}


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\begin{frame}
\frametitle{Restrict the sample space}
%\framesubtitle{} 
To condition on the event $B$, make $B$ the new, restricted sample space.  \pause

\begin{center}
\vspace{2mm}
\includegraphics[width=2in]{Venn2} \pause 
\vspace{2mm}
\end{center}

Express the probability of $A$ as a fraction of the probability of $B$\pause,  provided the probability of $B$ is not zero. \pause

\begin{displaymath}
    P(A|B) = \frac{P(A\cap B)}{P(B)}
\end{displaymath} 



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\begin{frame}
\frametitle{Definition: The probability of $A$ given $B$}
%\framesubtitle{} 
\begin{center}
{\Large
If $P(B)>0$, \pause $\displaystyle  P(A|B) = \frac{P(A\cap B)}{P(B)}$
} % End size
\vspace{2mm} \pause

\includegraphics[width=3in]{HIV} \pause 
\vspace{2mm} \pause

$P(A|F) = \frac{P(A\cap F)}{P(F)} \pause = \frac{0.01}{0.50} \pause = 0.02$
\end{center}
\end{frame}

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\begin{frame}
\frametitle{Multiplication Law} \pause
\framesubtitle{$P(A|B) = \frac{P(A\cap B)}{P(B)} \Rightarrow P(A\cap B) = P(A|B)P(B)$}  \pause
{\Large
\begin{displaymath}
    P(A\cap B) = P(A|B)P(B)
\end{displaymath} \pause
} % End size
\vspace{10mm}

Useful for sequential experiments. \pause A jar contains 15 red balls and 5 blue balls. What is the probability of randomly drawing a red and then a blue? \pause
\vspace{3mm}

$P(R_1 \cap B_2) \pause = P(R_1)P(B_2|R_1) \pause = \frac{15}{20} \, \frac{5}{19}
\pause = \frac{15}{76} \pause \approx 0.197$
\end{frame}

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\begin{frame}
\frametitle{Make a Tree}
\framesubtitle{Justified by the multiplication principle} 
%\begin{center}
\includegraphics[width=4.55in]{tree}
%\end{center}
\end{frame}

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\begin{frame}
\frametitle{Trees}
%\framesubtitle{} 
\begin{itemize}
    \item Can be extended to more than 2 stages. \pause 
    \item Are best for \emph{small} sequential experiments. \pause
    \item Can allow you to side-step two important theorems, if the problem is set up nicely for you. \pause
        \begin{itemize}
            \item The Law of Total Probability  \pause
            \item Bayes' Theorem.
        \end{itemize}
\end{itemize}
\end{frame}


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\begin{frame}
\frametitle{Law of Total Probability} \pause
%\framesubtitle{} 
Partition $\Omega$, into $B_1, B_2, \ldots B_n$, disjoint,  with $P(B_k)>0$ for all $k$.  \pause
\vspace{3mm}

\begin{columns}  
\column{0.5\textwidth}
\begin{center}
\includegraphics[width=2in]{TotalProb}
\end{center} \pause
\column{0.5\textwidth}
$A = \cup_{k=1}^n(A\cap B_k)$, disjoint
\begin{eqnarray*}
    P(A) & = & \sum_{k=1}^n P(A \cap B_k) \\ \pause
         & = & \sum_{k=1}^n P(A|B_k)P(B_k)
\end{eqnarray*} 
\end{columns}
\end{frame}

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\begin{frame}
\frametitle{Example}
\framesubtitle{Law of Total Probability: $P(A) = \sum_{k=1}^n P(A|B_k)P(B_k)$} 
In the 2 coins and one die example,  \pause
we got $P(2) = \frac{7}{18}$ from a tree.  \pause
\begin{eqnarray*}
    P(2) & = & P(2|\mbox{Coin 1}) \, P(\mbox{Coin 1}) \\
         & + & P(2|\mbox{Coin 2}) \, P(\mbox{Coin 2}) \\ 
         & + & P(2|\mbox{Die}) \, P(\mbox{Die}) \\ \pause
         & = & \frac{1}{2}\cdot\frac{1}{3} + \frac{1}{2}\cdot\frac{1}{3} + 
         \frac{1}{6}\cdot\frac{1}{3} \\ \pause
         & = & \frac{7}{18}
\end{eqnarray*} 
\end{frame}

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\begin{frame}
\frametitle{Law of Total Probability}
\framesubtitle{A general statement}  \pause
Let $\Omega = \cup_{k=1}^\infty B_k$, disjoint\pause, with $P(B_k)>0$ for all $k$. Then  \pause \vspace{5mm}

{\Large
\begin{displaymath}
    P(A) = \sum_{k=1}^\infty P(A|B_k)P(B_k)
\end{displaymath}  \pause
} % End size
\end{frame}

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\begin{frame}
\frametitle{Bayes Theorem: The idea}
%\framesubtitle{} 

{\Large
Bayes' Theorem allows you to turn conditional probability around, and obtain $P(B|A)$ from $P(A|B)$.
} % End size

\end{frame}


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\begin{frame}
\frametitle{Thomas Bayes (1701-1761)}
\framesubtitle{Image from the Wikipedia} 
\begin{center}
\includegraphics[width=2.8in]{Thomas_Bayes}
\end{center}
\end{frame}

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\begin{frame}
\frametitle{Bayes' Theorem}
\framesubtitle{One version of many}  \pause
Let $\Omega = \cup_{k=1}^\infty B_k$, disjoint\pause, with $P(B_k)>0$ for all $k$. Then  \pause \vspace{3mm}

{\Large
\begin{displaymath}
    P(B_j|A) = \frac{P(A|B_j)P(B_j)}{\sum_{k=1}^\infty P(A|B_k)P(B_k)}
\end{displaymath}  \pause
} % End size

\vspace{2mm}
An important special case is  \pause
%\vspace{2mm}

{\Large
\begin{displaymath}
    P(B|A) = \frac{P(A|B)P(B)}{P(A|B)P(B) + P(A|B^c)P(B^c)}
\end{displaymath}  
} % End size
\end{frame}



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\begin{frame}
\frametitle{Copyright Information}

This slide show was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistical Sciences, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:

\vspace{5mm}
\href{http://www.utstat.toronto.edu/~brunner/oldclass/256f18} {\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/256f18}}

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$ = \{\omega \in \Omega: \}$

\begin{frame}
\frametitle{} \pause
%\framesubtitle{} 
\begin{itemize}
    \item  \pause
    \item  \pause
    \item 
\end{itemize}
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\begin{frame}
\frametitle{Tree}
\framesubtitle{Useful for small sequential experiments} \pause
\begin{center}
\includegraphics[width=4.5in]{tree}
\end{center}
\end{frame}