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\title{MA162: Finite mathematics}
\subtitle{Conditional Probability}
\author{Paul Koester}
\institute{University of Kentucky}
\date{March 3 and March 5, 2014}

\begin{document}

\maketitle

\begin{frame}[t]{More Black Jack}
 \begin{itemize}
  \item The dealer draws two cards from a standard deck of 52 cards.  
  \item One of his cards is placed face up, so that all players can see it.
  \item The other card is placed face down, so only the dealer knows which card it is.
  \item What is the probability the dealer is dealt Black Jack? \vfill
  \item The dealer's face up card is an Ace.  What is the probability the dealer has Black Jack?
 \end{itemize}
\end{frame}

\begin{frame}[t]{Conditioning}
 \begin{itemize}
  \item The probability assigned to an event depends on  
    \begin{itemize}
      \item The underlying event
      \item Any relevant information we happen to have
    \end{itemize}
  \item \alert{Conditional Probabilities} give us the means to take the additional information into account.
  \item Given two events, A and B, \(P(A|B)\) is the probability of A, given that B occurs.  
  \item If \(P(A)\) and \(P(B)\) are nonzero, then \[P(A|B) = \frac{P(A\cap B)}{P(B)}\]
 \end{itemize}
\end{frame}

\begin{frame}[t]{Smoking and lung cancer}
Facts taken from 
\url{http://lungcancer.about.com/od/Lung-Cancer-And-Smoking/f/Smokers-Lung-Cancer.htm}
 \begin{itemize}  
  \item \(0.2\%\) probability of lung cancer for men who never smoked
  \item \(5.5\%\) probability of lung cancer for male former smokers
  \item \(15.9\%\) probability of lung cancer for current male smokers
  \item \(24.4\%\) probability of lung cancer for male ``heavy smokers'' defined as smoking more than 5 cigarettes per day
 \end{itemize}
 These are examples of conditional probabilities.  
 Determine a reasonable sample space, and determine events so that each of the above 
 probabilities can be expressed as a conditional probability.
\end{frame}

\begin{frame}[t]{Smoking and lung cancer}
A sample of 10,000 men.
 \begin{itemize}  
  \item 5680 of these men never smoked
  \item 2132 are former smokers
  \item 1737 are current ``light'' smokers
  \item The remaining men are ``heavy'' smokers.
 \end{itemize}

What is the probability a randomly chosen male from this group will develop lung cancer? 
\end{frame}

\begin{frame}[t]{More on Conditional Probability}
\begin{itemize}
  \item More often than not, \(P(A|B) = \frac{P(A\cap B)}{P(B)}\) is used to determine \(P(A\cap B)\).
  \item Thus, \[ P(A\cap B) = P(A|B)\cdot P(B)\]
  \item Furthermore, to find \(P(A)\), one often finds a disjoint partition of the sample space, 
        \[ S = B_1 \cup B_2 \cup B_3 \cup \ldots \cup B_n\]
        with \( B_i \cap B_j = \empty\) for \(i \neq j\)
  \item Then 
        \[ P(A) = \sum_{j=1}^{n} P(A|B_j) \cdot P(B_j) \]
\end{itemize}
\end{frame}

\begin{frame}[t]{Rolling coins}
Flip a fair two sided coin.  
 \begin{itemize}  
  \item If heads, roll a fair 6 sided die (with faces 1, 2, 3, 4, 5, 6).
  \item If tails, roll a fair 4 sided die (with faces 1, 2, 3, 4).
 \end{itemize}
What is the probability of rolling a number larger than 3?
\vfill
\end{frame}

\begin{frame}[t]{Flip you for it}
 A fair two sided coin is flipped twice.  
 \begin{itemize}  
  \item What is the probability of getting a heads on the second, given that the first was a heads? \vfill 
  \item What is the probability of getting a heads on the second? \vfill
 \end{itemize}
\end{frame}

\begin{frame}[t]{Independence}
 \begin{itemize}
  \item The conditional probability \(P(A|B)\) takes into account how the information in the event B affects the 
        likelihood of A.
  \item Some information may be totally irrelevant, like the probability the coin turns up heads, given that the 
        person flipping the coin is named Ashley. 
  \item In this case we would say the events \(A = \{\mbox{coin is flipped by someone named Ashley}\}\) and 
        \(B = \{\mbox{coin turns up heads}\}\) are \alert{independent}.
 \end{itemize}
\end{frame}

\begin{frame}[t]{Testing for independence}
Given that \(P(A)\) and \(P(B)\) are nonzero, all three of these hold when A and B are independent:
 \begin{itemize}
  \item \(P(A \cap B) = P(A)P(B)\)
  \item \(P(A|B) = P(A)\)
  \item \(P(B|A) = P(B)\)
 \end{itemize}
In fact, if any one of these hold, then all three hold.  
Thus, to verify that two events are independent, we only need to check that one of these are true.
\end{frame}

\begin{frame}[t]{Flip you for it}
 A fair two sided coin is flipped twice.  \(A\) is the event ``heads on the first flip.''  
 \(B\) is the event ``outcomes of the two flips are the same.''\\
 Are \(A\) and \(B\) independent or not?  
\end{frame}

\begin{frame}[t]{Independent Cards}
Standard deck of 52 cards.
\begin{itemize}
  \item \(A\) is event ``first card is Ace''
  \item \(B\) is event ``second card is Ace''
  \item \(C\) is event ``first card is Hearts''
  \item \(D\) is event ``second card is Spades''
\end{itemize}
For each pair of the above events, determine if the pair of events are independent.

\end{frame}


\end{document}
