\setcounter{ExampleCounter}{1}
We began by calculating the probabilities of single events occurring, and then we learned how to combine events using \emph{OR}. Now we ask a different question: suppose we know how to calculate the probability of $A$ and the probability of $B$ on their own; how can we calculate the probability that $A$ \emph{AND} $B$ both occur? To set this up, we'll look at two situations: flipping a coin twice and drawing two cards \emph{without replacement} (this will be important).
\paragraph{Flipping a coin twice} If\marginnote{Independent} we flip a coin twice in succession, the sample space is
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Now suppose we ask the following questions:
\begin{enumerate}
\item What is the probability that the first flip results in a head?
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\item What is the probability that the second flip results in a tail?
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\item What is the probability that the first flip results in a head \emph{AND} the second flip results in a tail?
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\end{enumerate}
Notice that the probability of both happening together is the probability of one times the probability of the other:
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Seeing this, and noting the title of the section, we may be tempted to jump to the conclusion that the probability of $A$ \emph{AND} $B$ is simply the probability of $A$ times the probability of $B$. However, the next scenario illustrates that we need to be a bit more careful.
Just as we found with the addition rule, there is a simple version that works if a certain condition is met, and if not, there is a more general version of the multiplication rule.
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\paragraph{Drawing two cards without replacement}
Suppose\marginnote{Not independent} we draw one card, and then \emph{without} placing it back and re-shuffling the deck, we draw a second card. What is the probability that we draw two Aces?\\
This situation is different from the previous one, because now what happens on the first draw affects the probabilities for the second draw. In other words, the probability of drawing an Ace the first time is 4/52. If we draw an Ace the first time, there are only 3 Aces left and 51 total cards left, so the probability of drawing an Ace the second time is 3/51. However, if we do not draw an Ace the first time, there are still 4 Aces in the deck, so the probability of drawing an Ace the second time is 4/51. We can illustrate this with a branching tree diagram.
\begin{center}
\begin{tikzpicture}
\draw [ultra thick,color=blue!40] (-3,0) -- (0,2);
\draw [ultra thick,color=blue!40] (0,2) -- (3,3);
\draw [ultra thick,color=blue!40] (0,2) -- (3,1);
\draw [ultra thick,color=blue!40] (-3,0) -- (0,-2);
\draw [ultra thick,color=blue!40] (0,-2) -- (3,-1);
\draw [ultra thick,color=blue!40] (0,-2) -- (3,-3);
\draw [yshift=-4cm,xshift=-2.5cm] node {First draw};
\draw [yshift=-4cm,xshift=2.5cm] node {Second draw};
\draw [yshift=2.5cm,xshift=0cm] node {Ace};
\draw [yshift=-2.5cm,xshift=0cm] node {Not Ace};
\draw [yshift=3cm,xshift=3.5cm] node {Ace};
\draw [yshift=1cm,xshift=4cm] node {Not Ace};
\draw [yshift=-1cm,xshift=3.5cm] node {Ace};
\draw [yshift=-3cm,xshift=4cm] node {Not Ace};
\end{tikzpicture}
\end{center}
Now the probability of drawing an Ace both times is the probability of drawing an Ace the first time multiplied by the probability of drawing an Ace the second time \textbf{given that we drew an Ace the first time}. Notice on the tree diagram that this corresponds to following the upward branch both times.
This is because \emph{only} if we draw an Ace the first time do we have any chance of fulfilling the scenario; if we fail to draw an Ace the first time, it doesn't matter what we do the second time--we've already failed.
Thus, the probability of drawing an Ace both times is
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This is what we call \emph{conditional probability}, and it's what we have to consider for the general multiplication rule.
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\subsection{Independence}
\begin{proc}{Independence}
\vspace{0.75in}
\end{proc}
Note that saying that two events are \emph{independent} is different than saying that two events are \emph{mutually exclusive}.
\begin{itemize}
\item \text{}
\vspace{0.5in}
\item \text{}
\vspace{0.5in}
\end{itemize}
\begin{example}{Independent events}
Determine whether these events are independent:
\begin{enumerate}
\item A fair coin is tossed two times. The two events are $A$ = first toss is Heads and $B$ = second toss is Heads.
\vspace{0.5in}
\item The two events $A$ = \emph{It will rain tomorrow in Frederick MD} and $B$ = \emph{It will rain tomorrow in Thurmont MD}
\vspace{0.5in}
\item You draw a red card from a deck, then draw a second card without replacing the first.
\vspace{0.5in}
\item You draw a face card from the deck, then replace it and re-shuffle the deck before drawing a second card.
\vspace{0.5in}
\end{enumerate}
\end{example}
Now we are ready to formally state the rule that we used in the first scenario at the beginning of the section.
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\pagebreak
\subsection{The Multiplication Rule for Independent Events}
\begin{formula}{Probabilities of independent events}
If $A$ and $B$ are independent, then the probability of both $A$ and $B$ occurring is
\vspace{0.75in}
We can generalize this to finitely many independent events $A_1, A_2, \dots, A_k$
\vspace{0.75in}
\end{formula}
\begin{example}{Coins and dice}
Suppose you flip a coin and roll a six-sided die once. What is the probability you get Tails and an even number? \\
\vspace{1.5in}
\end{example}
\begin{example}{Drawing Cards With Replacement}
Assume you have a 52 card deck, and you select two cards at random. Also
assume that you replace and reshuffle after each selection. Find the probability of drawing a king first and then a black card.\\
\vspace{0.75in}
\end{example}
\vfill
\pagebreak
\begin{example}{Left-handed population}
About 9\% of people are left-handed. Suppose 2 people are selected at random from the U.S. population. Because the sample size of 2 is
very small relative to the population, it is reasonable to assume these two people are independent. What is the probability that both are left-handed? \\
\vspace{0.75in}
\end{example}
\begin{example}{Boys and girls}
Assuming that probability of having a boy is 0.5, find the probability of a family having 3 boys. \\
\vspace{0.75in}
\end{example}
\subsection{Multiplication Rule for Dependent Events}
\paragraph{Conditional probability:} \text{}
\vspace{0.5in}
\paragraph{Notation:} \text{}
\vspace{0.5in}
\paragraph{Example:} Drawing two Aces:
\vspace{0.75in}
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\pagebreak
\begin{formula}{Multiplication formula for dependent events}
If events $A$ and $B$ are not independent, then
\vspace{0.75in}
\end{formula}
Note that this, like with the addition rule, is the general multiplication rule; if $A$ and $B$ are independent, $P(B|A)=P(B)$ (because the probability of $B$ is the same regardless of whether $A$ has occurred or not) and the general multiplication formula becomes the simpler form for independent events that we have already seen.
\begin{example}{Drawing cards without replacement}
If you pull 2 cards out of a deck, what is the probability that both are spades? \\
\vspace{1.5in}
\end{example}
\begin{example}{Sock Colors}
In your drawer you have 10 pairs of socks, 6 of which are white. If you reach in and
randomly grab two pairs of socks, what is the probability that both are white?\\
\vspace{1in}
\end{example}
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\pagebreak
\begin{example}{M\&M's}
A bag of M\&M's contains the following breakdown of colors:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|} \hline
Red & Yellow & Brown & Blue & Orange & Green \\ \hline
12 & 18 & 24 & 22 & 13 & 17 \\ \hline
\end{tabular}
\end{center}
Suppose you pull two M\&M's out of the bag (without replacing candy after each pull). Find the following probabilities:
\begin{enumerate}
\item The probability of drawing two red candies
\vspace{2in}
\item The probability of drawing a blue candy and then a brown candy
\vspace{2in}
\item The probability of not drawing 2 green candies
\vspace{2in}
\end{enumerate}
\end{example}
\vfill
\pagebreak
We'll conclude this section with an example of calculating conditional probability from a contingency table.
\begin{example}[https://www.youtube.com/watch?v=0oCoc5B1lVU]{Conditional Probability and Contingency Tables}
Again using the data regarding 130 FCC students, broken down by gender and dominant hand:
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
Gender & Right-handed & Left-handed & \textbf{Total} \\ \hline
Female & 58 & 13 & 71\\ \hline
Male & 47 & 12 & 59 \\ \hline
\textbf{Total} & 105 & 25 & 130 \\ \hline
\end{tabular}
\end{center}
\begin{enumerate}
\item What is the probability that a randomly chosen student is female, given that the student is left-handed?
\vspace{1.5in}
\item What is the probability that a randomly chosen student is right-handed, given that the student is male?
\vspace{1.5in}
\end{enumerate}
\end{example}
\begin{example}{Medical Test}
A certain disease infects 100 out of every 100,000 people. The test for this disease is correct 99\% of the time. If you get a positive result, what is the probability that you have the disease?
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\end{example}