\setcounter{ExampleCounter}{1}
In this section, we will focus on computing probabilities of events involving ``or'' as well as learning the concept of mutually exclusive events. We will also discuss complementary events and their probabilities.
To start, recall the experiment of drawing one card from a standard deck of cards. Let $J$ denote drawing a Jack, and $Q$ denote drawing a Queen. What is the probability of drawing a Jack? It is, of course, 4/52, and the same goes for the probability of drawing a Queen. Now, what is the probability of drawing a Jack \emph{OR} Queen? By looking back at the deck of cards, we can see that there are 8 cards that are either Jacks or Queens, so
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which happens to be the sum of their individual probabilities.
What about, though, if we wanted to find the probability of drawing a Jack or a diamond? Could we just add their individual probabilities (4/52 and 13/52, respectively)? Let's check by looking back at the cards and see which correspond to Jacks or diamonds.
\begin{center}
\includegraphics[width=0.8\textwidth]{playingcardsJD}
\end{center}
Notice that there are 16 cards that match that description, so the probability is 16/32, which ISN'T the sum of the individual probabilities. What went wrong?
\subsection{Mutually Exclusive Events}
The answer can be found by looking at the diagram above. Notice that if we add up the number of Jacks and the number of diamonds (for a total of 19), we \textit{double count} the Jack of diamonds. This brings us to an important definition that determines how we find the probability of one event OR another occurring: we need to find whether the events are \textbf{mutually exclusive} or \textbf{disjoint}. That is, can these two events happen at the same time?
\begin{proc}{Disjoint (mutually exclusive) outcomes}
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\end{proc}
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\pagebreak
Can we draw a card that is both Jack and Queen? Clearly, there is no such card, therefore these events are disjoint. Another familiar example of disjoint events would be getting an even or odd number when rolling a die. Each number is either even or odd, thus these two events are also mutually exclusive. Above, though, we showed that drawing a Jack and drawing a diamond are NOT mutually exclusive, since you can draw the Jack of diamonds.
Notice that the terms \textbf{disjoint} and \textbf{mutually exclusive}
are equivalent and interchangeable. The Venn diagram below illustrates the concept of mutually exclusive events: two events $A$ and $B$ do not overlap; they are disjoint.
\begin{center}
\begin{tikzpicture}
\draw [ultra thick,color=red, fill=red!20] (-1.5,0) circle (1.2cm);
\draw [yshift=-1.8cm,xshift=-1.5cm] node {\color{red}\Large $A$};
\draw [ultra thick,color=blue, fill=blue!20] (1.5,0) circle (1.2cm);
\draw [yshift=-1.8cm,xshift=1.5cm] node {\color{blue}\Large $B$};
\end{tikzpicture}
\end{center}
Before we formally define a formula for computing probabilities of disjoint events, let us solve some problems by using the rules we already know.
\begin{example}{Rolling a die}
Suppose you roll a fair six-sided die once. What is the probability of rolling a 6 or an odd number? \\
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\end{example}
\begin{formula}{Addition rule for mutually exclusive events}
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\end{formula}
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\pagebreak
\begin{example}{Drawing a card}
Suppose you draw one card from a standard 52-card deck.
\begin{enumerate}[(a)]
\item What is the probability that you get an Ace or a face card? \\
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\item What is the probability of getting a number or a red Jack?
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\item What is the probability of selecting a red card or a black card?
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\end{enumerate}
\end{example}
\begin{example}{Marbles}
A large bag contains 28 marbles: 7 are blue, 8 are yellow, 3 are white, and 10 are red. If one marble is randomly selected, what is the probability that it's either red or yellow? \\
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\end{example}
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\pagebreak
\subsection{Overlapping Events}
What if the events of interest are not mutually exclusive? How do we compute probabilities of events that are not disjoint? Pictorially, we can visualize this situation with the following diagram, where the red intersection of two circles represents all outcomes when two events both happen. For example, if we consider FCC students, selecting a female and selecting a full-time students would not be mutually exclusive events, since there are certainly female students who go to school full time.
\begin{center}
\begin{tikzpicture}
\draw [very thick,color=red, fill=red!20] (-0.8,0) circle (1.2cm);
\draw [yshift=-1.8cm,xshift=-0.8cm] node {\color{red}\Large $A$};
\draw [very thick,color=blue, fill=blue!20] (0.8,0) circle (1.2cm);
\draw [yshift=-1.8cm,xshift=0.8cm] node {\color{blue}\Large $B$};
\draw [ultra thick,color=blue!50!red, fill=blue!50!red] (0,-0.9) arc (-45:45:1.28cm);
\draw [ultra thick,color=blue!50!red, fill=blue!50!red] (-0.03,0.89) arc (135:225:1.24cm);
\draw [ultra thick,color=blue!50!red] (0,-0.9) -- (0,0.9);
\end{tikzpicture}
\end{center}
Let's go back to the deck of cards to see how to calculate probabilities in situations like this. We'll again use the example of drawing a Jack or a diamond.
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\begin{center}
\includegraphics[width=\textwidth]{playingcardsJD}
\end{center}
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As we noted already, these are not mutually exclusive events. Because of that, adding the probability of drawing a Jack (4/52) and the probability of drawing a diamond (13/52) gave an incorrect answer of 17/52, where the correct probability--as we noted earlier--is 16/52. Again, this is because we \textit{double counted} the Jack of diamonds, once when we calculated the probability of drawing a Jack and once when we calculated the probability of a diamond.
The way to correct for this double counting is to subtract off the overlap; thus, we'll add up the probability of drawing a Jack and the probability of drawing a diamond and then subtract the probability of drawing both together (i.e. of drawing the Jack of diamonds):
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In general, to calculate probabilities of compound events that are not mutually exclusive, we will use the General Addition rule:
\begin{formula}{General Addition rule}
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\end{formula}
Notice that this is a more general form of the addition rule we stated earlier, with mutually exclusive events. If two events are mutually exclusive, the probability of them occurring together is 0, so the general addition rule simplifies down in that case to the simpler addition rule.
\begin{example}{Drawing a card}
Suppose you draw one card from a standard 52-card deck.
\begin{enumerate}[(a)]
\item What is the probability that you get a King or a spade? \\
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\item What is the probability that you get a Queen or a face card? \\
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\end{enumerate}
\end{example}
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\pagebreak
\begin{example}{FCC students}
Consider the following information about a group of 130 FCC students:
\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}[(a)]
\item If one person is randomly selected from the group, what is the probability this student is female or left-handed? \\
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\item Compute the probability of selecting a male or a right-handed student. \\
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\end{enumerate}
\end{example}
\begin{example}{Speeding tickets and car color}
The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person has a red car \emph{or} got a speeding ticket
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
& Speeding ticket & No speeding ticket& \textbf{Total} \\ \hline
Red car & 15 & 135 & 150\\ \hline
Not red car & 45 & 470 & 515 \\ \hline
\textbf{Total} & 60 & 605 & 665 \\ \hline
\end{tabular}
\end{center}
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\end{example}
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\subsection{Complements}
The probability of an event not occurring can be just as useful as computing the probability of that event happening. The best way to introduce this concept is to consider an example. Let's revisit the standard 52-card deck, where we randomly select one card:
\begin{center}
\includegraphics[width=0.8\textwidth]{playingcards}
\end{center}
What is the probability of not drawing an Ace? Well, you know that there are 4 Aces in the deck, so $52 - 4 = 48$ cards that are not Aces. We compute:
\[ P(\mbox{not } \mbox{ Ace}) = \frac{48}{52} \approx 0.923 \]
Now, notice that
\[ \frac{48}{52} = 1 - \frac{4}{52}, \mbox{ where } P(Ace) = \frac{4}{52}\]
This is not a coincidence. If you recall the basic rules of probability, the sum of probabilities of all outcomes must be 1. In this case, the card you draw is either an Ace or it's not, so it makes sense that the probabilities of these two events add up to 1.
%%% NOT %%%
\begin{formula}{Complement of an event}
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\end{formula}
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\begin{example}{Not hearts!}
If you pull a random card, what is the probability it is not a
heart? \\
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\end{example}
\begin{example}{Multiple choice question}
A multiple choice question has 5 answers, and exactly one of them is correct. If you were to guess, what is the probability of not getting the correct answer? \\
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\end{example}
\begin{example}{FCC students' demographics}
According to the FCC website, female students make up 57\% of the Fall 2014 student body. If one student is randomly selected, what is the probability the student is not female? \\
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\end{example}
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