3.3: The Complement of an Event
Learning Objectives
 Know the definition of the complement of an event.
 Using the complement of an event to calculate the probability of an event.
 Understanding the complementary rule.
 Definition

The complement \begin{align*}A'\end{align*}
A′ of an event \begin{align*}A\end{align*}A consists of all the simple events (outcomes) that are not in the event \begin{align*}A\end{align*}A .
Let us refer back to the experiment of throwing one die. As you know, the sample space of a fair die is \begin{align*}S = \left \{1, 2, 3, 4, 5, 6 \right \}\end{align*}
\begin{align*}\text{A:} \left \{\text{observe an odd number}\right \}\end{align*}
Then, \begin{align*}A = \left \{1, 3, 5 \right \}\end{align*}
\begin{align*}A' = \left \{2, 4, 6 \right \}.\end{align*}
The Venn diagram is shown below.
This leads us to say that the event \begin{align*}A\end{align*}
The Complementary Rule
The sum of the probabilities of an event and its complement must equal \begin{align*}1\end{align*}
\begin{align*}P(A) + P(A') = 1\end{align*}
As you will see in the following examples below, it is sometimes easier to calculate the probability of the complement of an event rather than the event itself. Then the probability of the event, \begin{align*}P(A)\end{align*}
\begin{align*}P(A) = 1  P(A')\end{align*}
Example:
If you know that the probability of getting the flu this winter is \begin{align*}0.43\end{align*}
Solution:
First, ask the question, what is the probability of the simple event? It is
\begin{align*}P(A) = \left \{ \text{you will get the flu} \right \} = 0.43\end{align*}
The complement is
\begin{align*}P(A') = \left \{ \text{you will not get the flu} \right \} = 1  P(A) = 1  0.43 = 0.57\end{align*}
Example:
Two coins are tossed simultaneously. Here is an event:
\begin{align*}\text{A:} \left \{\text{observing at least one head }\right \}\end{align*}
What is the complement of \begin{align*}A\end{align*}
Solution:
Since the event \begin{align*}A\end{align*}
\begin{align*}A' = \left \{\text{observe no heads} \right \} = \left \{ TT \right \}\end{align*}
We can draw a simple Venn diagram that shows \begin{align*}A\end{align*}
The second part of the problem is to calculate the probability of \begin{align*}A\end{align*}
\begin{align*}P(A') = P(TT) = 1/4\end{align*}
and
\begin{align*}P(A) = 1  P(A') = 1  1/4 = 3/4.\end{align*}
Obviously, we could have gotten the same result if we had calculated the probability of the event of \begin{align*}A\end{align*}
Example:
Here is a new kind of problem. Consider the experiment of tossing a coin ten times. What is the probability that we will observe at least one head?
Solution:
Before we begin, we can write the event as
\begin{align*}\text{A} = \left \{\text{observe at least one head in ten tosses}\right \}\end{align*}
What are the simple events of this experiment? As you can imagine, there are many simple events and it would take a very long time to list them. One simple event may look like this: \begin{align*}HTTHTHHTTH,\end{align*}
To calculate the probability, each time we toss the coin, the chance is the same for heads and tails to occur. We can therefore say that each simple event, among \begin{align*}1024\end{align*} events, is equally likely to occur. So
\begin{align*}P(\text{any simple event among}\ 1024) = \frac{1} {1024}\end{align*}
We are being asked to calculate the probability that we will observe at least one head. You may find it difficult to calculate since the heads will most likely occur very frequently during \begin{align*}10\end{align*} consecutive tosses. However, if we calculate the complement of \begin{align*}A\end{align*}, i.e., the probability that no heads will be observed, our answer may become a little easier. The complement \begin{align*}A'\end{align*} is easy, it contains only one simple event:
\begin{align*}A' = \left \{ TTTTTTTTTT \right \}\end{align*}
Since this is the only event that no heads appear and since all simple events are equally likely, then
\begin{align*}P(A') = \frac{1} {1024}\end{align*}
Now, because \begin{align*}P(A) = 1  P(A'),\end{align*} then
\begin{align*}P(A) = 1  P(A') = 1  \frac{1} {1024} \approx 0.999 = 99.9 \%\end{align*}
That is a very high percentage chance of observing at least one head in ten tosses of a coin.
Lesson Summary
 The complement \begin{align*}A'\end{align*} of an event \begin{align*}A\end{align*} consists of all the simple events (outcomes) that are not in the event \begin{align*}A\end{align*}.
 The Complementary Rule states that the sum of the probabilities of an event and its complement must equal \begin{align*}1\end{align*}, or for an event \begin{align*}A,\end{align*} \begin{align*}P(A) + P(A') = 1.\end{align*}
Review Questions
 A fair coin is tossed three times. Two events are defined as follows: \begin{align*}& \text{A:} \left \{ \text{At least one head is observed} \right \} \\
& \text{B:} \left \{ \text{The number of heads observed is odd} \right \}\end{align*}
 List the sample space for tossing a coin three times
 List the outcomes of \begin{align*}A.\end{align*}
 List the outcomes of \begin{align*}B.\end{align*}
 List the outcomes of the events \begin{align*}A \cup B, A', A \cap B.\end{align*}
 Find \begin{align*}P(A), P(B), P(A \cup B), P(A'), P(A \cap B).\end{align*}
 The Venn diagram below shows an experiment with five simple events. The two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are shown. The probabilities of the simple events are: \begin{align*}P(1) = 1/10, P(2) = 2/10, P(3) = 3/10, P(4) = 1/10, P(5) = 3/10.\end{align*} Find \begin{align*}P(A'), P(B'), P(A' \cap B), P(A \cap B), P(A \cup B'), P(A \cup B), P [(A \cap B)']\end{align*} and \begin{align*}P[(A \cup B)'].\end{align*}
Review Answers

 all: \begin{align*}\left \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \right \} \end{align*}
 \begin{align*}\mathrm{A:} \left \{ HHH, HHT, HTH, THH, HTT, THT, TTH \right \}\end{align*}
 \begin{align*}\mathrm{B:} \left \{ HHH, HTT, THT, TTH \right \}\end{align*}
 \begin{align*}A \cup B\end{align*} same as \begin{align*}A,\end{align*} \begin{align*}A': \left \{ TTT \right \},\end{align*} \begin{align*}A \cap B\end{align*} same as \begin{align*}B\end{align*}
 \begin{align*}P(A) = P(A \cup B) = 7/8, P(B) = P(A \cap B) = 1/2, P(A') = 1/8\end{align*}
 \begin{align*}4/10, 2/10, 3/10, 5/10, 9/10, 7/10, 5/10, 1/10.\end{align*}
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