In this Concept, you will learn what is meant by the complement of an event, and you will be introduced to the Complement Rule. You will also learn how to calculate probabilities when the complement of an event is involved.

### Watch This

For an explanation of complements and using them to calculate probabilities
**
(1.0)
**
, see
jsnider3675, An Event's Complement
(9:40).

### Guidance

**
The Complement of an Event
**

The
*
complement
*
\begin{align*}A'\end{align*}
of the event
@$\begin{align*}A\end{align*}@$
consists of all elements of the sample space that are not in
@$\begin{align*}A\end{align*}@$
.

#### Example A

Let us refer back to the experiment of throwing one die. As you know, the sample space of a fair die is @$\begin{align*}S=\left \{1,2,3,4,5,6\right \}\end{align*}@$ . If we define the event @$\begin{align*}A\end{align*}@$ as observing an odd number, then @$\begin{align*}A=\left\{1,3,5\right\}\end{align*}@$ . The complement of @$\begin{align*}A\end{align*}@$ will be all the elements of the sample space that are not in @$\begin{align*}A\end{align*}@$ . Thus, @$\begin{align*}A' =\left \{2,4,6\right \}. \end{align*}@$

A
*
Venn diagram
*
that illustrates the relationship between
@$\begin{align*}A\end{align*}@$
and
@$\begin{align*}A'\end{align*}@$
is shown below:

This leads us to say that the sum of the possible outcomes for event @$\begin{align*}A\end{align*}@$ and the possible outcomes for its complement, @$\begin{align*}A'\end{align*}@$ , is all the possible outcomes in the sample space of the experiment. Therefore, the probabilities of an event and its complement must sum to 1.

**
The Complement Rule
**

The
*
Complement Rule
*
states that the sum of the probabilities of an event and its complement must equal 1.

@$\begin{align*}P(A)+P(A')=1\end{align*}@$

As you will see in the following examples, it is sometimes easier to calculate the probability of the complement of an event than it is to calculate the probability of the event itself. Once this is done, the probability of the event, @$\begin{align*}P(A)\end{align*}@$ , is calculated using the relationship @$\begin{align*}P(A)=1-P(A')\end{align*}@$ .

#### Example B

Suppose you know that the probability of getting the flu this winter is 0.43. What is the probability that you will not get the flu?

Let the event @$\begin{align*}A\end{align*}@$ be getting the flu this winter. We are given @$\begin{align*}P(A)=0.43\end{align*}@$ . The event not getting the flu is @$\begin{align*}A'\end{align*}@$ . Thus, @$\begin{align*}P(A')=1-P(A)=1-0.43=0.57\end{align*}@$ .

#### Example C

Two coins are tossed simultaneously. Let the event @$\begin{align*}A\end{align*}@$ be observing at least one head.

What is the complement of @$\begin{align*}A\end{align*}@$ , and how would you calculate the probability of @$\begin{align*}A\end{align*}@$ by using the Complement Rule?

Since the sample space of event @$\begin{align*}A=\left \{HT, TH, HH\right \}\end{align*}@$ , the complement of @$\begin{align*}A\end{align*}@$ will be all events in the sample space that are not in @$\begin{align*}A\end{align*}@$ . In other words, the complement will be all the events in the sample space that do not involve heads. That is, @$\begin{align*}A'=\left \{TT\right \}\end{align*}@$ .

We can draw a simple Venn diagram that shows @$\begin{align*}A\end{align*}@$ and @$\begin{align*}A'\end{align*}@$ when tossing two coins as follows:

The second part of the problem is to calculate the probability of @$\begin{align*}A\end{align*}@$ using the Complement Rule. Recall that @$\begin{align*}P(A)=1-P(A')\end{align*}@$ . This means that by calculating @$\begin{align*}P(A')\end{align*}@$ , we can easily calculate @$\begin{align*}P(A)\end{align*}@$ by subtracting @$\begin{align*}P(A')\end{align*}@$ from 1.

@$$\begin{align*}P(A')& =P(TT)=\frac{1}{4}\\ P(A)& =1-P(A')=1-\frac{1}{4}=\frac{3}{4}\end{align*}@$$

Obviously, we would have gotten the same result if we had calculated the probability of event @$\begin{align*}A\end{align*}@$ occurring directly. The next example, however, will show you that sometimes it is much easier to use the Complement Rule to find the answer that we are seeking.

#### Example D

Consider the experiment of tossing a coin ten times. What is the probability that we will observe at least one head?

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 be @$\begin{align*}HTTHTHHTTH\end{align*}@$ , another may be @$\begin{align*}THTHHHTHTH\end{align*}@$ , and so on. There are, in fact, @$\begin{align*}2^{10}=1024\end{align*}@$ ways to observe at least one head in ten tosses of a coin.

To calculate the probability, it's necessary to keep in mind that each time we toss the coin, the chance is the same for heads as it is for tails. Therefore, we can say that each simple event among the 1024 possible events is equally likely to occur. Thus, the probability of any one of these events is @$\begin{align*}\frac{1}{1024}\end{align*}@$ .

We are being asked to calculate the probability that we will observe at least one head. You will probably find it difficult to calculate, since heads will almost always occur at least once during 10 consecutive tosses. However, if we determine the probability of the complement of @$\begin{align*}A\end{align*}@$ (i.e., the probability that no heads will be observed), our answer will become a lot easier to calculate. The complement of @$\begin{align*}A\end{align*}@$ contains only one event: @$\begin{align*}A'=\left \{TTTTTTTTTT\right \}\end{align*}@$ . This is the only event in which no heads appear, and since all simple events are equally likely, @$\begin{align*}P(A')=\frac{1}{1024}\end{align*}@$ .

Using the Complement Rule, @$\begin{align*}P(A)=1-P(A')=1-\frac{1}{1024}=\frac{1023}{1024}=0.999\end{align*}@$ .

That is a very high percentage chance of observing at least one head in ten tosses of a coin.

### Guided Practice

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In a recent election, 35% of the voters were democrats and 65% were not. Of the democrats, 75% voted for candidate Z and of the non-Democrats, 15% voted for candidate Z. Define the following events:
*

A = voter is Democrat

B = voted for candidate Z

*
Find
@$\begin{align*}P(A)\end{align*}@$
and
@$\begin{align*} P(A^c)\end{align*}@$
*

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Solutions:
**

@$\begin{align*}P(A)=0.35\end{align*}@$

@$\begin{align*} P(A^c)=1-P(A)=1-0.35=0.65\end{align*}@$

### Explore More

For 1-5, a fair coin is tossed three times. Two events are defined as follows:

@$$\begin{align*}& A: {\text{at least one head is observed}}\end{align*}@$$

@$$\begin{align*}& B: {\text{an odd number of heads is observed}}\end{align*}@$$

- List the sample space for tossing the 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 following events: @$\begin{align*}A \cup B, A', \ A\cap B\end{align*}@$ .
- Find each of the following: @$\begin{align*}P(A), P(B), P(A \cup B), P(A'), P(A \cap B).\end{align*}@$

For 6-13, 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 as follows:

@$\begin{align*}P(1)=\frac{1}{10}, P(2)=\frac{2}{10}, P(3)=\frac{3}{10}, P(4)=\frac{1}{10}, P(5)=\frac{3}{10}.\end{align*}@$

- @$\begin{align*}P(A')\end{align*}@$
- @$\begin{align*}P(B')\end{align*}@$
- @$\begin{align*}P(A' \cap B)\end{align*}@$
- @$\begin{align*}P(A \cap B)\end{align*}@$
- @$\begin{align*}P(A \cup B')\end{align*}@$
- @$\begin{align*}P(A \cup B)\end{align*}@$
- @$\begin{align*}P(A \cap B')\end{align*}@$
- @$\begin{align*}P \left [(A \cup B)' \right ]\end{align*}@$

**
Keywords
**

Event

Intersection of events

Mutually exclusive

Sample space

Union of events

Venn diagram