# 3.3: The Complement of an Event

**At Grade**Created by: CK-12

## Learning Objectives

- Know the definition of the complement of an event.
- Use the complement of an event to calculate the probability of an event.
- Understand the Complement Rule.

## Introduction

In this lesson, 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.

### The Complement of an Event

The *complement* \begin{align*}A'\end{align*}

*Example*: 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*}

A *Venn diagram* that illustrates the relationship between \begin{align*}A\end{align*}

This leads us to say that the sum of the possible outcomes for event \begin{align*}A\end{align*}

### 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*}

*Example*: 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*}

*Example*: Two coins are tossed simultaneously. Let the event \begin{align*}A\end{align*}

What is the complement of \begin{align*}A\end{align*}

Since the sample space of event \begin{align*}A=\left \{HT, TH, HH\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)=\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*}

*Example*: 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.

## Lesson Summary

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

The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event \begin{align*}A, P(A)+P(A')=1\end{align*}.

## Multimedia Links

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

## Review Questions

- 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*}

- 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*}. Find each of the following: \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'), P \left [(A \cup B)' \right ]\end{align*}.

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