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3.3: The Complement of an Event

Difficulty Level: At Grade Created by: CK-12
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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.
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*}.

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

\begin{align*}\text{A:} \left \{\text{observe an odd number}\right \}\end{align*}

Then, \begin{align*}A = \left \{1, 3, 5 \right \}\end{align*}, which includes all the simple events of the set \begin{align*}S\end{align*} that are odd. Thus, the complement of \begin{align*}A\end{align*} is the set of simple events that will not occur in \begin{align*}A\end{align*}. So \begin{align*}A'\end{align*} will include all the elements that are not odd in the sample space of the set \begin{align*}S\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*} and its complement \begin{align*}A'\end{align*} are the sum of all the possible outcomes of the sample space of the experiment. Therefore, the probabilities of an event and its complement must sum to \begin{align*}1\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*}, is calculated using the relationship:

\begin{align*}P(A) = 1 - P(A')\end{align*}


If you know that the probability of getting the flu this winter is \begin{align*}0.43\end{align*}, what is the probability that you will not get the flu?


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*}


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*} and how would you calculate the probability of \begin{align*}A\end{align*} by using the complementary relationship?


Since the event \begin{align*}A\end{align*} is observing all simple events \begin{align*} A = \left \{HH, HT, TH \right \}\end{align*}, then the complement of \begin{align*}A\end{align*} is defined as the event that occurs when \begin{align*}A\end{align*} does not occur, namely, all the events that do not have heads, namely,

\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*} and \begin{align*}A'\end{align*} in the toss of two coins.

The second part of the problem is to calculate the probability of \begin{align*}A\end{align*} using the complementary relationship. Recall that \begin{align*}P(A) = 1 - P(A')\end{align*}. So by calculating \begin{align*}P(A')\end{align*}, we can easily calculate \begin{align*}P(A)\end{align*} by subtracting it from \begin{align*}1\end{align*}.

\begin{align*}P(A') = P(TT) = 1/4\end{align*}


\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*} occurring directly. The next example, however, will show you that sometimes it is easier to calculate the complementary relationship to find the answer that we are seeking.


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?


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*} another \begin{align*}THTHHHTHTH,\end{align*} etc. Is there a way to calculate the number of simple events for this experiment? The answer is yes but we will learn how to do this later in the chapter. For the time being, let us just accept that there are \begin{align*}2^{10} = 1024\end{align*} simple events in this experiment.

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

  1. 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*}.
  2. 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

  1. 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*}
    1. List the sample space for tossing a coin three times
    2. List the outcomes of \begin{align*}A.\end{align*}
    3. List the outcomes of \begin{align*}B.\end{align*}
    4. List the outcomes of the events \begin{align*}A \cup B, A', A \cap B.\end{align*}
    5. Find \begin{align*}P(A), P(B), P(A \cup B), P(A'), P(A \cap B).\end{align*}
  2. 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

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

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