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3.3 Complementary Events

A very useful concept in the calculation of event probabilities is the notion of complementary events:

The complement of an event A is the event that A does not occur—that is, the event consisting of all sample points that are not in event A. We denote the complement of A by $A^{c} .$ $A^{c} .$

An event A is a collection of sample points, and the sample points included in $A^{c}$ $A^{c}$ are those not in A. Figure 3.9 demonstrates this idea. Note from the figure that all sample points in S are included in either A or $A^{c}$ $A^{c}$ and that no sample point is in both A and $A^{c}$ $A^{c}$ . This leads us to conclude that the probabilities of an event and its complement must sum to 1:

Figure 3.9
Venn diagram of complementary events

Rule of Complements

The sum of the probabilities of complementary events equals 1; that is,

[&P|pbo|A|pbc||+|P|pbo|A^{c}|pbc||=|1. &]

P (A) + P (A^{c}) = 1.

$P (A) + P (A^{c}) = 1.$

In many probability problems, calculating the probability of the complement of the event of interest is easier than calculating the event itself. Then, because

[&P|pbo|A|pbc||+|P|pbo|A^{~it~c}|pbc||=|~normal~1~norm~ &]

P (A) + P (A^{c) = 1}

$P (A) + P (A^{c) = 1}$

we can calculate P(A) by using the relationship

[&P|pbo|A|pbc||=|1|-|P|pbo|A^{~it~c}|pbc|~normal~.~norm~ &]

P (A) = 1 - P (A^{c) .}

$P (A) = 1 - P (A^{c) .}$

Example 3.10 Probability of a Complementary Event—Coin-Toss Experiment

Problem

Consider the experiment of tossing fair coins. Define the following event: $A : {Observing at least one head} .$ $A : {Observing at least one head} .$
1. Find P(A) if 2 coins are tossed.
2. Find P(A) if 10 coins are tossed.

Solution

When 2 coins are tossed, we know that the event $A : {Observe at least one head .}$ $A : {Observe at least one head .}$ consists of the sample points

[&A: |cbo|HH, HT, TH|cbc| &]
$A : {H H, H T, T H}$ $A : {H H, H T, T H}$

The complement of A is defined as the event that occurs when A does not occur.

Therefore,

[&A^{c}: |cbo|~rom~Observe no heads~normal~.|cbc||=||cbo|*N*[-1%0]TT*N*[-0.5%0]|cbc| &]
$A^{c} : {Observe no heads .} = {T T}$ $A^{c} : {Observe no heads .} = {T T}$

This complementary relationship is shown in Figure 3.10. Since the coins are balanced, we have

Figure 3.10
Complementary events in the toss of two coins

[&P|pbo|A^{c}|pbc||=|~normal~P|pbo|TT|pbc||=|*frac*{1}{4}~norm~ &]
$P (A^{c}) = P (T T) = \frac{1}{4}$ $P (A^{c}) = P (T T) = \frac{1}{4}$

and

[&P|pbo|A|pbc||=|1|-|P|pbo|A^{c}|pbc||=|~normal~1|-|*frac*{1}{4}|=|*frac*{3}{4}.~norm~ &]
$P (A) = 1 - P (A^{c}) = 1 - \frac{1}{4} = \frac{3}{4} .$ $P (A) = 1 - P (A^{c}) = 1 - \frac{1}{4} = \frac{3}{4} .$
We solve this problem by following the five steps for calculating probabilities of events. (See Section 3.1.)

Step 1 Define the experiment. The experiment is to record the results of the 10 tosses of the coin.
Step 2 List the sample points. A sample point consists of a particular sequence of 10 heads and tails. Thus, one sample point is HHTTTHTHTT, which denotes head on first toss, head on second toss, tail on third toss, etc. Others are HTHHHTTTTT and THHTHTHTTH. Obviously, the number of sample points is very large—too many to list. It can be shown (see Section 3.7) that there are $2^{10} = 1, 024$ $2^{10} = 1, 024$ sample points for this experiment.
Step 2 List the sample points. A sample point consists of a particular sequence of 10 heads and tails. Thus, one sample point is HHTTTHTHTT, which denotes head on first toss, head on second toss, tail on third toss, etc. Others are HTHHHTTTTT and THHTHTHTTH. Obviously, the number of sample points is very large—too many to list. It can be shown (see Section 3.7) that there are $2^{10} = 1, 024$ $2^{10} = 1, 024$ sample points for this experiment.
Step 3 Assign probabilities. Since the coin is fair, each sequence of heads and tails has the same chance of occurring; therefore, all the sample points are equally likely. Then

[&P|pbo|~rom~Each sample point|pbc||=|~normal~*frac*{1}{1,024} &]
$P (Each sample point) = \frac{1}{1, 024}$ $P (Each sample point) = \frac{1}{1, 024}$
Step 4 Determine the sample points in event A. A sample point is in A if at least one H appears in the sequence of 10 tosses. However, if we consider the complement of A, we find that

[&A^{c}|=||cbo|~rom~No heads are observed in 10 tosses~normal~.*N*[-1.5%0]|cbc| &]
$A^{c} = {No heads are observed in 10 tosses .}$ $A^{c} = {No heads are observed in 10 tosses .}$

Thus, $A^{c}$ $A^{c}$ contains only one sample point:

[&A^{c}: |cbo|TTTTTTTTTT|cbc| &]
$A^{c} : {T T T T T T T T T T}$ $A^{c} : {T T T T T T T T T T}$

and $P (A^{c}) = \frac{1}{1, 024}$ $P (A^{c}) = \frac{1}{1, 024}$
Step 5 Now we use the relationship of complementary events to find P(A):

[&P|pbo|A|pbc||=|1|-|P|pbo|A^{c}|pbc||=|1|-|*frac*{1}{1,024}|=|*frac*{1,023}{1,024}|=|.999 &]
$P (A) = 1 - P (A^{c}) = 1 - \frac{1}{1, 024} = \frac{1, 023}{1, 024} = .999$ $P (A) = 1 - P (A^{c}) = 1 - \frac{1}{1, 024} = \frac{1, 023}{1, 024} = .999$

Look Back

In part a, we can find P(A) by summing the probabilities of the sample points HH, HT, and TH in A. Many times, however, it is easier to find $P (A^{c})$ $P (A^{c})$ by using the rule of complements.

Look Forward

Since $P (A) = .999$ $P (A) = .999$ in part b, we are virtually certain of observing at least one head in 10 tosses of the coin.

Now Work Exercise 3.47e–f

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Table of Contents for 3.3 Complementary Events

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Table of Contents for
3.3 Complementary Events