Lesson Explainer: Complementary Events Mathematics

In this explainer, we will learn how to find the probability of complementary events.

In the context of probability, we recall that an event is a set of outcomes. For example, if we think about the days of the week, then every day in the calendar is 1 of 7 days of the week. In many countries, the daysΒ  Saturday and Sunday are classified as the weekend, and an event might be β€œborn on the weekend.” There are two possible outcomes in this set: {Saturday, Sunday}.

If we assume that births are equally likely to occur on any given day of the week, the probability that a person chosen at random was born on the weekend is the number of days in the weekend divided by the total number of days in the week: 𝑃()=27.bornontheweekend

We might then ask, what is the probability of not being born on the weekend? We can see in the above diagram that the days Β Monday–Friday are the weekdays, and so 𝑃()=57.notbornontheweekend

We call this event the complement of the event β€œborn on the weekend” since it is equivalent to the event not occurring. The complement of an event and the event itself cannot happen at the same time. In our example, we know that it cannot be both a weekday and the weekend at the same time. We recall that this means that the events β€œweekday” and β€œweekend” are mutually exclusive: there is no overlap between them. This is true in general, an event and its complement are always mutually exclusive.

We can define the concept of a complementary event more formally as follows.

Definition: Complementary Event

The complement of an event 𝐴 (written 𝐴) is equivalent to the event 𝐴 not occurring.

Let’s now discuss the probability of complementary events. We can note that the union of an event and its complement is the entire sample space. This is because the complement of an event is equivalent to the event not occurring, so saying 𝐴 occurs or does not occur must be the entire sample space: 𝐴βˆͺ𝐴=𝑆.

We can combine this result, the fact that 𝐴 and 𝐴 are mutually exclusive, and the addition rule for probability to prove a useful result.

First, 𝑃(𝑆)=𝑃(𝐴βˆͺ𝐴).

We know that 𝑃(𝑆)=1, and we can rewrite the right-hand side of the equation by using the addition rule of probability to get 1=𝑃(𝐴)+𝑃(𝐴)βˆ’π‘ƒ(𝐴∩𝐴).

Since 𝐴 and 𝐴 are mutually exclusive, we have 𝐴∩𝐴=βˆ…οŽ˜ and 𝑃(𝐴∩𝐴)=0, so 1=𝑃(𝐴)+𝑃(𝐴).

We can rearrange this equation in the following two ways: 𝑃(𝐴)=1βˆ’π‘ƒ(𝐴),𝑃(𝐴)=1βˆ’π‘ƒ(𝐴).

These allow us to use the probability of an event to determine the probability of its complement or vice versa.

We have shown the following result.

Property: Probability of Complementary Events

If 𝐴 and 𝐴 are complementary events, then

  • 𝑃(𝐴)=1βˆ’π‘ƒ(𝐴),
  • 𝑃(𝐴)=1βˆ’π‘ƒ(𝐴),
  • 𝑃(𝐴βˆͺ𝐴)=1,
  • 𝑃(𝐴∩𝐴)=0.

Let’s now see an example of how to use these formulas to determine the probability of a complement.

Example 1: Determining the Probability of a Complement of a Given Event

If the probability of an event occurring is 1336, what is the probability that it does not occur?

Answer

We first recall that the event not occurring is called its complement and that the probability of an event added to the probability of its complement is 1. In particular, we have 𝑃(𝐴)=1βˆ’π‘ƒ(𝐴).

In this case, we have 𝑃(𝐴)=1βˆ’1336=2336.

In our next example, we will apply this formula to a word problem involving complementary events.

Example 2: Finding the Probability of a Complementary Event in a Given Context

If the probability that a student passes in mathematics is 0.7, what is the probability that the student fails?

Answer

We note that the event of not passing is equivalent to saying that the student fails. In other words, the events β€œpass” and β€œfail” are complementary. We recall that we can find the probability of a complementary event by subtracting the probability of the event occurring from 1; hence, 𝑃()=1βˆ’π‘ƒ()=1βˆ’0.7=0.3.failpass

In our next example, we will use the formulas involving probability of complementary events and the definition of a probability to determine the number of nonred balls in a bag given the probability of picking out a red ball and the total number of balls in the bag.

Example 3: Solving a Word Problem Using Complementary Events

A box contains 56 balls. The probability of selecting at random a red ball is 57. How many balls in the box are not red?

Answer

We can start by noting that the color of the ball will not affect the chance of selecting the ball from the bag and that the event of choosing a nonred ball is the complement of choosing a red ball.

Since the chance of choosing any ball from the bag is the same, we have 𝑃()==56.nonrednumberofnonredballstotalnumberofballsnumberofnonredballs

Multiplying the equation through by 56 gives numberofnonredballsnonred=56×𝑃().

Therefore, we can determine the number of nonred balls in the bag by finding the probability of choosing a nonred ball from the bag.

Since the events are complementary, we have 𝑃()=1βˆ’π‘ƒ()=1βˆ’57=27.nonredred

Substituting this value into the equation gives numberofnonredballs=56Γ—27=16.

We can check this answer by noting that numberofredballstotalnumberofballsnumberofnonredballs=βˆ’=56βˆ’16=40.

Then, 𝑃()==4056=57.rednumberofredballstotalnumberofballs

Since this agrees with the given information, we have confirmed there are 16 nonred balls in the bag.

In our next example, we will use a frequency table to show two different methods for calculating the probability of a complementary event.

Example 4: Using a Frequency Table to Find the Probability of a Complementary Event

The table represents the data collected from 200 conference attendees of different nationalities.

Only Speak ArabicOnly Speak EnglishOnly Speak FrenchTotal Sum
Man453545125
Woman4030575
Sum856550200

Find the probability that a randomly selected participant does not speak English.

Answer

We first note that each person in attendance only speaks one language. There are two ways to find the probability that a participant chosen at random does not speak English.

The first method for finding the probability that a participant does not speak English is to note that, in this case, speaking English and not speaking English are complementary events.

Therefore, 𝑃()=1βˆ’π‘ƒ().notspeakEnglishspeakEnglish

Then, the probability of choosing an English speaker is the number of English speakers divided by the total number of people. We can see both of these totals in the table.

Hence, there are 65 English speakers and 200 people in the group, so the probability that a participant chosen at random does not speak English is 𝑃()=1βˆ’π‘ƒ()=1βˆ’65200=135200=0.675.notspeakEnglishspeakEnglish

The second method is to work out the total number of participants who do not speak English and then divide this by the total number of attendees. We can find this information from the table.

The number of participants who do not speak English is 85+50=135. The total number of participants is 200. Hence, 𝑃()==135200=0.675.notspeakEnglishnumberofnon-Englishspeakersnumberofpeopleinthegroup

Before we continue with more examples, it is worth noting that the properties of complementary events allow us to determine complementary events from Venn diagrams. We know that complementary events are mutually exclusive (so they have no overlap), and we know that together they make up the entire sample space.

In our final two examples, we will use this fact about complementary events in Venn diagrams to solve problems involving complementary events.

Example 5: Finding the Probability of a Complementary Event Using a Venn Diagram

The days in a certain month are classified as rainy days, hot days, both rainy and hot days, or neither rainy nor hot days. Let 𝑅 denote rainy days, and let 𝐻 denote hot ones. Use the Venn diagram below to calculate the probability that a day is not rainy.

Answer

There are two ways we can calculate this probability.

In the first method, we note that 𝑃()=.notrainynumberofrainydaystotalnumberofdaysinthemonth

We can find the number of days in the month by adding all of the data to get totalnumberofdaysinthemonth=9+4+2+15=30.

For a day to be not rainy, it cannot be in the event 𝑅. This is the complementary event π‘…οŽ˜ and is given by the following Venn diagram.

We can see that there are 15 days that were hot but not rainy, and 9 days that were neither hot nor rainy, so numberofnonrainydays=15+9=24.

Hence, 𝑃()=2430=45.notrainy

In the second method, we note that rainy and nonrainy days are complementary events, so 𝑃()=1βˆ’π‘ƒ().notrainyrainy

We can determine the number of rainy days from the Venn diagram.

There are 4 days that were rainy but not hot and 2 days that were rainy and hot, so numberofrainydays=4+2=6.

We note as before that there are 30 days in the month, so 𝑃()=1βˆ’π‘ƒ()=1βˆ’=1βˆ’630=45.notrainyrainynumberofrainydaystotalnumberofdaysinthemonth

In our final example, we will use a given Venn diagram to determine the probability of a complementary event.

Example 6: Finding the Probability of a Complementary Event Using a Venn Diagram

In a music class, students learn to play the piano, the guitar, and the drums. Some students play 2 instruments, some play all 3 instruments, and some play none of the instruments. Let 𝑃 denote those who play the piano, 𝐺 those who play the guitar, and 𝐷 those who play the drums.

Using the given diagram, calculate the probability that a student does not play the piano.

Answer

There are two methods we can use to determine the probability that a student does not play the piano.

In the first method, we note that 𝑃()=.notplaythepianonumberofstudentswhodonotplaythepianototalnumberofstudents

Adding all of the data from the Venn diagram, we get totalnumberofstudents=2+3+2+1+5+2+2+14=31.

We can determine the number of students who do not play the piano from the Venn diagram. The event of choosing a student who does not play the piano is the complement of 𝑃, so we can draw this on our Venn diagram as everything not in 𝑃.

We can then add all of the data not in 𝑃 to get numberofstudentswhodonotplaythepiano=2+1+5+14=22.

Hence, 𝑃()=2231.notplaythepiano

In the second method, we note that not playing the piano is a complementary event of 𝑃, so 𝑃()=1βˆ’π‘ƒ().notplaythepianopiano

We note that 𝑃()=.pianonumberofstudentswhoplaythepianototalnumberofstudents

We can determine the number of students who play the piano by adding all of the data in 𝑃 from the Venn diagram.

This gives numberofstudentswhoplaythepiano=3+2+2+2=9.

Hence, 𝑃()=1βˆ’931=2231.notplaythepiano

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The complement of an event 𝐴 (written 𝐴) is equivalent to the event 𝐴 not occurring.
  • Any event is mutually exclusive with its complement (i.e., 𝐴∩𝐴=βˆ…οŽ˜).
  • In a sample space 𝑆, since 𝐴 is equivalent to the event 𝐴 not occurring, and 𝐴 is equivalent to 𝐴 occurring, one of these events must occur, so 𝐴βˆͺ𝐴=π‘†οŽ˜ and 𝑃(𝐴βˆͺ𝐴)=1.
  • Applying the addition rule for probability to two complementary events tells us 𝑃(𝐴βˆͺ𝐴)=𝑃(𝐴)+𝑃(𝐴)βˆ’π‘ƒ(𝐴∩𝐴)=𝑃(𝐴)+𝑃(𝐴). Hence, 𝑃(𝐴)+𝑃(𝐴)=1.
  • The probability of the complement, 𝐴, of event 𝐴, or β€œnot 𝐴,” is given by 𝑃(𝐴)=1βˆ’π‘ƒ(𝐴). This can be rewritten as 𝑃(𝐴)=1βˆ’π‘ƒ(𝐴).
  • In a Venn diagram, 𝐴 and 𝐴 have no overlap and combine to make the entire sample space.

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