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In this explainer, we will learn how to use Venn diagrams to organize information and calculate probabilities.

In probability, a Venn diagram is a figure with one or more circles inside a rectangle that describes logical relations between events. The rectangle in a Venn diagram represents the sample space or the universal set, that is, the set of all possible outcomes. A circle inside the rectangle represents an event, that is, a subset of the sample space.

We consider the following Venn diagram involving two events, 𝐴 and 𝐵.

In the diagram above, we have two events 𝐴 and 𝐵 within the sample space (or universal set) 𝑆. Sometimes, the sample space is denoted by 𝜎 or 𝜉 instead of 𝑆. Colored regions in this Venn diagram represent the following events: GreenandpurpleregionsBlueandpurpleregionsPurpleregionGreen,purple,andblueregionsYellowregionalternatively,𝐴,𝐵,𝐴𝐵,𝐴𝐵,𝐴𝐵,(𝐴𝐵).

Definition: Two-Event Venn Diagrams

Let 𝐴 and 𝐵 be events described in a Venn diagram. Then,

  • the circles do not overlap if 𝐴 and 𝐵 are mutually exclusive events, that is, 𝐴𝐵=;
  • the circles overlap if 𝐴𝐵, in which case the intersection 𝐴𝐵 is represented by the overlapping region;
  • the region outside both circles but within the rectangle represents the complement of the union of both events, that is, 𝐴𝐵 or, alternatively, (𝐴𝐵).

Within each divided region of a Venn diagram, we can add data in one of the following ways:

  • the outcomes of the event,
  • the number of outcomes in the event,
  • the probability of the event.

In out first example, we will use a Venn diagram to organize our data and use it to compute the probability of an event.

Example 1: Organizing Data using Venn Diagrams to Find Probabilities

A class contains 100 students; 70 of them like mathematics, 60 like physics, and 40 like both. If a student is chosen at random, using a Venn diagram, find the probability that they like mathematics but not physics.

Answer

We begin by drawing an empty Venn diagram to represent this example.

We know that the overlapping region in a Venn diagram represents the intersection of the events. We are given that 40 students belong to this intersection, since they like both mathematics and physics.

70 students like mathematics, and 40 of them also like physics. This tells us that the number of students who like mathematics only is 7040=30. Similarly, we can deduce that 6040=20 students like physics only. This leads to the following Venn diagram.

We note that the number 10 outside is to ensure that the sum of all values within the Venn diagram is equal to 100, since the class contains 100 students total. We are looking for the probability that a randomly chosen student likes mathematics but not physics. The region of the Venn diagram representing this event is highlighted below.

From the Venn diagram, we note that 30 students like mathematics but not physics. Since a student is randomly selected, we can obtain the probability of this event when dividing this number by the total number of students, which is 100.

Hence, the probability that a randomly selected student likes mathematics but not physics is 30100=0.3.

In our next example, we will use a Venn diagram to compute the probability of an event.

Example 2: Using Venn Diagrams to Calculate Probability

Yara has drawn this Venn diagram to record the result of randomly selecting a number between 1 and 12.

  1. What is the probability of selecting a number that is a factor of 20? Give your answer as a fraction in its simplest form.
  2. What is the probability of selecting a number that is a factor of 20 and a multiple of 3? Give your answer as a fraction in its simplest form.
  3. What is the probability of selecting a number that is not a multiple of 3? Give your answer as a fraction in its simplest form.

Answer

In the given Venn diagram, two events are represented as circles: “Factor of 20” and “Multiple of 3.” The sample space of this Venn diagram is the set of integers between 1 and 12. In other words, the sample space (or the universal set) is given by 𝑆={1,2,3,,12}.

Part 1

Since we are randomly selecting a number between 1 and 12, the probability of an event can be obtained from dividing the size of the event by the size of the sample space, which is 12. In other words, 𝑃(20)=2012.factorofnumberoffactorsof

We can see from the Venn diagram that there are 5 different factors of 20 between 1 and 12.

Hence, the probability of selecting a number between 1 and 12 that is a factor of 20 is 512.

Part 2

We note that the two events are mutually exclusive since the circles in the Venn diagram do not overlap. In other words, there is no number between 1 and 12 which is both a factor of 20 and a multiple of 3.

Hence, the probability of this event is equal to 0.

Part 3

From the Venn diagram, the numbers between 1 and 12 that are not a multiple of 3 lie outside of the circle. We can see that there are 8 numbers that belong to this case.

Hence, the probability that a randomly selected number between 1 and 12 is not a multiple of 3 is 812, which can be simplified to 23.

We can also use a Venn diagram to compute conditional probabilities. In our next example, we will organize the given data using a Venn diagram and use it to compute the probability of an event given a condition.

Example 3: Finding the Conditional Probability of an Event

In the final exams, 40% of the students failed chemistry, 25% failed physics, and 19% failed both chemistry and physics. What is the probability that a randomly selected student failed physics given that he passed chemistry?

Answer

Let us begin by drawing an empty Venn diagram.

We can use the given information to add the relative proportion of each region. Since 40% of the students failed chemistry while 19% failed both chemistry and physics, we can deduce that 40%19%=21% of the students only failed chemistry. Similarly, we can obtain that 25%19%=6% of the students only failed physics. Then, we have the following data:

  • 21% of the students failed chemistry but not physics;
  • 6% of the students failed physics but not chemistry;
  • 19% of the students failed both physics and chemistry.

Now, the remaining relative proportion is for students who passed both subjects. Since the sum of all relative proportions must equal 100%, the percentage of students who passed both subjects is given by 100%21%6%19%=54%.

We add these percentages onto the Venn diagram, as follows.

We want to find the probability that a randomly selected student failed physics given that he passed chemistry. The percentage of students who passed chemistry is given outside of the circle labeled “Failed chemistry” in the Venn diagram above. This is the region shaded in purple below. Also, we know that the region shaded in green represents the students who failed physics and passed chemistry.

Notice that we are given that the student passed chemistry. This means that we are selecting a student from the group of students who passed chemistry rather than from the entire class. In other words, our sample space (or universal set) is the purple-shaded region above, not the rectangle.

The percentage of students who passed chemistry is given by summing the percentages within the purple region, which leads to 6%+54%=60%. Among the students who passed chemistry, the percentage of students who failed physics is found in the green-shaded region, which is 6%. So, the probability of randomly selecting a student who failed physics from the group of students who passed chemistry is given by percentageofstudentswhofailedphysicsbutpassedchemistrypercentageofstudentswhopassedchemistry=6%60%=0.1.

Let us check this answer using the formula for the conditional probability for event 𝐴 given event 𝐵: 𝑃(𝐴𝐵)=𝑃(𝐴𝐵)𝑃(𝐵).

In our example, this formula can be written as 𝑃()=𝑃()𝑃().studentswhofailedphysicsstudentswhopassedchemistrystudentswhofailedphysicsandpassedchemistrystudentswhopassedchemistry

From the green- and purple-shaded regions of the Venn diagram above, we can obtain 𝑃()=0.06,𝑃()=0.06+0.54=0.60.studentswhofailedphysicsandpassedchemistrystudentswhopassedchemistry

Substituting these values into the equation above, we obtain 𝑃()=0.060.60=0.1.studentswhofailedphysicsstudentswhopassedchemistry

Hence, the probability that a randomly selected student failed physics given that he passed chemistry is equal to 0.1.

In our previous example, we computed a conditional probability using a Venn diagram. In general, the conditional probability of event 𝐴 given another event 𝐵 can be computed using a Venn diagram and following similar steps.

How To: Computing Conditional Probabilities Using Venn Diagrams

To compute the conditional probability of event 𝐴 given another event 𝐵 using a Venn diagram, we need to

  1. identify the region representing event 𝐵 from the Venn diagram and compute the probability of 𝐵,
  2. identify the region representing the intersection 𝐴𝐵 and compute the probability of 𝐴𝐵,
  3. compute 𝑃(𝐴𝐵)𝑃(𝐵).

We note that this process leads to the standard formula 𝑃(𝐴𝐵)=𝑃(𝐴𝐵)𝑃(𝐵), where 𝑃(𝐴𝐵) is the conditional probability of event 𝐴 given another event 𝐵.

We can also use a Venn diagram to describe relationships between three events. In two of the special circumstances listed below, we have a special representation of a three-event Venn diagram.

Let us consider an example where we use a three-event Venn diagram, where one of the events is mutually exclusive to the other two events.

Example 4: Using the Venn Diagram to Determine the Probability of the Union of Two Events

Use the diagram of the sample space 𝑆 to determine 𝑃(𝐵𝐶).

Answer

From the given Venn diagram, we can tell that event 𝐶 is mutually exclusive to events 𝐵 and 𝐶. We recall that the probability of an event can be obtained using a Venn diagram by computing the relative proportion of the event with respect to the sample space.

In the Venn diagram, the sample space is denoted 𝑆 and contains 10 different outcomes. Thus, we can compute the probability of our event by dividing the number of outcomes in the event by 10. Event 𝐵𝐶 is outlined in purple below.

From the diagram above, we note that there are 6 distinct outcomes in the region representing 𝐵𝐶.

Thus, the probability is given by 𝑃(𝐵𝐶)=610=35.

Unless one of the three events is mutually exclusive to the other two events, we use the following convention for a three-event Venn diagram.

We note that all three circles overlap over the middle of the standard three-event Venn diagram. This region represents the intersection of all three events, 𝐴𝐵𝐶.

Let us consider an example of computing a probability using a standard three-event Venn diagram.

Example 5: Using a Venn Diagram to Determine an Intersection of Events

271 students voted for the types of music they wanted at the school dance. The results are shown in the Venn diagram. Find the probability that a randomly selected student voted for rock and not jazz.

Answer

In this example, we are given a three-event Venn diagram. Since a student is randomly selected, we know that the probability of an event is given by dividing the number of students for the event by the total number of students, which is 271. In other words, 𝑃()=271.eventnumberofstudentsbelongingtotheevent

So, we need to find the number of students who voted for rock but not jazz. From the given Venn diagram, the region representing this event is outlined in black:

Using the Venn diagram, we can see that the number of students for this event is 70+7=77. Dividing this number by 271 gives the probability of this event occurring.

Hence, the probability that a randomly selected student voted for rock and not jazz is 77271.

In our final example, we will organize data involving three events using a standard three-event Venn diagram and use it to find a probability.

Example 6: Organizing Data using Venn Diagrams to Find Probabilities

In a sample of 100 students enrolling in a university, a questionnaire indicated that 45 of them studied English, 40 studied French, 35 studied German, 20 studied both English and French, 23 studied both English and German, 19 studied both French and German, and 12 studied all three languages.

Using a Venn diagram, find the probability that a randomly chosen student studied only one of the three languages.

Answer

We begin by drawing an empty standard three-event Venn diagram.

To fill in this diagram, we begin at the middle where all three circles overlap. The middle of the three-event Venn diagram represents the intersection of all three events. In our example, this represents the students who studied all three languages, of which there are 12.

Next, we consider the other overlapping regions. We know that 20 students studied both English and French, but we also know that 12 of these students studied all three subjects, which means they also studied German. From this, we can deduce that 2012=8 students studied English and French, but not German. Similarly, we can gather the following data:

  • Students who studied English and German but not French: 2312=11 students,
  • Students who studied French and German but not English: 1912=7 students.

We now add these figures in their corresponding regions.

We proceed to obtain the values for the part of each circle not overlapping the other two circles. We are given that 45 students studied English, but 8+12+11 of them also studied French or German. So, the number of students who studied English but not French or German is given by 4581211=14.

Likewise, we obtain the following data:

  • Students who studied French but not English or German: 408127=13 students,
  • Students who studied German but not English or French: 3571211=5 students.

We now add these figures to the Venn diagram.

In this Venn diagram, the region representing the students who studied only one of the three languages is shaded in yellow below.

Then, the number of such students is given by 13+14+5=32.

Since the student is randomly selected, we can obtain the probability of our event by dividing 32 by the total number of students, which is 100.

The probability that a randomly chosen student studied only one of the three languages is 32100=0.32.

Let us summarize a few important concepts from this explainer.

Key Points

  • Each of the divided regions in a Venn diagram can contain one of the following data:
    • outcomes of an event,
    • number of outcomes in an event,
    • probability or relative proportion of an event.
  • A two-event Venn diagram describes the relationship between two events in the following ways:
    • If the two events are mutually exclusive, then the circles representing each event do not overlap.
    • If the two events are not mutually exclusive, then the two circles overlap. The overlapping region represents the intersection of the two events.
  • To compute the conditional probability of event 𝐴 given another event 𝐵 using a Venn diagram, we need to
    • identify the region representing event 𝐵 from the Venn diagram and compute the probability of 𝐵,
    • identify the region representing the intersection 𝐴𝐵 and compute the probability of 𝐴𝐵,
    • compute 𝑃(𝐴𝐵)𝑃(𝐵).
  • In a three-event Venn diagram, we can follow the convention of a two-event Venn diagram when one event is mutually exclusive to the other two events. Otherwise, we use the following standard three-event Venn diagram.
  • The middle of a standard three-event Venn diagram where all three circles overlap represents the intersection of all three events. When organizing data into a standard three-event Venn diagram, we can begin by identifying the middle.

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