Lesson Explainer: Operations on Events | Nagwa Lesson Explainer: Operations on Events | Nagwa

Lesson Explainer: Operations on Events Mathematics • Third Year of Preparatory School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

In this explainer, we will learn how to find the probabilities of the intersection and union of events.

Before we begin discussing some of the different operations we can apply to events, let’s start by recapping how we calculate the probability of a simple event and what this probability tells us.

Key Term: Probability of Simple Events

If 𝐴 is an event in a sample space 𝑆, then the probability of event 𝐴 occurring is 𝑃(𝐴)=𝑛(𝐴)𝑛(𝑆), where 𝑃(𝐴) represents the probability of event 𝐴, 𝑛(𝐴) represents the number of elements in event 𝐴, and 𝑛(𝑆) represents the number of elements in the sample space 𝑆 assuming each outcome is equally likely.

Since 𝐴 is a subset of 𝑆, we must have 𝑛(𝐴)≀𝑛(𝑆). Combining this with 𝑛(𝐴)β‰₯0, we get 0≀𝑛(𝐴)𝑛(𝑆)=𝑃(𝐴)≀1.

Hence, the probability of an event occurring lies between 0 and 1 and the value of this probability tells us the likelihood of the event occurring. In particular, we have the following:

  • If 𝑃(𝐴)=0, the event is impossible; it will never occur.
  • If 𝑃(𝐴)=1, the event is certain to happen.
  • The higher the probability of an event, the more likely it is to occur.

For example, we can use this to determine various probabilities of events involving a fair six-sided die. The faces of this die will be numbered from one to six as shown.

We can calculate the probability of rolling an even number, 𝑃()even, by using the formula 𝑃()=.evennumberofevenfacestotalnumberoffaces

We can see that there are three even faces:

Furthermore, there are 6 faces in total, so 𝑃()=36=12.even

We can use this same process for more complicated events. For example, what is the probability of rolling a face whose number is even and a face whose number is divisible by 3?

We have already seen that there are three faces with even numbers, and we can also show that there are two faces with numbers divisible by 3:

We can do this directly from the lists. We can see that only is in both lists. However, to generalize this process, we recall that we can write events as sets, which means we can apply any set operations to these events.

Now, we note that for a face to be both even and divisible by 3, it must be in both events. If we introduce sets for each event, say,,𝐴=and,𝐡=,then saying the face is both even and divisible by 3 means that it is in both sets, so it is in their intersection 𝐴∩𝐡. We have 𝐴∩𝐡=,which we can write formally as follows.

Definition: Intersection of Events

The intersection of events 𝐴 and 𝐡, denoted by 𝐴∩𝐡, is the collection of all outcomes that are elements of both of the sets 𝐴 and 𝐡; this is equivalent to both events occurring.

This allows us to determine the probability of rolling a face that is even and divisible by 3. We have ,,,𝑃(𝐴∩𝐡)=π‘ƒο€½ο­οΉβˆ©ο­οΉο‰=𝑃.

Then, the probability is the number of elements in 𝐴∩𝐡 divided by the number of elements in the sample space, so 𝑃(𝐴∩𝐡)=16.

This process is quite difficult when working with sets since we need to list all of the elements in 𝐴 and 𝐡 and then find their intersection. It is much easier to follow this process visually. To do this, we will introduce the idea of a Venn diagram.

Definition: Venn Diagram

A Venn diagram is a visual representation of a sample space 𝑆 and a finite number of subsets. Every subset is represented by a shape (usually a circle), everything within the shape is an element of the subset (or sometimes the number of elements of that subset), and everything outside the shape is not an element of the subset.

Venn diagrams usually show all possible relations between the subsets by overlapping the different circles to represent elements belonging to multiple subsets.

To see an example of a Venn diagram, let’s create a Venn diagram for the event in the example above. Our sample space will be all of the possible faces, so ,,,,,𝑆=.We have shown that ,,𝐴=,,𝐡=, and 𝐴∩𝐡=.So, in our Venn diagram, needs to be inside both the circles for 𝐴 and 𝐡, and will be in 𝐴 but not in 𝐡, is in 𝐡 but not in 𝐴, and and are not in 𝐴 or 𝐡 but are in 𝑆. We get the following diagram.

We can use this diagram to determine many different probabilities. For example, for the face to be both even and divisible by 3, it must lie inside both circles. We can see this set on the diagram as follows.

The intersection of the events is all of the elements in both circles as highlighted. We can see this is only one element out of the six total elements, so 𝑃(𝐴∩𝐡)=16.

This is not the only probability we can determine from this diagram. We can see that there are three elements in event 𝐴, so 𝑃(𝐴)=36=12, and there are 2 elements in event 𝐡, so 𝑃(𝐡)=26=13. We can also use this diagram to answer a more complicated question: What is the probability of rolling an even face or a face divisible by 3?

This is the same as asking for the probability of 𝐴 or 𝐡; therefore, we need to determine the elements in either 𝐴 or 𝐡. We can recall that this is the union of the sets 𝐴βˆͺ𝐡.

Definition: Union of Events

The union of events 𝐴 and 𝐡, denoted by 𝐴βˆͺ𝐡, is the collection of all outcomes that are elements of either set 𝐴, set 𝐡, or both sets; this is equivalent to either of the events occurring.

In the diagram, for a face to be even or divisible by 3, it can lie in either 𝐴 or 𝐡, so it can be any element inside those circles, as shown.

There are 4 elements in either 𝐴 or 𝐡, so 𝑃(𝐴βˆͺ𝐡)=46=23.

Before we move on to our first example, it is worth noting there can be any number of sets included in a Venn diagram and these do not need to overlap. This is because there may be no elements in both events. For example, on a six-sided die, there is no face that is both even and odd. The Venn diagram of the events of rolling an even and an odd number can be drawn as follows.

We can see in the diagram that there is no overlap between the two events. This means that the event of both occurring is impossible since 𝑛(∩)=0evenodd. To help us understand this possibility, we will introduce the idea of the empty set, which is the set with no elements.

Definition: Empty Set and Impossible Events

If an event 𝐴 in a sample space 𝑆 cannot occur, then its probability is 0. Since 𝑃(𝐴)=𝑛(𝐴)𝑛(𝑆)=0, we must have 𝑛(𝐴)=0. In other words, there are no elements in 𝐴. We call the set with no elements the empty set and denote this set by βˆ….

If an event is empty, then it is impossible, and if an event is impossible, then it is empty.

Let’s now see an example of using a given Venn diagram to determine a probability.

Example 1: Finding the Probability of the Intersection of Events from a Venn Diagram

The diagram represents the sample space 𝑆 and the events 𝐴, 𝐡, and 𝐢. Determine 𝑃(𝐴∩𝐡).

Answer

We recall that a Venn diagram is a visual representation of a space 𝑆 and a finite number of subsets. Every subset is represented by a shape (usually a circle), everything within the shape is an element of the subset, and everything outside the shape is not an element of the subset.

We want to use the given Venn diagram to determine 𝑃(𝐴∩𝐡). To do this, we first need to recall that the probability of an event is the number of possibilities where the event occurs divided by the total number of outcomes in the same space.

We also recall that 𝐴∩𝐡 is the intersection of both events, so it contains all of the elements that lie in both events at once. In the Venn diagram, this will be all of the elements inside shapes 𝐴 and 𝐡. We can highlight this on the diagram.

The only element in both 𝐴 and 𝐡 is 6, so 𝐴∩𝐡={6}.

The sample space is every element in the diagram, that is, 𝑆={4,6,8,16,18,19}.

Thus, the sample space has 6 elements, and hence, 𝑃(𝐴∩𝐡)=𝑛(𝐴∩𝐡)𝑛(𝑆)=16.

In our next example, we will determine the probability of the union of two events from a given Venn diagram.

Example 2: Using a Venn Diagram to Determine the Union of Two Events

A survey was conducted on a group of 263 children to determine their favorite superheroes. The results are shown in the Venn diagram. Find 𝑃()WonderWomanorBatman.

Answer

We recall that a Venn diagram is a visual representation of a sample space 𝑆 and a finite number of subsets. Every subset is represented by a shape (usually a circle). In this Venn diagram, each number represents the number of students who selected the given hero as their favorite superhero.

We want to use the given Venn diagram to determine 𝑃()WonderWomanorBatman. We can note that this is equivalent to 𝑃(βˆͺ)WonderWomanBatman. To find this probability, we first need to recall that the probability of an event is the number of possibilities where the event occurs divided by the total number of outcomes in the same space.

We also recall that 𝐴βˆͺ𝐡 is the union of both events, so it contains all of the elements that lie in either event. In a Venn diagram, this will be all of the elements inside either shape 𝐴 or shape 𝐡. It also includes all elements in the overlap of both shapes.

In this case, these are the circles labeled Wonder Woman or Batman, so we highlight both of these together on the diagram.

We can determine the number of elements in this shape by adding the values together, so we have 𝑛(βˆͺ)=64+8+8+9+14+75=178.WonderWomanBatman

The sample space is all of the students questioned, and we are told that this was a group of 263 children, so 𝑛(𝑆)=263.

Hence, 𝑃(βˆͺ)=𝑛(βˆͺ)𝑛(𝑆)=178263.WonderWomanBatmanWonderWomanBatman

We can use a similar reasoning to the two questions above to construct a general formula for the union of two events. Consider the following Venn diagram of events 𝐴 and 𝐡.

We want to determine 𝑃(𝐴βˆͺ𝐡) in terms of more simple probabilities, so we can start by highlighting 𝐴βˆͺ𝐡 on the Venn diagram. It is all of the elements in either 𝐴 or 𝐡.

We can note this is almost the same as the sum of 𝑃(𝐴) and 𝑃(𝐡). We have the following.

Adding these probabilities will count any element in both 𝐴 and 𝐡 twice, so we need to remove the probability of the intersection. We have

We call this the addition rule for probability, and we can write this result as follows.

Definition: Addition Rule for Probability

𝑃(𝐴βˆͺ𝐡)=𝑃(𝐴)+𝑃(𝐡)βˆ’π‘ƒ(𝐴∩𝐡).

In our next example, we will need to construct a Venn diagram of the information given in order to determine various probabilities.

Example 3: Finding Probabilities of Intersection and Union of Events in a Word Problem

Students at a school must wear a sweatshirt or a blazer and are allowed to wear both. In one class of 32 students, 12 students wear blazers and 4 of the students who wear a blazer also wear a sweatshirt. Let 𝐴 be the event of randomly selecting a student from the class who wears a blazer and let 𝐡 be the event of randomly selecting a student from the class who wears a sweatshirt.

  1. Find 𝑃(𝐴), giving your answer as a fraction in its simplest form.
  2. Find 𝑃(𝐡), giving your answer as a fraction in its simplest form.
  3. Find 𝑃(𝐴∩𝐡), giving your answer as a fraction in its simplest form.
  4. Determine the value of 𝑃(𝐴βˆͺ𝐡), giving your answer as a fraction in its simplest form.

Answer

To answer this question, it can help to sketch a Venn diagram of the information given. We are told that 4 students wear both a blazer and a sweatshirt, so the intersection of 𝐴 and 𝐡 has 4 students. Next, 12 students wear blazers. This includes the 4 who also wear a sweatshirt, so the remaining part of 𝐴 will have 12βˆ’4=8 students. Finally, all the remaining students must wear only a sweatshirt. This is 32βˆ’12=20. This gives us the following.

Part 1

We recall that the probability of an event occurring is the number of possibilities where the event occurs divided by the total number of outcomes in the same space, so 𝑃(𝐴)=𝑛(𝐴)𝑛(𝑆), where 𝑛(𝐴) will be the number of students wearing a blazer and 𝑛(𝑆) will be the number of students in total. We are told that 12 students wear blazers, so 𝑛(𝐴)=12, and we know that there are 32 students in the class, so 𝑛(𝑆)=32.

Hence, 𝑃(𝐴)=1232=38.

Part 2

We have 𝑃(𝐡)=𝑛(𝐡)𝑛(𝑆), where 𝑛(𝐡) will be the number of students wearing a sweatshirt. We can find 𝑛(𝐡) from our diagram; it is the number of all students wearing a sweatshirt. We have 𝑛(𝐡)=4+20=24.

Hence, 𝑃(𝐡)=2432=34.

Part 3

We have 𝑃(𝐴∩𝐡)=𝑛(𝐴∩𝐡)𝑛(𝑆), where 𝑛(𝐴∩𝐡) will be the number of students wearing a blazer and a sweatshirt. We are told that there are 4 students wearing both, so 𝑃(𝐴∩𝐡)=432=18.

Part 4

There are three methods we can use to find 𝑃(𝐴βˆͺ𝐡).

First, we recall that 𝑃(𝐴βˆͺ𝐡)=𝑃(𝐴)+𝑃(𝐡)βˆ’π‘ƒ(𝐴∩𝐡). We have found these three values in the previous part, so we can substitute these value into the formula to get 𝑃(𝐴βˆͺ𝐡)=38+34βˆ’18=1.

Second, we recall that 𝐴βˆͺ𝐡 is the event of either 𝐴 or 𝐡 happening, so this is picking a student in either a blazer or a sweatshirt. In the Venn diagram, this is given by any student in the following diagram.

This includes all 32 students, so 𝑃(𝐴βˆͺ𝐡)=3232=1.

Third, we are told in the question that all students must wear a sweatshirt or a blazer, so we know the probability of choosing a student wearing one of these items is guaranteed. Hence, it must have a probability of 1.

In our next two examples, we will apply the formula for the probability of a union of events to determine the probabilities of compound events.

Example 4: Using the Addition Rule to Find the Probability of Union of Two Events

Denote by 𝐴 and 𝐡 two events with probabilities 𝑃(𝐴)=0.2 and 𝑃(𝐡)=0.47. Given that 𝑃(𝐴∩𝐡)=0.18, find 𝑃(𝐴βˆͺ𝐡).

Answer

We recall that 𝑃(𝐴βˆͺ𝐡)=𝑃(𝐴)+𝑃(𝐡)βˆ’π‘ƒ(𝐴∩𝐡). We can substitute the given values of 𝑃(𝐴), 𝑃(𝐡), and 𝑃(𝐴∩𝐡) into the formula to get 𝑃(𝐴βˆͺ𝐡)=0.2+0.47βˆ’0.18=0.49.

Example 5: Using the Addition Rule to Find the Probability of Intersection of Two Events

Denote by 𝐴 and 𝐡 two events with probabilities 𝑃(𝐴)=0.58 and 𝑃(𝐡)=0.2. Given that 𝑃(𝐴βˆͺ𝐡)=0.64, find 𝑃(𝐴∩𝐡).

Answer

We recall that 𝑃(𝐴βˆͺ𝐡)=𝑃(𝐴)+𝑃(𝐡)βˆ’π‘ƒ(𝐴∩𝐡). We can substitute the given values of 𝑃(𝐴), 𝑃(𝐡), and 𝑃(𝐴∩𝐡) into the formula to get 0.64=0.58+0.2βˆ’π‘ƒ(𝐴∩𝐡).

Adding 𝑃(𝐴∩𝐡) to both sides of the equation and subtracting 0.64 from both sides yields 𝑃(𝐴∩𝐡)=0.58+0.2βˆ’0.64=0.14.

In our final example, we will construct a Venn diagram and use the addition rule of probability to solve a word problem.

Example 6: Finding the Probability of the Intersection of Events in a Word Problem

Out of a group of 100 people, 46 have dogs, 41 have cats, and 28 have rabbits. 12 of the people have both dogs and cats, 10 have both cats and rabbits, and 9 have both dogs and rabbits. 8 of the people have dogs, cats, and rabbits.

  1. Find the probability of randomly selecting a person who has dogs, cats, and rabbits. Give your answer as a fraction in its simplest form.
  2. Find the probability of randomly selecting a person who only has dogs and rabbits. Give your answer as a fraction in its simplest form.
  3. Find the probability of selecting a person who has pets. Give your answer as a fraction in its simplest form.
  4. Find the probability of selecting a person who does not have pets. Give your answer as a fraction in its simplest form.

Answer

To answer this question, let’s sketch a Venn diagram of the group of people. Let’s call the group of 100 people the sample space 𝑆, the group of people who own a dog D, the group of people who own a cat C, and the group of people who own a rabbit R.

We know that 8 of the people have dogs, cats, and rabbits, so the intersection of all three sets will have 8 members. Hence, 𝑛(∩∩)=8DCR. We can add this information to get the following Venn diagram.

We also know the following:

  • 9 people have both dogs and rabbits. Removing the 8 people who have all three animals, we have 1 person with dogs and rabbits but no cats.
  • 10 people have both cats and rabbits. Removing the 8 people who have all three animals, we have 2 people with cats and rabbits but no dogs.
  • 12 people have both dogs and cats. Removing the 8 people who have all three animals, we have 4 people with dogs and cats but no rabbits.

We can use this to update the Venn diagram.

We can follow this process again:

  • 46 people have dogs. Removing the 1+8+4=13 people we have already accounted for, this leaves 46βˆ’13=33 people who only have dogs.
  • 41 people have cats. Removing the 4+8+2=14 people we have already accounted for, this leaves 41βˆ’14=27 people who only have cats.
  • 28 people have rabbits. Removing the 1+8+2=11 people we have already accounted for, this leaves 28βˆ’11=17 people who only have rabbits.

Finally, we can see that there are 33+4+27+1+8+2+17=92 people accounted for in the Venn diagram. These are the people with any of these animals. The remaining 100βˆ’92=8 people must not have any of the given animals. We write this number outside of the circles to get the following:

Part 1

To determine the probability of choosing a person who has all three animals, we recall that the probability will be the number of people with all three animals from the group divided by the total number of people from the group.

We are told that 8 of the people have dogs, cats, and rabbits, and we know there are 100 people in the group, so 𝑃(∩∩)=8100=225.DCR

Part 2

To find the probability of randomly selecting a person who only has dogs and rabbits, we need to determine the number of people in the group who have dogs and rabbits but do not have cats. This must lie in the overlap between the circle for dogs and the circle for rabbits but not the circle for cats. We can highlight this as follows.

We can see there is only one such person, so the probability is 1100.

Part 3

We saw previously that there were 92 people with animals in the group, so the probability of choosing a person with animals is 92100=2325.

Part 4

We saw previously that there were 8 people with none of the given animals in the group, so the probability of choosing a person with none of these animals is 8100=225.

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

Key Points

  • The intersection of events 𝐴 and 𝐡, denoted by 𝐴∩𝐡, is the collection of all outcomes that are elements of both of the sets 𝐴 and 𝐡; this is equivalent to both events occurring.
  • The union of events 𝐴 and 𝐡, denoted by 𝐴βˆͺ𝐡, is the collection of all outcomes that are elements of either set 𝐴, set 𝐡, or both sets; this is equivalent to either of the events occurring.
  • If a set has no elements, we call this the empty set, and it is denoted by βˆ…. If the collection of outcomes of an event is the empty set, then the event cannot occur; hence, 𝑃(βˆ…)=0.
  • The addition rule of probability states that 𝑃(𝐴βˆͺ𝐡)=𝑃(𝐴)+𝑃(𝐡)βˆ’π‘ƒ(𝐴∩𝐡).

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy