Lesson Video: Operations on Events | Nagwa Lesson Video: Operations on Events | Nagwa

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

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

15:37

### Video Transcript

In this video, we will learn how to find the probabilities of the intersection and union of events. We will begin by considering a definition of the union and intersection of events and how they can be represented on a Venn diagram.

The intersection of events π΄ and π΅, denoted π΄ intersect π΅, is the collection of all outcomes that are elements of both the sets π΄ and π΅. In other words, π΄ intersect π΅ is the event that both events π΄ and π΅ occur. This can be represented on a Venn diagram, as shown. We shade the region that is in circle π΄ and in circle π΅. The union of events π΄ and π΅, denoted π΄ union π΅, is the collection of all outcomes that are elements of one or the other of sets π΄ and π΅ or of both of them. This means that π΄ union π΅ is the event that π΄ occurs, π΅ occurs, or they both occur. We can demonstrate π΄ union π΅ on a Venn diagram by shading everything in circle π΄ or in circle π΅. We will now look at a couple of examples where we can use Venn diagrams to determine the intersection and union of events.

The diagram represents the sample space π and the events π΄, π΅, and πΆ. Determine the probability of π΄ intersection π΅.

We recall that the intersection of two events π΄ and π΅ are those outcomes that occur in event π΄ and in event π΅. On a Venn diagram, this is represented by the overlap of circle π΄ and circle π΅. We can see from the Venn diagram that the sample space π contains the numbers four six, eight, 16, 18, and 19. This is a total of six outcomes. Of these six outcomes, only one appears in the intersection of event π΄ and event π΅, the number six. We know that the probability of an event occurring can be written as the number of successful outcomes over the number of possible outcomes. This means that the probability of π΄ intersection π΅ is one out of six. There is a one in six chance of selecting an outcome that is in event π΄ and in event π΅.

Our next question is a probability problem in context.

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 nine have both dogs and rabbits. Eight of the people have dogs, cats, and rabbits. There are four parts to this question. Find the probability of randomly selecting a person who has dogs, cats, and rabbits. Find the probability of randomly selecting a person who has only dogs and rabbits. Find the probability of selecting a person who has pets. And find the probability of selecting a person who does not have pets.

In all four parts, weβre asked to give our answer as a fraction in its simplest form. The first two parts can be answered directly from the question. We are told that the group contains 100 people, and eight of them have dogs, cats, and rabbits. This means that the probability of randomly selecting a person who has dogs, cats, and rabbits is eight out of 100. Dividing both the numerator and denominator by four, this simplifies to two out of 25 or two twenty-fifths. We are also told in the question that nine of the people have both dogs and rabbits. This is a subset of dogs, cats, and rabbits, those that have all three animals. And we know that eight people have dogs, cats, and rabbits. We can therefore subtract eight from nine to calculate the number of people that have just dogs and rabbits. This is equal to one. And we can therefore conclude that the probability of randomly selecting a person who has only dogs and rabbits is one out of 100.

An alternative way to visualize this would be to use a Venn diagram. We will now clear some space to draw this. Our Venn diagram has three sections, one for dogs, one for cats, and one for rabbits. The parts of the Venn diagram where these sections overlap represent the people with two or more pets. As eight people have all three pets, we can place an eight in the intersection of the dogs, cats, and rabbits. As nine people have both dogs and rabbits and eight of these have all three pets, only one person has only a dog and a rabbit. In the same way, we are told 10 people have both cats and rabbits. This means that two people have only cats and rabbits. Likewise, since a total of 12 people have both dogs and cats, four of these will have only dogs and cats.

Next, we see that 28 people have rabbits. This means that the sum of the numbers in the rabbit section must equal 28, and there are therefore 17 people who have only rabbits. Repeating this for cats and dogs, we see that 27 people have only cats and 33 people have only dogs. In order to calculate the number of people who have pets, we can find the sum of the seven numbers in the Venn diagram currently. This is equal to 92, and this means that the probability of randomly selecting a person who has pets is 92 out of 100. As the numerator and denominator are both divisible by four, this simplifies to 23 out of 25. The probability of selecting a person who has pets is twenty-three twenty-fifths.

As the sum of all of our numbers in the Venn diagram must equal 100, we can calculate the number that lies outside the three sections by subtracting 92 from 100. There are eight people in the group who do not have pets. Therefore, the probability of randomly selecting a person who does not have pets is eight out of 100. Once again, this fraction can be simplified by dividing the numerator and denominator by four. The probability of randomly selecting a person who does not have pets is two twenty-fifths.

This question shows us that drawing a Venn diagram can be useful when calculating the probability of events. Before looking at any more examples, we will consider how the definition of the union and intersection of events leads us to a general rule. The addition rule of probability states that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ intersection π΅. This can also be shown using Venn diagrams. When adding the probability of event π΄ to the probability of event π΅, we add the intersection twice. This means that by subtracting the probability of the intersection of events π΄ and π΅ from this gives us the probability of the union of events π΄ and π΅. Letβs now consider how this rule is used in practice.

Denote by π΄ and π΅ two events with probabilities the probability of π΄ is 0.2 and the probability of π΅ is 0.47. Given that the probability of π΄ intersection π΅ is 0.18, find the probability of π΄ union π΅.

In order to answer this question, we recall the addition rule of probability, which links all four of these events. This states that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ intersection π΅. Substituting in the values from this question, we have the probability of π΄ union π΅ is equal to 0.2 plus 0.47 minus 0.18. This is equal to 0.49. It is important to note that we could also represent this using Venn diagrams, as shown.

In our final question, we will use the addition rule of probability to solve a problem in context.

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 four of the students who wear a blazer also wear a sweatshirt. Let π΄ be the event of randomly selecting a student from the class who wears a blazer and let π΅ be the event of randomly selecting a student from the class who wears a sweatshirt. There are four parts to this question. Find the probability of π΄, find the probability of π΅, find the probability of π΄ intersection π΅, and find the probability of π΄ union π΅.

In all four cases, weβre asked to give our answer as a fraction in its simplest form. We could work out the first and third parts directly from the question. We are told that 12 out of the 32 students wear blazers; therefore, the probability of π΄ is 12 out of 32. As both the numerator and denominator are divisible by four, this simplifies to three out of eight or three-eighths. We are also told that four of the students who wear a blazer also wear a sweatshirt. This means that the probability of π΄ intersection π΅ is four out of 32, as this is the probability of selecting a student who wears a blazer and a sweatshirt. Once again, we can divide the numerator and denominator by four, giving us one-eighth.

A key word in the question is βmust,β as weβre told that students must wear a sweatshirt or a blazer. We can use this fact to write down the probability of π΄ union π΅. As all of the students must wear a sweatshirt or a blazer or both, the probability of π΄ union π΅ is 32 over 32, which is equal to one. It is certain that a randomly selected student will be wearing at least one of a sweatshirt or a blazer.

This leaves us with the second part of the question, calculating the probability of π΅, the event of randomly selecting a student who wears a sweatshirt. We can answer this by using the addition rule of probability, which states that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ intersection π΅. Substituting in the values we already know, we have one is equal to three-eighth plus the probability of π΅ minus one-eighth. The right-hand side simplifies to two-eighths plus the probability of π΅, which in turn is equal to a quarter plus the probability of π΅. Subtracting one-quarter from both sides of this equation gives us the probability of π΅ is equal to three-quarters.

We now have answers to all four parts of this question. They are three-eighths, three-quarters, one-eighth, and one, respectively. We could also represent the information on a Venn diagram, where the numbers shown are the number of students in each section. There were 12 students that wear blazers, 24 students that wear sweatshirts, and four students that wear both. The three numbers sum to give us a total of 32 students.

We will now summarize the key points from this video. The intersection of events π΄ and π΅ is the collection of all outcomes that are elements of both of the sets π΄ and π΅. The union of events π΄ and π΅ is a collection of all outcomes that are elements of one or the other of the sets π΄ and π΅ or of both of them. These definitions are linked together by the addition rule of probability, which states that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ intersection π΅. This can be represented using Venn diagrams as shown.

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