### 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 πΆ. 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, weβre 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΅. Letβs 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, weβre 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
βmust,β as weβre 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.