### Video Transcript

In this video, we will learn how to use Venn diagrams to organize information and calculate probabilities. Venn diagrams are made of two or more circles that sometimes overlap. They are common when dealing with sets and probability. We will begin by looking at what a Venn diagram looks like and some of the notation involved.

As mentioned, Venn diagrams usually contain two or more circles as shown. In the majority of cases, the circles will overlap. The two circles represent two events, which we will call π΄ and π΅. Each section of the Venn diagram will contain a number. This can sometimes be a fraction or decimal. In an example like this where we have integers, the total number of items in the sample is the sum of the integers. To calculate the total, we need to add three, one, nine, and seven. This is equal to 20, so the majority of our probability questions will be out of 20.

Letβs know consider some of the notation that is used in different questions. π of π΄, where π΄ is in parentheses or brackets, means the probability of event π΄ occurring. This means that we want all the values that are inside circle π΄. As three plus one is equal to four, the probability of event π΄ is four out of 20. This can be simplified to one-fifth or the decimal 0.2. For the time being, we will leave it as four out of 20.

π of π΅ means the probability of event π΅ occurring. This would be the sum of all the values contained in circle π΅. One plus nine is equal to 10. So the probability of event π΅ occurring is 10 out of 20. Once again, this could be simplified to one-half or 0.5.

The probability of π΄ bar, also sometimes written as π΄ prime, is the probability of event π΄ not occurring. To calculate this, we firstly need to find the sum of all the numbers outside of circle π΄. These numbers are nine and seven. Nine plus seven is equal to 16. So the probability of π΄ bar or π΄ not occurring is 16 out of 20.

We notice here that the probability of π΄ and the probability of π΄ not occurring sum to one. Four out of 20 plus 16 out of 20 is equal to 20 out of 20, or one. This leads us to the formula the probability of π΄ bar, π΄ not occurring, is equal to one minus the probability of π΄. Knowing this formula will save us some time with some of our working.

The next notation we will look at is the intersection notation. This is denoted by a lowercase n. In terms of our Venn diagram, this is the probability of event π΄ and event π΅ occurring. We need to find the part of our diagram that is in circle π΄ and in circle π΅. As the word βintersectionβ suggests, this is the overlap of the two circles. As the number in this section is a one, the probability of π΄ and π΅ both occurring is one out of 20.

Next, we will look at the union. This is written as a lowercase u. In terms of our Venn diagram, this is the probability of event π΄ or event π΅ occurring. As a result, weβre looking for all the numbers in circle π΄ and numbers in circle π΅. We need to find the sum of three, one, and nine. This is equal to 13. The probability of π΄ or π΅, or π΄ union π΅, is 13 out of 20.

A common mistake here is to add the probability of π΄ to the probability of π΅, four out of 20 and 10 out of 20. This would give us 14 out of 20, which is not the correct answer. The reason for this is that we have counted the one in the intersection twice. This leads us to another key formula. 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΅. In this question, 13 out of 20 is equal to four out of 20 plus 10 out of 20 minus one out of 20.

The last three pieces of notation are less commonly used; however, theyβre still important when dealing with Venn diagrams. Letβs firstly consider the probability of π΄ intersection not π΅. This means that we want event π΄ to occur but not event π΅. There is a total of four inside circle π΄. The number one, however, is also in circle π΅. This means that the only value that is in circle π΄ and not in circle π΅ is three. The probability of π΄ and not π΅ is three out of 20. This could also be calculated using the formula the probability of π΄ minus the probability of π΄ intersection π΅. Four out of 20 minus one out of 20 is equal to three out of 20.

Next, we have the probability of not π΄ and not π΅. We want all the values that are not in circle π΄ and also not in circle π΅. There is only one such number, the seven that is outside both of the circles. The probability of not π΄ and not π΅, or not π΄ intersection not π΅, is seven out of 20. We notice here that this is one minus the union; it is everything other than the union of π΄ and π΅. One minus 13 over 20 is seven out of 20.

Our final bit of notation is the βgiven thatβ notation, the probability of π΄ given π΅ occurs. There is a formula for this, but we will begin by looking at the Venn diagram and circle π΅. Inside circle π΅, we have a total of 10. And we are told that this occurs. We want to calculate the probability of π΄ given that we are in this circle. The only part that is in circle π΄ and π΅ is the intersection, denoted by one. The probability of π΄ given π΅ is, therefore, one out of 10. Out of the 10 items in circle π΅, one of them is also in circle π΄.

This can also be calculated using the formula probability of π΄ intersection π΅ divided by the probability of π΅. In this case, we need to divide one out of 20 by 10 out of 20. This is equal to the correct answer of one out of 10 or one-tenth. An advantage of not canceling our original fractions means that we can cancel the 20s in this case.

We will now look at a combination of using some of our formulas and also Venn diagrams to solve some problems involving probability.

The figure shows a Venn diagram with the probabilities given for events π΄ and π΅. Work out the probability of π΄. Work out the probability of π΄ intersection π΅. Work out the probability of π΅ given π΄.

We recall that in any Venn diagram, the probabilities must sum to one. 0.3 plus 0.2 plus 0.4 plus 0.1 is equal to one. The first part of our question asks us to calculate the probability of π΄. This is equal to the sum of all the probabilities inside circle π΄. We need to add 0.3 and 0.2. This is equal to 0.5. The probability of event π΄ occurring is 0.5.

The second part of our question wants us to calculate the probability of π΄ and π΅. This is known as the intersection. It is the part of the Venn diagram where both π΄ and π΅ occur. The probability in the overlap of the two circles is 0.2. This means that the probability of π΄ intersection π΅ is 0.2.

The final part of our question wants us to calculate the probability of π΅ given that π΄ occurs. The notation in this question means given that. We recall that the probability of π΅ given π΄ is equal to the probability of π΅ intersection π΄ divided by the probability of π΄. We have already worked out both of these answers. Itβs important to note that the probability of π΅ intersection π΄ is the same as the probability of π΄ intersection π΅. The probability of π΅ given π΄ is, therefore, equal to 0.2 divided by 0.5. This is equal to two-fifths or 0.4. The probability of π΅ given π΄ is 0.4.

We could also have worked this out from our Venn diagram. As we want the probability of π΅ given that π΄ happens, we begin with the 0.5 inside circle π΄. Which of these probabilities is also in circle π΅? This is the 0.2. So once again we have 0.2 out of or divided by 0.5. Our three answers in this question are 0.5, 0.2, and 0.4. The 0.1 that was on the outside of both circles is the probability that neither event π΄ nor event π΅ occur.

We will now look at another question involving Venn diagrams with a more specific example.

Isabella has drawn this Venn diagram to record the result of randomly selecting a number between one and 12. What is the probability of selecting a number that is a factor of 20? What is the probability of selecting a number that is a factor of 20 and a multiple of three? What is the probability of selecting a number that is not a multiple of three?

We note that our Venn diagram is made up of two circles, firstly the numbers that are factors of 20. We will call this event π΄. Secondly, we have the circle that contains all the multiples of three. We will call this event π΅. The numbers seven, eight, and 11 that are outside of both circles are the numbers between one and 12 that are not factors of 20 and also not multiples of three.

The first part of our question wants us to calculate the probability of selecting a number that is a factor of 20. This is the probability of event π΄ occurring. We note that the numbers one, two, four, five, and 10 are all factors of 20. This means that five of the 12 numbers between one and 12 are factors of 20. The probability of selecting a number that is a factor of 20 is, therefore, equal to five-twelfths or five out of 12. As five and 12 have no common factor apart from one, this fraction is in its simplest form.

The second part of our question asks us to calculate the probability of selecting a number that is a factor of 20 and a multiple of three. This is the probability of event π΄ and event π΅ occurring, written as probability of π΄ intersection π΅. There are no numbers in the intersection of the two circles. This means that the probability of π΄ intersection π΅ or π΄ and π΅ is zero out of 12 or zero. The probability of selecting a number that is a factor of 20 and a multiple of three is zero. This is because there are no numbers between one and 12 that satisfy both of the criteria.

This leads us to an important fact when dealing with probability. If the intersection of two events is equal to zero, then the two events themselves are mutually exclusive. In this case, we could actually draw the two circles separately as there doesnβt need to be any intersection. There is no value that is a factor of 20 and a multiple of three.

The final part of our question wants us to calculate the probability of selecting a number that is not a multiple of three. This can be written as the probability of π΅ bar or the probability of π΅ prime. The probability of any event not occurring is equal to one minus the probability of that event occurring. We know that there are four numbers that are multiples of three: three, six, nine, and 12. This means that eight of the numbers between one and 12 will not be multiples of three. These are the numbers one, two, four, five, and 10 β that are factors of 20 β and the numbers seven, eight, and 11 β that are neither factors of 20 nor multiples of three.

The probability of selecting a number that is not a multiple of three is eight out of 12 or eight-twelfths. Both of these numbers have a factor of four, so we can divide the numerator and denominator by four. The fraction eight-twelfths simplifies to two-thirds. This means that the probability of selecting a number that is not a multiple of three in its simplest form is two-thirds.

We will now summarize the key points from this video. We found out in this video that Venn diagrams are made of two or more circles that sometimes overlap. If we have two circles that donβt overlap, the two events are said to be mutually exclusive. This occurs when the intersection or the probability of π΄ and π΅ is zero. Some key formulas that help us solve problems involving Venn diagrams are shown. These include the union, intersection, and given that notation. All of these can be proven by looking at the values or probabilities in a Venn diagram.