# Lesson video

In progress...

Hi.

I'm Mrs. Dennett.

And in this lesson, we're going to be finding probabilities from Venn diagrams containing two sets.

We'll start by recapping some basic set notation and then move on to some more complex notation.

In this example, 50 children are placed in groups A, B, both, or neither.

The information is given in the Venn diagram.

We need to work out some probabilities, firstly, the probability that a child selected at random is in set A.

A child that is in set A, or group A, can also be in set B, or group B, so we need to include the overlap.

We need all the children in the circle labelled A, so we add up 14 and 18 to give us 32.

There are 50 children in total, so this gives us 32 out of 50, which simplifies to 16 out of 25.

Now we need to work out the probability that a child is not in group B.

This is the complement of B.

So what we're going to do is we're going to ignore anything that's in circle B.

There we go.

So we've got 14 and three, which is 17 children, so this gives us 17 out of 50.

We now want to work out the probability that a child is in group A or group B.

This includes the overlap.

We say the probability that the child is in group A union group B.

So we use the word union to represent this particular probability.

So the probability of A union B is 47 out of 50.

Now we have to find the probability the child is in group A and group B.

We look at the overlap, or the intersect, of A and B.

We call this the probability of A intersect B.

There are 18 children here, so the probability of A and B is 18 out of 50, which simplifies to nine out of 25.

Here is a question for you to try.

Pause the video to complete the task, and restart when you are finished.

Nothing too complicated here, but if you are struggling, just go back and watch the previous example again before we move on to some further set notation.

Now let's take a look at some further set notation.

We have the same Venn diagram as in the first example.

We want to find the probability that a child is not in A and B.

The easiest way to think of this is to shade anything in the complement of A, anything not in A, and then shade anything in B.

You can see this section here, part of circle B which has been double-shaded, So we want the intersection, or overlap, of the part that has been shaded twice.

So we need this section.

There are 15 children here, so the probability that they are not in A and in B is 15 out of 50.

For part b, we want the probability A union not B, so the probability of A or not B.

So we shade in the probability of getting A, so that's all of circle A, and then shade in the complement of B, so everything not in circle B.

Now remember that this time we want the union.

So notice that we shaded in these sections.

We don't need the white section.

We don't want the section with 15 in it.

So we want 14 add 18 add three, which gives us a probability of 35 out of 50.

Finally, we want the probability that a child is not in A or not in B, not A union not B.

That is everything outside A or everything outside B, which you can see here.

This is the same as the complement of A union B.

So we want everything outside of the circles.

So we have three children outside, so we get the probability that a child is not in A union not B is three out of 50.

Here is a question for you to try.

Pause the video to complete the task, and restart when you are finished.

For part a, work out the complement of B.

This is anything not in circle B.

For the next question, shade in the complement of A first in one colour, and then shade in circle B in another colour.

We want the intersect of these sections, so we look for the sections of being shaded twice.

This is the outer part of the circle B, not including the overlap.

So this will be the following numbers.

Four, six, eight, 10, 12, and 14.

This is six out of 15 members of the universal set.

So the probability is 6/15.

For parts c and d, shade in the respective sections.

We're asked for the union of these in both questions.

A or not B is nine out of 15.

And for part d, not A or not B is three out of 15.

So we're now going to look at some conditional probabilities, and we're going to use the same Venn diagram as in my previous examples.

This time we want the following probabilities.

So firstly, we want the probability of A intersect B, given that a child is in B.

So we have a line here which represents given that the child is in B.

This tells us that the child selected is already in set B, and we can see there are 18 plus 15 children altogether in set B.

That's 33 children in group B.

Now we'll look for A intersect B.

This is the overlap.

There are 18 children here.

So the probability of A and B, given that they're already in B, is 18 out of 33.

For part i, we want a child in group A and group B, given that we're told they're already in group A.

So we can see that there are 32 children in group A and 18 in the overlap.

That's A and B.

So the probability is 18 out of 32, which simplifies to nine out of 60.

Here is a question for you to try.

Pause the video to complete the task, and restart when you are finished.

Let's look more closely at part c and d, as these are the conditional probabilities.

For part c, we want the intersection, which is seven students out of the 11 students in set D.

For part d, we want the intersection again, but this time given that the student is selected from set C.

There are 14 students in circle C who have cats.

So we get seven out of 14 for the probability that the student has a cat and a dog, given that they already have a cat.

Here is a final question for you to try.

Pause the video to complete the task, and restart when you're finished.