Lesson video

Hello there.

My name is Dr.

Rowlandson, and I'll be guiding you through this lesson.

Let's get started.

Welcome to today's lesson from the unit of conditional probability.

This lesson is called "Checking and securing, calculating probabilities from diagrams," and by the end of today's lesson, we will be able to calculate probabilities from probability trees and Venn diagrams. Here are some previous keywords that will be useful during today's lesson, so you might want to pause the video if you need to remind yourself what any of the words mean and press Play when you're ready to continue.

The lesson is broken into two learning cycles.

In the first part of lesson, we're going to be looking at Venn diagrams, and the second part of lesson, we'll look at probability trees.

Let's start with probability on Venn diagrams. Here we have a spinner with some numbers on, and we have Jacob, who says, "I can split the outcomes of this spinner into two events, prime numbers and even numbers, and then represent each group on a Venn diagram." Let's construct that Venn diagram together.

The space inside one circle, or circular shape, represents an event.

For example, it could represent the prime numbers.

The space inside the other circle, or circular shape, represents a second event, for example, the even numbers.

You can sometimes have a third circular shape in a Venn diagram, but not always.

They're usually enclosed inside a rectangle.

All the space inside the rectangle represents all possible outcomes in our sample space, and that's denoted by the Greek symbol, Xi, and that symbol represents the universal set, in other words, every single possible outcome that could happen.

So all the numbers from that spinner will be represented somewhere inside that rectangle.

Some of them will be inside the circle, which represents the prime numbers.

Some will be inside the circle represents the even numbers, and some will be in both of those circles.

Let's fill in some of it now.

The number, 2, is both a prime number and an even number.

So, because 2 satisfies both events, we place it in the centre of the Venn diagram, in the intersection where the region for prime and even overlap.

The number 3 is a prime number but not an even number, so it goes in the left-hand crescent of this Venn diagram, that is in the region which is for prime numbers but is not overlapped by the region for even numbers.

The number 4 is an even number but not a prime number, so we place it in the right-hand crescent of the Venn diagram, that is the region which is for even numbers but is not overlapped by the region for prime numbers.

Where should other numbers on this spinner go? We're going to fill it in together shortly, but perhaps pause the video and have a go at filling it in yourself now, and then press Play when you're ready to continue.

Let's fill the rest of this in now.

The numbers 5, 7, and 11 are all just prime numbers, so they can go in the left-hand crescent, which is just for prime numbers.

The numbers 6 and 20 are both just even numbers, so they can go in the right-hand crescent, which is just for even numbers.

And then we have some numbers that are neither prime or even, so they do need to go in this Venn diagram somewhere because they are outcomes and they do belong to a universal set, but they need to go in a region that is not inside either these circles.

So they would go on the outside of the circles but still inside the rectangle, which is for the sample space.

So let's check what we've learned.

We have a universal set that includes integers from 20 to 40, including the numbers 20 and 40, and then we have two events.

Event A includes all numbers with 3 as a digit.

Event B includes all odd numbers, and we have a Venn diagram that has not yet been completed.

One the regions that Venn diagram is labelled w, could you please give an example of a number that would go in that region? Pause video while you do it, and press Play when you're ready for some answers.

Well, there are multiple possible answers you can give, let's see if you've got any.

The numbers 23, 31, 33, 35, 37, and 39 are all numbers which are odd and have 3 as a digit.

We have a different region now labelled with x, could you please give an example of a number that would go in that region? Pause while you do it and press Play for answers.

Once again, there are multiple answers you can have here, and here they are: 30, 32, 34, 36, and 38, all have 3 as a digit and they are even, or in other words, they are not odd.

How about this region which is labelled y? Give an example of a number that would go in that region.

Pause the video while you do it and press Play when you're ready for answers.

Once again, there are multiple numbers that would go in that region.

Here they are: 21, 25, 27, and 29.

These are all numbers that do not have a 3 in them and they are odd.

Did you get any of those? And one last one, the region on the outside of the circles, but still inside the rectangle is labelled z.

Could you please give an example of a number that would go in that region? Pause video while you do it and press Play for answers.

This region is for numbers that do not have 3 as one of its digits and are not odd.

So they would be 20, 22, 24, 26, 28, and 40.

Here we have another Venn diagram.

Now, we don't necessarily know what the context is behind this Venn diagram.

That doesn't really matter because the Venn diagram has already been completed for us.

We can see includes lots of numbers, but there are some numbers missing that doesn't necessarily matter.

What we know here is all the numbers that are inside the rectangle belong to the universal set, and we can see how they're categorised.

Some are prime, some are even, some are both prime and even, and that is the number 2, and there are some numbers that are neither prime or even, and we can see those on the outside of the circle but still inside the rectangle.

Now we have Jacob who says, "Gosh, there are so many outcomes on this Venn diagram.

Do I need to show them all?" Well, we can have different types of Venn diagrams. This Venn diagram shows us the outcomes, but we could have a Venn diagram that shows us the frequencies of the outcomes instead, and that will look a little bit neater, but we would lose some information along the way.

Let's take a look together.

Rather than a Venn diagram showing each individual outcome of a trial, we could create a Venn diagram that shows the frequency of each event occurring.

So, on the left we have an outcome Venn diagram that shows each individual outcome.

We can see that in some cases, there are lots of them.

On the right, we have a frequency Venn diagram which looks very similar to the outcome Venn diagram, but the difference will be in how we complete it.

Rather than writing on the individual outcomes themselves, we'll write the frequency of outcomes which satisfy each event.

So for example, the left-hand crescent, which is for prime, but not even numbers, there are 14 of those which satisfy that event.

So we'd write the frequency of 14 there.

We haven't written down what those outcomes are, but we know there are 14 outcomes which satisfy that event.

In the intersection, the numbers which are prime and even, there is only one of those in that left-hand outcome Venn diagram, it's the number 2, but in our frequency Venn diagram, we just put the number 1 because that's the frequency.

In the right-hand crescent, which is for even numbers which are not prime, there are 12 outcomes which satisfy that event, so we put 12 there.

And in the outside region, for numbers which are not prime and not even, we can see that there are eight numbers that satisfy that event, so we put an 8 there.

The frequency Venn diagram on the right doesn't tell us what the individual outcomes are, which are prime, even, both, or neither.

What it does tell us though is how many outcomes there are, which are prime, even, both, or neither.

Let's check what we've learned.

On the left we have an outcome Venn diagram, on the right we have a frequency Venn diagram.

The frequency Venn diagram is made from the outcomes on the outcome Venn diagram.

You can see on the frequency Venn diagram, there are regions labelled x, y, z, and w.

Could you please find the values of those unknowns? Pause the video while you do that and press Play for answers.

Here are the answers.

There are seven outcomes Which are consonants, which have a line of symmetry, and they are B, F, G, J, L, N, and P.

Now we don't actually need to know what they are.

What we just need to know is what is the frequency, which is 7.

There are 3 of those outcomes which are consonants and have a line of symmetry, there are 5 which have lines of symmetry but are not consonants, and there are 0 outcomes which are neither.

Let's take a look at another example: A survey asks people whether they lived in Oakfield or elsewhere, and whether they shopped more often by online delivery or by going in-store, and the Venn diagram shows results.

For example, we can see that there are 160 people who live in Oakfield and do also shop in-store.

We can see that 90 people who live in Oakfield do not shop in-store, and we can see that 100 people who shop in-store are not from Oakfield.

And we can see that there are 50 people who are not from Oakfield and do not shop in-store, but were included in the survey.

A random person is selected from those surveyed to win a prize.

We can calculate probabilities for this by looking at the frequencies in the Venn diagram.

The most important thing here is to start by looking at the total frequency in the Venn diagram.

How many people were surveyed altogether? That would be 90 + 160 + 100 + 50, which is equal to 400.

So 400 people took part in a survey.

Now, we know that, we know what the denominator will be in our fractions for probabilities.

So, let's work out some probabilities.

Calculating the probability for the person who's from Oakfield means identifying the frequency of everyone from Oakfield.

That includes people who shop in-store and don't shop in-store.

In other words, everyone that's inside this left-hand circle.

We can do 90 + 160 to get that there are 250 people in the survey from Oakfield.

So the probability of randomly choosing a person from this survey from Oakfield will be 250/400.

To find the probability that the person chosen shops in-store but is not from Oakfield, we would need to look at the frequency in this right-hand crescent here.

This is everyone who shops in-store but is not from Oakfield.

That means we'd get 100/400.

So let's check what we've learned.

Here, we have another Venn diagram where the members of Oakfield sport centre either have a weekday pass or a weekend pass, they can't have both.

And some members play football and some don't.

The Venn diagram shows us the frequencies of those events.

A member is chosen at random to be surveyed.

Could you please find some probabilities about this person who is randomly chosen? Those probabilities are labelled from A to D.

Pause video while you work these out, and press Play when you're ready for answers.

Okay, let's go through some answers.

The total frequency in this Venn diagram is 880, so there must be 880 members of Oakfield sports centre, and that will be the denominator of our fractions.

The probability that the person visits on a weekday will be 550/880.

Now, we can't see the number 550 in our Venn diagram, but we can get it by adding together 204 and 346, that is a total number of people who visit on a weekday, including whether they play football or not.

For part B, the probability that the person doesn't play football and visits on a weekday will be 346/880, and the probability that the person doesn't play football and visits on the weekend will be 184/880.

So far, we've looked at Venn diagrams that show individual outcomes and Venn diagrams that show frequencies of outcomes, and we can see on the screen here a frequency Venn diagram, that is one that shows the frequencies of outcomes.

Let's now consider something else that a Venn diagram could show you.

If a Venn diagram is being used to calculate probabilities, then it is also possible to construct a probability Venn diagram.

A probability Venn diagram has the probability of each event written instead of the frequencies.

It looks something a bit like this.

Once again, it looks the same as a Venn diagram on the left, but the way we fill it out will be different.

Let's fill this one in together.

Imagine choosing someone at random from the Venn diagram on the left.

The probability that they are from Oakfield and in-store would be 160/400.

The probability that they'll be from Oakfield but do not shop in-store will be 90/400.

The probability that they shop in-store but are not from Oakfield would be 100/400, and the probability that they are neither from Oakfield and don't shop in-store would be 50/400.

So let's check what we've learned.

1,000 people were asked whether they owned a driver's licence or a bus pass, and it's possible to own both.

A random person is chosen, and the Venn diagram on the bottom left of the screen shows a probability that that person satisfies each of the different four events there.

Could you please find the probabilities of three events shown on the screens labelled A, B, and C.

Pause the video while you do it, and press Play when you're ready for answers.

Okay, the answer to part A will be 338/1,000.

Part B will be 87/1,000.

And part C will be 352/1,000.

Now, 352 cannot be seen in the Venn diagram.

We can get it by adding together the probability of 87/1,000 and 265/1,000.

That gives the total probability of choosing someone with a bus pass.

Okay, it's over to you now for task A.

This task includes four questions and here is question one.

Pause the video while you do it, and press Play for question two.

And here is question two.

Pause the video while do this, and press Play for questions three and four.

And here are questions three and four.

Pause while you do this and press Play for some answers.

Here are the answers to question one.

Pause while you check these against your own and press Play for more answers.

Here are the answers to question two.

Pause while you check this and press play when you're ready for more answers.

Here he answers to question three, pause while you check and press Play for more.

And here he answers to question four.

Pause while you check this, and press Play for the next part of today's lesson.

Well done so far, let's now move on to the next part of this lesson, which is looking at two-stage trials on probability trees.

Here we have the spinner with some numbers on and the spinner is going to be spun once.

We can show the probability of each outcome occurring on a one-layer probability tree, it looks something a bit like this.

Each branch represents an outcome for this particular trial.

One of them shows number 1, because 1 is one of the outcomes on the spinner, and so on with 3, 6, and 9.

We label the probability of each outcome occurring on their branches.

So for example, the probability that the spinner lands on 1 is 2/10, because two of the sectors of the spinner show 1, and there are 10 sectors altogether.

And we can apply the same reason for the probabilities of each of the other outcomes.

It is also possible to construct a probability tree that shows the probability of events instead.

This reduces the number of branches needed.

For example, rather than each individual outcome, we might think about events such as getting a square number and not getting a square number.

Each branch represents an event, and probability can be written on in a similar way.

The probability of getting a square number will be 3/10 because there are three sectors on that spinner that have a square number on them, and the probability of not getting a square number will be 7/10 because there are seven sectors on that spinner but do not have a square number on.

If the probability tree has been constructed correctly, each group of branches that meet at a point should sum to 1 whole.

For the probability tree on the left, we can see that 2/10 + 4/10 + 3/10 + 1/10 is 1 whole.

So all four of those probabilities sum to 1.

With the probability tree on the right, we can see that 3/10 + 7/10 is also equal to 1 whole, so those probabilities sum to 1.

Let's check what we've learned.

Here, we have a probability tree that shows the probability of one of two events occurring.

We can see that probability is 17/30 for event A.

Could you please find the value of x, which is the probability for event B? Pause the video while do that, and press Play for an answer.

Well, those two probabilities should sum to 1 whole.

So the value of x is 13/30, which you can get from doing 1 - 17/30.

Here we have a spinner with some numbers on and a probability tree.

You can see that there are some unknowns in that probability tree, labels from A to D.

Some of the unknowns are for probabilities and some are for the actual outcomes.

Could you please find the values of A to D? Pause while you do it and press Play for an answer.

Well, here they are: A is 2/6, that's a probability.

B was one of the outcomes, which is 10, C was 1/6, that's a probability, and D was an outcome, which is 6.

Here we have a spinner and coin, and Jun says, "What if I wanted to look at two stages of a trial, like spinning the spinner and flipping a fair coin at the same time?" He says, "Can I still represent this on a probability tree?" Hmm, let's take a look at this together.

We could start by drawing something a bit like this.

This probability tree here shows the outcomes of spinning the spinner could either get A or B.

The probability of it lands on A is 5/8, the probability it lands on B is 3/8.

So, this layer of branches represents one stage of the trial, and that is spinning the spinner.

So what about flipping the coin? Well, we could create another layer of branches for that, but each branch will come from one of the outcomes of this stage here.

So for example, if the spinner lands on A, and then we flip the coin, one of two things could happen, we get heads or tails.

The probability of getting heads is 1/2, and the probability of getting tails is 1/2.

What this set of branches shows is the possible outcomes of the coin flip if the spinner lands on A.

So with that in mind, how could we show the outcomes of the coin flip if the spinner lands on B? Perhaps pause the video and think about this, and press Play when you're ready to continue.

Well, we could show it here.

This pair of branches represents the possible outcomes of the coin flip if the spinner lands on B.

The spinner lands on B, we could get heads on the coin or we can get tails on the coin, and the probability of each of those is 1/2.

This entire layer of branches represents the other stage of the trial, and that is flipping the coin.

Every group of probabilities that meet at a point are exhaustive and mutually exclusive, so they sum to 1.

For example, these two probabilities sum to 1, these two probabilities sum to 1, and so do these.

These two connected branches represent the outcomes that the spinner lands on B and the coin lands on heads.

And this is the sample space of the entire two-stage trial, spinning the spinner and flipping the coin.

We can see that sample space includes A in heads, A in tails, B in heads, and B in tails.

So, let's check what we've learned.

We've got two spinners and we've got a probability tree with unknowns on it labelled a, b, x, and y.

Could you please find the values of those unknowns? Pause while you do it, and press Play for answers.

Here are the answers.

With spinner A, the probability of getting a prime number is 6/10, the probability of not getting a prime is 4/10.

And with spinner B, the probability of getting a prime is 2/6, and not getting a prime is 4/6.

Here we have two connected branches which are now highlighted.

Could you please write a sentence explaining what these two marked branches represent? Pause while you do it and press Play for an answer.

Here's the answer.

It's the events that spinner A lands on a non-prime and B lands on a prime.

Here's another probability tree.

You can see it has two probabilities labelled on it, and you have two more probabilities labelled with unknowns, a and b.

Could you please find the values of A and B? Pause while you do it and press Play for answers.

The answers are 3/8 and 0.

34.

You get those answers by subtracting the other probability from 1 whole.

So here we have a probability tree for a two-stage trial.

One stage involves spinning a spinner and the event involves flipping a coin.

We can see it's now all completed with the outcomes listed and also the probability labelled on the branches.

And we have Jun.

Jun says, "Probability trees sure are a nice way of listing all outcomes, but isn't that what an outcome tree is for?" Hmm, he says, "What's actually the purpose of writing down the probabilities on the branches?" Let's take a look.

We know the probabilities of each outcome in the sample space for stage one of our trial, they are 5/8 and 3/8.

We know the probabilities of each outcome in the sample space for stage two of our trials, they are 1/2 and 1/2.

We can also calculate the probability of each outcome in the whole trial.

In other words, once completed both stages.

To find the probability of both the spinner landing on A and the coin landing on heads, we can multiply the probabilities on the branches of the outcomes at each stage of the trial.

In this case, we would do 5/8, which is a probability that the spinner lands on A, multiplied by 1/2, which is the probability that the coin lands on heads, and we'd get 5/16.

And that is a probability of both the spinner landing on A and the coin landing on heads.

Jun says, "So the probability of landing on B and landing on tails is 3/8 times 1/2, which is 3/16." Yes, that's correct.

And we can do the same for the other outcomes.

The probability that the spinner lands on A and the coin lands on tails would be 5/8 multiplied by 1/2, which is 5/16.

The probability that the spinner lands on B and the coin lands on heads is 3/8 multiplied by 1/2 which is 3/16.

Now, you may sometimes be able to simplify these fractions, but you don't necessarily have to, and it can be helpful to have the same denominator in all of your fractions because the probabilities of all outcomes in a sample space must sum to 1, and this can be a good way to check that all the calculations in your probability tree are correct.

We can do it by adding together our four answers there.

5/16 + 5/16 + 3/16 + 3/16 is equal to 16/16.

And we know that 16/16 is equal to 1 whole, therefore the sample space is likely to be correct.

So let's check what we've learned.

Here, we've got a probability tree for a two-stage trial.

One stage is spinning spinner 1 and the other is spinning spinner 2.

Each spinner is spun once.

The probability tree models this two-stage trial, and we can see we've got the probabilities labelled on the branches.

And the box in the right-hand side of this probability tree will be for writing the outcomes for the overall trial such as getting X and getting P, and writing down the probabilities, for example, 0.

09.

Could you please write down what outcome would go in place of a in this sample space? Pause the video while do that and press Play for an answer.

The answer is Y and P, that is the event of getting a Y on spinner 1 and a P on spinner 2.

What is the probability that will go in place of B for the sample space? Pause while you write down the probability and plus Play for an answer.

The answer is 0.

36.

You get that by multiplying the probability for Y by the probability for P, 0.

8 multiplied by 0.

45 is 0.

36.

So could you please complete the remaining sections, part C and part D? Write down what the overall outcome is and its probability in each case.

Pause while you do that and press Play for answers.

Box C represents the outcome of getting X in spinner 1 and Q in spinner 2, and its probability is 0.

11, that's 0.

2 multiplied by 0.

55.

Box D represents the outcome of getting a Y on spinner 1 and a Q on spinner 2, and its probability is 0.

44.

That is from doing 0.

8 multiplied by 0.

55.

What should all four of these decimals sum to? Pause while you write it down, and press Play for an answer.

They should all sum to 1, and if you add them up, they do.

Okay, it's over to you now for task B.

This task contains four questions, and here is question one.

Pause the video while you do it and press play for question two.

And here is question two.

Pause while you do this, and press play for question three.

Here is question three.

Pause while you do it and press Play for question four.

And finally here is question four.

Pause while you do this and press Play for some answers.

Here are the answers to question one.

Pause while you check this against your own and press Play for more answers.

Here are the answers to question two.

Pause while you check and press Play to continue.

Here are the answers to question three.

Pause while you check, and press Play to continue.

And question four, here are the answers to the "Dragon's breath" part of this question.

Pause while you check against your own, and press Play for the rest of the answers for this question.

And then here the answers for the "Unicorn's strike" part of this question.

Pause while you check this against your own and press Play for conclusion for today's lesson.

Fantastic work today.

Now, let's summarise what we've learned.

Two pairs of events from a trial can be shown on an outcome or a frequency Venn diagram.

Probabilities of these events can be calculated and shown on a probability Venn diagram.

Outcomes or events for each stage of a trial can be shown as a layer of branches on a probability tree.

Those outcomes or events are mutually exclusive and exhaustive.

The probability of a combination of outcomes can be calculated by multiplying the probabilities of one outcome on each layer of branches.

Well done today, have a great day.