Lesson video

Hello there and welcome to today's lesson.

My name is Dr.

Rowlandson and I'll be guiding you through it.

Let's get started.

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

This lesson is called: Comparing Multiple Representations to Calculate Conditional Probabilities.

And by the end of today's lesson, we will be able to compare and contrast the usefulness of different representations when calculating conditional probabilities.

Here are some previous keywords that will be useful during today's lesson.

So you may want to pause the video if you need to remind yourself what any of these words mean, and then press play when you're ready to continue.

The lesson is broken into two learning cycles.

We're gonna start by looking at two-stage trials on probability diagrams. The combined outcomes and probabilities of a two-stage trial can be represented on different diagrams. Let's take this scenario as an example.

We have two spinners: spinner A and spinner B.

On spinner A, there are two possible outcomes.

It could either land on J or it can land on K, and spinner B could either land on X or Y.

And in this particular case, we define a trial as being both of these spinners being spun at once.

We can represent the outcomes of this in multiple different ways.

For example, we could represent it on a table that looks something a bit like this.

The columns of the table shows the outcomes of spinner A.

The row shows the outcomes of spinner B.

And the cells on the table show the sample space for this two-stage trial.

They look something like this.

These are all the possible combined outcomes when these two spinners are spun.

We could also look at this as a frequency tree.

So far, the frequency tree shows us that there are two possible outcomes to spinner A: can land an either J or K.

And for each of those, there are two possible outcomes to spinner B: could land an either X or Y.

If we compare this to the two-way table above it, we can see that there are 24 outcomes in the sample space altogether.

Of those outcomes, 9 of them contain a letter J and 15 contain a letter K.

And of the outcomes that contain a letter J, 6 also have X and 3 have Y.

And of the 15 outcomes that contain K, 10 of them have X and 5 of them have Y.

We could also represent this on a probability tree.

So far, this layer of branches shows us the possible outcomes of spinner A: could either land on J or K.

We write the probability of these outcomes on the branches.

So probability that the spinner lands in A is 3/8.

The probability that lands in K is 5/8.

And for each of those outcomes, there are two possible outcomes for spinner B: X and Y.

And we can write the probabilities on those branches as well.

And when it comes to think about the outcomes of the entire trial, we can get the probabilities by multiplying the probability of each individual outcome on the relevant branches.

So for example, the probability that the outcome is J on one spinner and X in the other spinner is calculated by doing 3/8 multiplied by 2/3 which is 6/24.

And we can see that for the other possible outcomes as well.

So as we can see here, we can represent a scenario using multiple different representations and can draw similar conclusions from all of them.

For example, each of these representations show that there are 6 outcomes that contain J and X from a total of 24 outcomes.

In the two-way table, we can see that there are 6 cells that have letters JX in out of 24 possible outcomes altogether.

In the frequency tree, we can see on the far left we start with 24 outcomes altogether.

And by the time we sort them into ones of J and then ones of X, we can see that there are 6 of them with J and X.

And in the probability tree, we can see that in the probability for J and X occurring.

We can also work out probabilities from each of these representations.

For example, the probability that one spinner lands on Y is 8/24.

That's because there are 8 outcomes out of a total of 24 outcomes altogether that have a Y in them.

We can see in the probability table, it is the bottom row of that two-way table.

They all contain a letter Y out 24 outcomes altogether.

In the frequency tree, we can get the 8 by adding together the two frequencies that contain the letter Y.

And in the probability tree, we can get 8/24 by adding together the two probabilities that have Y as one of the outcomes.

So let's check what we've learned.

A spinner is spun once and a fair coin is flipped once.

The outcomes on the spinner are the numbers 4, 9, 11, and the outcomes of the coin is it could either land showing +10 or land showing -10.

The outcomes are added together and you can see the outcomes on the outcome table.

Here is a frequency tree for the same two-stage trial.

Could you please work out what is the value of A in that frequency tree? Pause the video while you work it out and press play when you're ready for an answer.

A is equal to 20 because there are 20 combined outcomes in the outcome table.

So how many of these 20 outcomes are made from the spinner landing on a 4, a 9, or 11? Or in other words, what are the values of b, c, and d in the frequency tree? Pause the video while you work them out and press play When you're ready for answers, The answers are 6, 10, and 4.

Now the frequency tree is fully completed.

Please could you use it to find the probability of a negative sum? In other words, what's a probability that the final result of this trial is a negative number? Pause the video While you do that and press play when you're ready to see the answer.

Well, there are two combinations of outcomes that lead to a negative number as the result, getting a 4 and -10 or getting a 9 and -10.

And we can see the frequency of outcomes for each of these are 3 and 5.

So if we add those together, we get 8 out of a possible 20 outcomes.

We can also see that on the outcome table, there are 8 cells that contain a negative number out of 20 cells altogether inside the outcome table.

Let's now think about conditional probabilities.

Each of these diagrams can also be used to identify conditional probabilities.

For example, if we want to work out the probability that one spinner shows Y given that we already know the other spinner has shown J, we can see that in each of these three representations.

Firstly, we don't wanna look at the entire sample space in this scenario because we know J has already happened.

So we don't need to look at all the outcomes that don't include J.

We only wanna look at the outcomes that include J and there are 9 outcomes.

In the outcome table, we can see those here.

These are 9 outcomes all include a letter J.

In the frequency tree, we can see it here at the end of the branch for J.

And in the probability tree, we can see it in these two possible combined outcomes.

We can get J and X or we can get J and Y.

So there are 9 outcomes that include letter J.

Of those, 3 of them also include the letter Y.

In the outcome table, we can see it here.

So the probability of Y given J will be 3/9.

In the frequency tree, we can see it here.

Out of the 9 occasions where J happens, 3 of them result in Y.

So the probability would be 3/9 there.

In the probability tree, it presents itself in a slightly different way.

We know that J has happened.

So if we look at J on the probability tree and look at what the probability that Y happens next, it'll be 1/3 and that is equivalent to 3/9.

So in whichever way we look at it, the probability of Y giving J will be 3/9 or 1/3.

So let's check what we've learned.

Here you've got an outcome table for a trial involves spinning two spinners.

Spinner 1 has outcomes A, B, and C and spinner 2 has outcomes A and B.

Now you might be looking at a table and thinking, why has it got 0s on 1s as the outcomes rather than As, Bs, and Cs? That's because the way the events are defined.

If both outcomes from the spinner are the same, then the result is 1.

If both outcomes are different, then the result is 0.

So use that information.

Could you please find the three probabilities that you can see on the screen for A, B, and C? Pause the video while you do it and press play when you're ready for answers.

Okay, let's take a look at the answers.

For part a, the probability that the final result is 1, given that spinner 1 lands on B, you can get that by looking at this part of the outcome table and the probability will be 4/16 or 1/4.

For part B, the probability that the final result is 0, given that spinner 2 lands on A, that would be 18/27 or 2/3.

You can get that by looking at this part of the outcome table.

And for part c, the probability that spinner 2 lands on A, given that spinner 1 has landed on A, that would be 9/12 or 3/4, which you can see from this part of the outcome table here.

Now it's worth noting that because spinner 1 and spinner 2 are independent of each other, this means this probability is the same as a probability that spinner 2 lands on A in general.

Let's now start transferring some this information onto a probability tree.

You can see the probability tree has been partially completed with three probabilities that are known labelled x, y, and z.

Could you please use the outcome table to work out the values of x, y, and z? Pause while you do it and press play for answers.

Okay, let's take a look.

We can See the first layer of branches must represent spinner 1.

It's got three outcomes and the second layer of branches must represent spinner two because they've got two outcomes.

Therefore, extra probability that spinner 2 shows B when we know that spinner A has already shown one.

We can see that there are 12 outcomes where spinner 1 shows A.

And of those, 3 of them have B for spinner 2.

So probability would be 3/12 or 1/4.

What is the probability of that spinner 2 lands on A given that spinner 1 has landed on C? We can see from the table that there are 8 outcomes where spinner 1 lands on C.

And of those, there are 6 of them where spinner 2 lands on A.

So the probability would be 6/8 or 3/4.

And then z, that's the probability that both of the spinners land on B, and that's out of the entire sample space.

We can see that there are four outcomes that those spinners land on B out of 36 outcomes altogether, so that'd be 4/36 or 1/9.

We can now see four probabilities on a tree that are labelled A to D.

Which of them is a probability for the events shown in the brackets on the bottom left of the screen? That's the probability that spinner 2 lands on A given that spinner 1 has landed on B.

Pause while you work it out and press play when you're ready for an answer.

The answer is b.

Could you now please use either of these two representations to find the probability, but the final result is 0 given that either spinner has landed on A? Pause the video while you do that and press play when you're ready for answers.

The answer is 21/30.

What I want you to do now is consider which of those two diagrams did you find most useful for getting the results and why? There's no right or wrong answer for this, but perhaps just think about it and if there's anyone you can discuss it with, discuss it with them.

Pause while you do that and press play when you're ready to continue.

Well, if we look at the outcome table, we can see that there are 30 outcomes where one of the spinner has landed on A.

And of those, we can see that 21 of them have a result of 0, so looking at the probability of 21 out of 30 that way.

When it comes to the probability tree, we can work this probability out but it's not quite as easy.

It takes a little bit more reasoning to get our answer.

We can start by thinking about all the outcomes that include at least one A 'cause those are the ones that we wanna focus on.

And if we add together those four probabilities, we'd get 30 out of 36.

Now the numerator of that fraction tells us how many outcomes in this scenario result in at least one A.

Those are the only outcomes we really care about when it comes to writing this conditional probability.

So that would be the denominator of our fraction.

And then out of those 30 outcomes, we can see that there are 21 of them that result in 0.

And that's because those are the ones where the same letter has not appeared twice.

Like I said, same answer but it's a bit more complicated using that particular representation.

Let's now look at a different scenario.

It is possible to show the outcomes and probabilities on both an outcome table and probability tree for a two-stage trial where two objects are removed randomly from a group and not replaced.

For example, here we have a jar that contains 6 marbles.

4 are yellow and 2 are blue.

2 marbles are chosen randomly from this jar at the same time.

Now when it says at the same time, practically the way that works is you choose one marble, and then immediately afterwards, choose another.

So you can still think of this as two stages where one happens before the other and you need to remember that when the outcome of stage one happens, when the first marble is picked, it's not replaced back into the jar.

So you have fewer marbles to choose from for your second pick.

Let's take a look at what this would look like in a probability tree.

For the first marble that's chosen, there are two possible outcomes, yellow or blue, and the probabilities are 4/6 and 2/6.

If the first marble was a yellow one, there are two possibilities for the second marble, yellow and blue, but the probabilities are different.

There are only 5 marbles remaining in the jar for the second pick which is why the denominator is 5.

There are only 3 yellow marbles left in the jar because one has been taken out, but there are still 2 blue marbles left in the jar.

If the first marble was a blue one, there'll be still two possibilities for the second marble, yellow and blue, and the probabilities will be different again.

The denominator's 5 because there are 5 marbles left, but there are 4 of 'em which are yellow and only 1 now which is blue.

And then once we've got this, we can work out the probabilities for the combined outcomes of the two picks, and that's what we can see on the screen here.

Now this can also be represented in a two-way, outcome table a bit like this.

The column show us the choice of the first marble and the row show us the choice of the second marble.

The thing we need to bear in mind when we fill out this table is that once one marble is chosen, it cannot be chosen again.

Yes, we can have the same colour twice, but we can't have the same individual marble twice, so we should black out that particular outcome 'cause it can't happen.

And the same with all situations where the same individual marble is chosen twice.

They cannot happen, so we black those out on the table.

Once we do that, we can see that there are 30 outcomes altogether.

We can see it in the two-way table and we can see it in the denominators of the probability tree.

Now we've done that, we can start thinking about conditional probabilities and how each representation might help us.

For example, let's say we want to find the probability but the second marble is yellow given that the first marble is blue, and maybe wanna use the probability tree to help us do that.

Well, we can go to the situation where the first marble is blue, and then look at what is the probability that the second marble is yellow in that case, and it would be 4/5.

Let's now find a conditional probability using the outcome table.

How about this one here? The probability that we get the same colour marble, given that we know at least one of the marbles is yellow.

Well, we can look at all the outcomes where at least one marble is yellow and there are 28 of them.

And of those, there are 12 of them were they are the same colour, they are all the ones which are yellow-yellow.

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

On the left, you have an outcome table for drawing a first marble and a second marble from a back.

And you also have a probability tree that is partially completed on the right.

There are three unknown probabilities labelled x, y, and z.

Please could you use the outcome table to find the values of x, y, and z? Pause while you do it and press play for answers.

Okay, let's go through some answers.

X is the probability that the first marble chosen is a yellow marble which is 4/7.

Y is the probability that the second marble is blue, given that the first marble is blue, that would be 2/6.

You can also see it as being 6/18.

And z is a probability that we get a blue marble and a yellow marble.

That would be 12/42.

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

This task has three questions and here is question 1.

Pause while you do it and press play for question 2.

And here is question 2.

Pause while you do this and press play for question 3.

Here is question 3.

Pause while you do it and press play to see some answers.

Okay, here are the answers to question 1.

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

Here are the answers to question 2.

Pause while you check these and press play for more answers.

And here are the answers to question 3.

Pause while you check these 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 where we're going to look at conditional events on probability diagrams. We can take the outcomes from a trial or experiment and group them into sets of events.

And then, we can represent the frequency or probability of these events on different representations.

For example, we can represent them on a Venn diagram that can be used either for frequencies or probabilities.

We could represent them in some kind of tree, whether that be a frequency tree or a probability tree, or we can represent them in a two-way table.

Let's take a look at an example of this.

A population of 1200 birds are tested for a virus called avian pox.

An investigation is conducted to see whether birds who are tested positive for the virus were more likely to show scarring on them than birds who don't have the virus.

We can display the results from this investigation on a frequency Venn diagram.

It would look something a bit like this.

We have two events.

One is that the bird has the virus and the other is that the bird has scarring.

And these two events can both happen at the same time which is why those regions overlap.

And once the data from this investigation was collected, it was filled in on the Venn diagram a bit like this.

These same results can also be represented on a frequency tree such as the one you can see on the screen here.

This frequency tree has two layers of branches.

The first sort the birds out into those who have the virus and those who don't have the virus.

And for each of those groups, they are then sort out in the second layer into those who have scarring and those who do not have scarring.

Let's now look at how we can transfer the information from the Venn diagram onto the frequency tree.

You may notice that the frequency tree has 7 empty cells in it, but there are only 4 numbers in the Venn diagram.

So we're going to need to do some calculations along the way as well.

But let's start off with the numbers that we know.

We know that there are 189 birds who have the virus and also have scarring.

There are 261 birds who have the virus but do not have scarring.

There are 135 birds who do not have the virus but do have scarring, and there are 615 birds who do not have the virus and do not have scarring.

Now we've got those cells filled, we can start thinking about the cells to the left of them and what they would be.

And we can get those by using addition and it'll be 450 and 750.

And then, we could work out the remaining cell by adding together those last two numbers to get 1200, which is also the population of birds altogether.

Now we have these two representations, we can start thinking about probabilities.

For example, from this population of birds, we know that the probability that a bird has the virus is 450 over 1200.

In the Venn diagram, we get that by adding together the two frequencies for birds that do have the virus.

That'll be 261 plus 189, and that will be out of 1200 because that's the population.

In the frequency tree, that addition has already been done for us and we can see that straight away as 450.

We can also work out the probability that a bird has scarring and that would be 324 over 1200.

Now in this case, no matter which representation you use, you'll have to perform an addition of 189 and 135.

And that's because with the frequency tree, we've separated the birds out first according to whether or not they have the virus.

So those who have scarring are separated into two groups.

We can also use both representations to calculate conditional probabilities.

For example, if we wanted to work out the probability that a bird has scarring given that we know it has the virus, then we could look at where we can find those numbers in each representation.

In the Venn diagram, we'd look only at the bird who have the virus, and that would be the sum of 261 and 189, and that will be our denominator of 450.

And out of those, 189 of them have scarring.

So the probability will be 189/450.

In the frequency tree, we can see those numbers more explicitly.

We can see that out of the 450 birds that have the virus, 189 have scarring.

And if we wanted to work out the probability that a bird has scarring given that it does not have the virus, we could do it in a very similar way.

With a Venn diagram, we need to work out the number of birds that do not have the virus.

By doing 135 plus 615 to get a denominator of 750.

But that is already done for us in the frequency tree.

We can see that already.

In either case, we can see 135 of those do have scarring.

Now that we've calculated these probabilities, let's go back to what it was we're investigating in the first place.

The investigation was conducted to see whether birds who tested positive for the virus were more likely to show scarring than birds who do not have the virus.

Well, we have two probabilities now that help us see that.

We have the probability that a bird has scarring with the virus and we have a probability that a bird has scarring without the virus.

So we want to compare these probabilities to see which one's more likely to happen.

Now, we could use a common denominator but it might be easier to use decimals instead.

And now we have these, we can see that a bird is more likely to have scarring with the virus than have scarring without the virus.

So we can conclude that these two diagrams show the probability of a bird having a scar given they are infected by a virus is 42%, and that is greater than the probability of a bird having a scar and not having the virus which is 18%.

So let's check what we've learned.

With a slightly different scenario, but we can use a similar process.

An investigation into 1640 rhubarb plants aims to see if the application of a special fertiliser increases the chance of a plant yielding rhubarb.

Let's call the event of the special fertiliser being used F and the event of a plant yielding rhubarb R.

Now it's represented in a Venn diagram and a two-way frequency table, but what you can see is that they are both partially completed and there are some unknowns in there.

Please could you find the values of a to e? Pause the video while you do it and press play for some answers.

Okay, let's go through some answers.

a would be 680 which you can read from the table and write into your Venn diagram.

c would be 390 which you can work out by doing 1070 subtract 680.

b would be the same as c.

It'll be 390.

d would be 300 which you can see from the Venn diagram.

And you can write into the table.

e would be 270.

You get that from doing 1640 subtract each of 300, 390, and 680.

Okay, now we have all those frequencies, could you please work out these two probabilities? One is a probability that a plant yields rhubarb given that the fertiliser is used and the other is a probability that the plant yields rhubarb given that the fertiliser is not used.

And please give your answers as decimals to two decimal places.

Pause while you do it and press play for some answers.

Okay, let's go through some answers.

The probability that a plant yields rhubarb given that fertiliser is used would be 0.

69.

The probability that a plant yields rhubarb given that fertiliser is not used would be 0.

59.

So now we can see the probability that the plant yields rhubarb is increased by the use of fertiliser.

Let's take a look at one more scenario together.

A spinner is spun 2500 times and the outcomes are categorised into three events.

The outcome could be a square number or not a square number.

It could be an even number or not an even number.

And it could be two digits or not two digits.

Now we can't see the spinner so we can't work out the theoretical probabilities for each of these events, which is why an experiment was conducted.

It was spun 2,500 times and what we're going to see are the results from this experiment.

We can display these three events of this experiment on a probability tree and also on a Venn diagram.

Let's take each of these events and turn.

The outcome could either be square or not square.

And what you can see on the branches here are probabilities that are derived from the results of this experiment.

Those are the proportions of times that the outcome landed on a square number and not a square number.

Of those outcomes that were a square number, one proportion of them were even, and the other proportion were not even.

And those are proportions, you can see on the screen there, as those experimental probabilities.

Of all the occasions where it was not a square number, then one proportion were even, and another proportion of them were not even.

And those are the probabilities you can see on those branches.

And then for each of these scenarios, one proportion had two digits and another proportion did not have two digits.

And we have experimental probabilities for each of these situations as well.

Now all of these can also be represented on a Venn diagram and it could look something a bit like this.

Let's transfer the information from the probability tree into the Venn diagram.

The height region here on the Venn diagram shows all the outcomes where the outcome was square, and that is highlighted as well in the probability tree.

Let's take a look at the intersection first.

We can calculate the probability that all three of those events happen, square, even, and two digits by multiplying those probabilities together and then put that in the intersection of all three regions on the Venn diagram.

This bit here is the probability that it's a square number, an even number, but not two digits, and that would be 0.

084, which would go here in the Venn diagram.

This probability is for the event that it's a square number, not an even number, but has two digits, and that would be 0.

052, which would go here in a Venn diagram.

And then we can do the same with the rest of them.

Can you think, where would these remaining probabilities go in the Venn diagram? Pause video why you consider that and press play when you're ready to continue.

Well, let's put 'em in now.

Now we've these diagrams completed.

Let's think about some conditional probabilities.

Because different conditional probabilities are easy to spot on different diagrams. For example, if we wanted to work out the probability that the number was even given that it was square, how can we get these from these two representations? With the probability tree, it's quite straightforward, because we've put the layer of branches for square numbers before we've put the layer branches for even numbers.

So we can travel across the branch, which says it's a square number, and then look at what is the probability it's even on the next one.

Should be 0.

6.

So that would be our probability.

With a Venn diagram, this is a little bit more complicated.

We need to look at all of the outcomes, which are a square number and the probabilities of them all.

The total probability would be the sum of those four numbers, and that'll make the denominator of our fraction.

And then out of those, we wanna see which ones are an even number.

And add those two probabilities together, and that'll make our numerator.

That gives us 3/5, which simplifies as well to 0.

6.

Same answer, but a little bit more complicated.

However, if we want to work out this probability instead, the probability at the number is square, given that we know it's a two-digit number.

Either these representations will be quite tricky to use.

The reason why the probability tree isn't as easy this time is because we have our layer branches for D after the layer branches for S.

So we can't just travel across one and see what the next probability is, like we could in the previous question.

In each case, we need to look at all of the outcomes where D has occurred, which two digits, add together those probabilities to find our denominator.

And then see which of those outcomes include S and add together those probabilities as our numerator, and it would give us approximately 0.

216.

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

Here you have the same probability tree and Venn diagram.

Could you work out the three probabilities showed on the screen? Pause way to do it and press play for answers.

Okay, let's go through some answers.

The probability that E happens, given that S has not happened, will be 0.

5.

You can see it in the probability tree on the left, and you can see it in the Venn diagram in the middle.

For b, the probability that D happens, given that S has not happened, that would be 0.

4, which you can see in either the Venn diagram or the probability tree.

And for c, the probability that S happens, given that E has not happened, that would be 0.

17 round to two decimal places, and you can see it in these two places on each representation.

Please explain why a two-way table would not be appropriate to show three groups of events.

Pause while you write something down and press play for an answer.

A two-way table can only show two events, horizontally and vertically.

There is no space for a third group of events.

You would need a three-way table for three groups events, and that would involve going into 3D.

So it is technically possible, but very, very difficult to construct.

Okay, it's over to you then for task b.

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

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

Here is question two.

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

And here is question three.

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

Okay, let's take a look at some answers.

For question one, here are the answers to parts a, b, and c.

Pause while you check these and press play for more.

And here's to answer to question d.

Pause while you check these and press play for more.

Then question two.

Here answers to parts a, b, and c.

Pause while you check and press play to continue.

And here's the answer to part d.

Pause while you check and press play to continue.

And then question three.

Here your answers.

Pause while check this and press play for a summary of today's lesson.

Fantastic work today.

Now let's summarise what we've learned.

The outcomes and probabilities of a two-stage trial can be shown on an outcome table, frequency tree, or a probability tree.

Conditional probabilities can be identified in different ways on each diagram.

Groups of events from a trial or experiment can be shown on a two-way table, frequency tree, probability tree, or Venn diagram.

And different conditional probabilities can be more easily found on some diagrams than others.

Well done today.

Have a great day.