Lesson video

Hello, you made a great choice with this lesson today.

It's gonna be a good un.

My name is Dr.

Olsen and I'll be guiding you through it.

So let's get started.

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

This lesson is called "Frequency Trees" and by the end of today's lesson, we'll be able to draw and use a frequency tree to calculate probabilities.

Here are some previous keywords that will be useful during today's lesson, so you might want to pause the video if you want to remind yourself what any of them mean and then press play when you're ready to continue.

The lesson is broken into three learning cycles and we're going to start with introducing the frequency tree.

When conducting an experiment, we may collect data from a large number of trials.

We can sort the results into groups based on an event, for example, splitting a population by a particular characteristic.

We can categorise outcomes by more than one event and display the number of results in each category on a frequency tree, for example, it might look something a bit like this.

When constructing a frequency tree, the total number of trials, the population in this case, is the single value that is usually shown on the far left of the tree.

In this case it's 78.

The first layer of branches shows the first set of subgroups that we can split the population into.

Now, this could be multiple branches or it could be two branches to represent an event and its compliment, for example A and not A.

Either way, the branches represent mutually exclusive and exhaustive outcomes, therefore, the sum of all the branches in this first layer should sum to the total population.

In this case, we can see 48 plus 30 is 78.

The second layer of branches shows the population being split up further into more subgroups.

Each of those first groups has been split into two subgroups in this case.

A set of branches that meet at a point should always sum to the frequency at that point.

For example, we can see that the group with a frequency of 48 was split up into two subgroups, one with a frequency 11 and one with a frequency of 37.

The sum of 11 and 37 is 48.

So let's check what we've learned.

An experiment involves studying a large population.

The results are displayed on this frequency tree.

What is the total size of the population? Pause while I write it down and press play for an answer.

The answer is 1050.

So what frequency would go here in the space that has been highlighted? Pause the video while I write it down and press play for an answer.

The answer is 400, which you get from doing the sum of 210 and 190.

It is possible to construct and complete a frequency tree if we're given enough information about the population.

For example, the attendance of 1040 students at Oakfield Academy was surveyed.

So we know that the population in this situation is 1040.

820 students attended on that day and that's now represented in our tree.

If 820 of the 1040 students attended, how many students didn't attend, in other words, were absent? Well, we could do this by doing 1040 subtract 820 and that will get 220 students who were absent, and we can represent it in our tree like this.

Let's now split these groups into further subgroups.

Of those that attended, 500 were in key stage three and the rest were in key stage four.

And we can see the frequency of students who attended in key stage three is now shown in our table.

To find the number of key stage four students that attended, we could calculate 820 subtract 500.

That is the total number of students who attended subtract the number of students who attended who are in key stage three.

And that would give 320.

Let's now look at those 220 students who were absent.

Of those who were absent, an equal number were in key stage three and in key stage four.

So we could take the total frequency of the students who were absent and half it and that would give 110 students in key stage three and 110 in key stage four who were absent.

A good way to check that the frequency tree is correct is to sum all the values in the right-most layer of the branches, because they should sum the total population.

And in this case, they do.

500 plus 320, plus 110, plus 110 is 1040.

So we can be confident that our frequency tree is correct, because this sum equals the total frequency.

So let's check what we've learned.

Alex's dad drove to work on 140 days and each day he took one of three different routes.

And we can see that information shown in the frequency tree, but we some unknowns in there.

Alex's dad drove on the main road on 42 days.

On 48 days, he took the B road.

On the remaining days, he took the country road.

Could you please calculate the values of the unknowns A, B and C in this frequency tree? Pause while you do this and press play for answers.

Here they are.

The value of A is 42, the value of B is 48 and the value of C is 140 subtract the other two values, which gives you 50.

So let's split these up into further subgroups.

On the days the main road was taken, he was on-time 28 times, whilst he was late the rest of the time.

On the days the B road was taken, he was late 15 times, while he was on-time the rest of the time.

You've got some unknowns in that frequency tree.

Could you please calculate the values of D and E? Pause the video while you do this and press play for answers.

Here are the answers.

The answer to D is 14, that is 42 subtract 28.

And the answer to E is 33, that is 48 subtract 15.

So let's now look at the country road and split that frequency up into on-time and late.

Alex's dad was late a total of 60 days altogether.

Could you please use that information and the rest of the frequency tree to calculate the values of F and G? Pause the video while you do this and press play when you're ready for answers.

Well, we can work out the value of G by taking the sum of 14, 15 and G and that should be equal to 60.

We can create an equation of that information and rearrange it to get G equals 31.

And now you know that G is equal to 31, F will be equal to 50 subtract 31, which is 19.

Frequency trees can also be used to present scaled versions of the collected data in order to present the information in a way that others can understand.

For example, data on hundreds of thousands of driving tests was collected.

This frequency tree we're about to draw now will show for every 1000 driving tests taken whether people passed or failed and what the number of attempts the test was.

So here we have 1000 as our total population and we can see that is branching off already into pass one and fail one.

Of the 1000 that took their first driving test, 680 passed.

The rest failed.

So we can put 680 on our tree and then we could calculate the amount that failed, which would be 1000 subtract 680, that's 320.

Now, before we draw any more branches on this frequency tree, let's take a second to consider the context that we're dealing with here and what implications that might have for how we construct our tree.

If someone passed their driving test, would they take a second attempt? No, they wouldn't.

So this means we only need to branch off the 320, the people who failed the first attempt.

We don't need to branch off the people who passed the first attempt.

So if we branch off the 320, we're going for passing the second time and failing a second time.

So let's take a look at those numbers.

Of the people who took their second driving test, half of them passed, the rest failed.

Well, half of 320 is 160, which means 160 must've passed and 160 must've failed.

And then we are not going to draw any more branches off pass two, because they will not take a third attempt.

We will however draw some more branches off fail two.

So we're looking here at the third attempt.

Of those that failed the second driving test, 45 passed their third test, whilst the remaining failed again.

So let's put that in our tree.

45 passed and the remaining will be 160 subtract 45, which is 115.

Now, if we had more data, we could continue to branch off those failed attempts each time, but let's just stop it here for now and check what we've got so far.

In order to check that all our calculations are correct, we sum the frequencies at the end of the right-most branches at each part of the tree.

So in other words, we find the sum of 680, 160, 45 and 115 and that would give 1000, which is the total population.

So we can be quite confident that this is correct.

Let's check what we've learned.

Driving test results in a different part of the country were collected and what we can see here, is some of the information is shown in a frequency tree, but we have three unknowns, A, B and C.

Could you calculate their values? Pause while you do this and press play for answers.

Here are the answers.

Working from right to left, A would be 130, B would be 240 and C would be 500.

Could you please complete this sentence? For every 500 people that take the driving tests, blank will fail twice before passing their third test.

Pause while you write it down and press play when you're ready for an answer.

The answer is 45, which we can see from the frequency tree.

Could you please complete this sentence now? For every blank people that fail their driving test once, blank will pass on their next attempt.

Pause the video while you write down the answers for those two blanks and press play when you're ready for answers.

Okay, let's take a look.

For every 240 people that fail their driving test once, 110 will pass on their next attempt.

Now, you could also put other numbers here as well, but not numbers we can see directly from this tree.

Scaled frequencies here would also be correct, for example, you could say, "For every 1000 people, then 1375 will pass." What we've done there is we've taken the number 240 and 110 and multiplied them both by the same thing to create two more numbers, which are scaled frequencies.

In this case, we multiplied them both by 12.

5.

We can multiply them by anything so long as you multiply them both by the same thing.

What fraction of people fail their driving test three times? The answer is 85 500ths or anything that's equivalent to that.

What fraction of people who fail their driving test twice also fail a third time? Pause the video while you do that and press play when you're ready for answers.

Well, we're not looking for a fraction of the population for this data, we're looking for a fraction of the people who fail their driving test twice.

So we're looking for a fraction of 130 people and that would 85 130ths.

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

This task contains two questions and here is question one.

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

And here is question two.

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

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

Here are the answers to question one.

Pause while you check these against your own and press play for answers of question two.

And here are the answers to question two.

Pause while you check this and press play when you're ready for the next part of the lesson.

You're doing great so far.

Now let's move on to the second part of this lesson, which is frequency trees with proportion.

Sometimes we may have to construct a frequency tree if given proportional information alongside some frequencies.

For example, a tech repair shop predicts that over the next week, they will have 250 customers.

They want to see where they can improve their customer service the most.

So let's construct a frequency tree with the information that's going to be gradually given to us over this problem.

So far we can put the number 250 in our frequency tree as the total population.

Based on the previous weeks, 70% of the customers will ask for a phone to be repaired.

The rest will ask for laptops to be repaired.

So let's do some calculations of that information.

70% of the total population will be 70% of 250, which is 175.

They will ask for their phones to be repaired.

So we can put 175 in our tree here.

And the rest will ask for their laptops to be repaired.

So we can do 30%, which is the remaining percent of 250 and that would give 75.

And we can put that in our tree here.

Alternatively, you could've done 250 subtract 175 to get the same result.

Let's get some more information now.

Based on previous weeks, four fifths of the customers who asked for their laptops to be fixed were happy with the results.

That's great, however, only three fifths of all customers were happy with the results.

Let's break this information down.

We have four fifths of the customers who asked for their laptops to be fixed.

That's four fifths of 75.

If we do four fifths of 75, we get 60.

So that means 60 goes in this box here.

And then we could do 75 subtract 60 to find 15 for this space here.

And then we have the information that three fifths of all customers were happy with the results.

Now, you may be tempted here to find three fifths of 175, because you're thinking about how to split the 175 into two subgroups, but it doesn't say three fifths of those who asked for their phones to fixed.

It says three fifths of all customers.

So that means we need to find three fifths of 250, which is 150.

But that 150 is not going to go in any of those blanks spaces, because is says 150 people overall were happy with results, which means if we look at those people who were happy, it'd be those who asked for their laptops to get fixed were happy and those who asked for phones to be fixed in a happy.

So the sum of those two numbers will be 150.

So 60 plus our unknown number will be 150.

So we could do 150 subtract 60 to get 90 and then we can look at just the people who asked for their phones to be fixed, there are 175 altogether, 90 of them are happy, which means 85 of them will be sad.

175 subtract 90 is 85.

And if you want to, you could check that this information is correct by adding together 60, 15, 90 and 85 and check if you get 250, which you do.

So what conclusions could we draw from this? A majority of customer who asked for their laptops to be fixed were happy.

However, nearly half of the customers who asked for their phones to be fixed were sad.

So the shop should improve the customer service when fixing phones in particular.

So let's check what we learned.

A spinner with outcomes A, B and C is spun 560 times.

The spinner landed on A, B and C in the ration five to four to seven.

This information is shown in a frequency tree, but you've got four unknowns, W, X, Y and Z.

Could you please find the values of those unknowns? Pause while you do it and press play when you're ready for answers.

Well, the easiest thing to work out here will be W.

That will be 560, 'cause it's the total population, which in this context is the total number of times the spinner was spun.

And then to work out the values of X, Y and Z, we need to divide 560 in the ratio five to four to seven.

Now, there are different ways you can do that, but one way is to use an equation, for example, 5p plus 4p, plus 7p equals 560.

You can use any letter instead of P, but don't use X, Y or Z, because those are different unknowns.

Once you do that, you would get p is equal to 35 and then we can work out X, Y and Z by multiplying that 35 by five, four and seven and you'd get 175, 140 and 245.

Following each spin, a coin is flipped and data about those results are also collected.

When the spinner lands on A, three sevenths of the coin flips landed on heads.

When the spinner landed on B, seven tenths of the coin flips landed on tails.

You've got now our frequency tree with six unknowns.

Could you please find the value of as many of those frequencies as you possibly can? You might not be able to get more, but work out as many as you can do based on this information.

Pause the video while you do that and press play when you're ready for answers.

Well, D would be three sevenths of 175, which is 75.

And E would be 175 subtract 75, which is 100.

Then we can work out G and G is seven tenths of 140, which is 98, which means F would be 140 subtract 98, which is 42.

H and I cannot be worked out based on this information, so let's get some more information.

60% of all coin flips landed on tails.

So that should be enough information now to work out the values of H and I.

Please pause while you do that and press play when you're ready for answers.

Okay, let's take a look.

60% of all coin flips landed on tails, so 60% of 560 will be 336, which means we can do 336 subtract 100, subtract 98 to get 138 for the value of I and then we know with the value of I, we can find the value of H by doing 245 subtract 138 and that would give 107.

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

This task contains two questions and here is question one.

Pause while you do it and press play when you're ready for question two.

And here is question two.

Pause while you do this and press play when you're ready for some answers.

Okay, here are the answers to question one.

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

Okay, and now let's go through question two.

Here are the answers to that.

Please pause the video while you check them against your own and press play when you're ready for the next part of the lesson.

Well done so far.

Let's now move on to the third and the final part of this lesson, which is looking at probability and proportion of frequency trees.

Here we have a probability tree and a frequency tree.

Let's now compare these two side by side and consider how they are similar and how they are different.

Frequency trees and probability trees both show events.

That could be either from a trial or an experiment.

So that's one way that they are similar, but how are they different? Well, the nodes of the frequency tree show the frequencies of these events or groups while the branches of a probability tree describe the probabilities of the event occurring.

It is possible to calculate probabilities from a frequency tree.

For example, here's a frequency tree that shows some information about 250 customers.

One of the 250 customers is chosen at random.

Let's work out some probabilities about that chosen customer.

What is the probability of the customer having their laptop fixed and also being happy? Well, let's take a look at the tree.

There are 60 people who fit that category who had their laptop fixed and are happy, so that would be the numerator of our probability, and there are 250 people altogether in this frequency tree.

So it will be 60 250ths.

That could be simplified or it could be written as a decimal or percentage.

What is the probability of the customer being sad? Now, it doesn't say whether they had their laptop fixed or phone fixed.

It's just any of the customers being a sad customer.

Well, we could find the number of people who were sad altogether, just 15 plus 85.

We could add those together to get the numerator of our fraction, which would be 100 and out of 250 still.

So it's 100 250ths or anything that's equivalent.

Let's check what we've learned.

350 people from either Oakfield or Elmsleigh have their vehicle fixed.

A random person is chosen for a quality assurance survey.

You've got information about this on the frequency tree on the right side of the screen and there are two questions to answer about this.

For A, could you please find the probability that the person was from Oakfield and had their car fixed? And for part B, could you find the probability that the person had their van fixed? Pause while you do this and press play when you're ready for answers.

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

For part A, it'd be 130 350ths.

That is our 130 people who are from Oakfield and had their car fixed and there are 350 people altogether in this survey.

For part B, it would be 145 350ths.

The 145 is not anywhere in this frequency tree, but that comes from doing the sum of 80 and 65.

The other people who had their vans fixed from either of the two places.

It is also possible to calculate proportions involving subgroups of the population.

This means that we may be looking at a smaller total than the whole population.

For example, if we want to only consider the subgroup of people who asked for their laptops to be fixed in this frequency tree, then our total would be 75 people, not the 250 people from the entire population, but we're only concerned with those who asked for their laptops to be fixed.

And there were 75 people who asked for their laptops to be fixed.

Here's an example of a question like that.

What fraction of the customers who asked for their laptops to be fixed were happy with the results? Now, we're not looking at a fraction of the overall population here.

We're only looking for a fraction of those people who asked for their laptops to be fixed.

In other words, we're asking for a fraction of these 75 people here.

So our denominator is going to be 75, because we only care about the group of people with laptops.

And then our numerator will be 60, because 60 of these 75 people with laptops were happy.

So answer is 60 75ths.

Now, that question is subtly different to this next question, which is what fraction of the customers are people who had laptops and were happy? With this question, we are looking for the fraction of customers altogether.

So that'd be 250, and of the customers, 60 of them have laptops and are happy.

So it will be 60 out of 250.

Here's another pair of questions that are subtly different to each other in the way that they are worded, but the difference between them will affect the numbers that we use.

One question is, what fraction of people who were sad asked to get their phones fixed? And the other is, what fraction of people were sad and asked to get their phones fixed? Very subtly different.

What is the difference between those two questions and how will that difference affect the numbers that we use in each fraction? Perhaps pause the video while you read those questions again and think about that and press play when you're ready to continue.

Okay, let's take a look at this together.

With question one, we're looking for the fraction of the people who were sad.

That's not the fraction of the people altogether, just the fraction of the people who were sad.

So we only care about the group of people who were sad here.

We can ignore a lot of the tree and just focus on those particular numbers.

The number of people who were sad was 15 plus 85 which is 100.

So out of those people, how many of them asked to get their phones fixed? Well, that would be 85 people who were sad asked to get their phones fixed.

So it would be 85 100ths.

Then the second question said, what fraction of the people were sad and asked to get their phones fixed? So here we're looking for the fraction of the people, which is 250.

So in this one, we want to put 85 250ths, because we care about all 250 people in the population.

Let's check what we've learned.

An experiment is made from repeating a trial 560 times and the trial has two stages to it.

Stage one, a spinner with outcomes A, B and C is spun and stage two, a coin is flipped.

The frequency tree shows results.

What fraction of the trials resulted in the outcome B and heads? Pause the video while you write it down and press play when you're ready for an answer.

The answer is 42 560ths.

What fraction of the trials where the spinner landed on B also resulted in heads? Pause the video while you do that and press play when you're ready for an answer.

The answer is 42 140ths.

What fraction of the trials where the coin landed on heads did the spinner also land on B? Pause the video while you do that and press play when you're ready for an answer.

The answer is 42 224ths.

The 224 is from finding the sum of all the occasions where the coin landed on heads.

75 plus 42 plus 107 and the 42 are the occasions where it landed on heads and the spinner landed on B.

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

This task contains one question and here it is.

Pause while you do it and press play for answers.

Okay, let's go through some answers.

For part A, the probability of wearing a football shirt is 15,500 over 26,000 and the probability of not wearing a football shirt and being from Rowanwood is 5000 over 26,000.

For part B, what fraction of people from Rowanwood are not wearing a football shirt? That's 5000 out of 11,500.

And part C, what fraction of people wearing a football shirt are from Oakfield? That's 9000 over 15,500.

Fantastic work today.

Now let's summarise what we've learned.

Frequency trees can be constructed to break down a population into different groups and each of those groups can be split into subgroups as well.

It is possible to fully complete a frequency tree if given incomplete information about the population.

You just need enough information in order to be able to work out the rest.

Frequency trees can be constructed if given either frequency data or proportional data.

Probabilities can be found from a frequency tree and proportions can be calculated from subgroups of a frequency tree.

Well done today, have a great day.