Lesson video

Hi, my name's Mrs. Wheelhouse and welcome to today's lesson on using an outcome tree to display outcomes for more than two events.

This lesson falls within our unit on probability: possible outcomes.

Now I love probability, so I'm really looking forward to doing this lesson with you today.

Let's get started.

By the end of today's lesson, you'll be able to systematically find all the possible outcomes for more than two events by using an outcome tree diagram.

Now, here are some keywords that we're gonna be using in our lesson today.

And they should be familiar to you.

But by all means, feel free to pause the video, have a read through, and just check that you are familiar with what these words mean.

Pause and do this now if you wish.

Our lesson today has two parts, and we're going to start with the first part on outcome trees For multi-stage trials.

Jun will spin the spinner twice.

He represents the possible outcomes for this two-stage trial using two different representations.

The first is an outcome table, and you can see here that of his two spins, he could win win, lose win, win lose, or lose lose.

He also uses an outcome tree, and you can see here that the first layer of branches refers to our first spin where we could win or lose, and then we have our second layer of branches where again, we could win or lose.

You can see when we created the sample space at the end of the outcome tree diagram, we got the same four outcomes as we saw in our outcome table so we could win win, win lose, lose win or lose lose.

Now Jun says, I wonder whether it's possible to list all the outcomes in an outcome table or outcome tree if I spin the spinner three times.

Well, Sofia has a good point here.

She says, I'm pretty sure you can't show a three-stage trial in an outcome table.

The rows show one spin and the columns show the other.

It's a good point, Sofia and Jun says, yeah, you have nowhere to record the third spin.

But what about with an outcome tree? And Sofia points out, well, the first layer of branches was the first spin, and the second layer of branches was the second spin, meaning that it's possible to draw more branches to the right to represent the outcomes of a third spin.

I think you might be right there, Jun, and this is exactly what it would look like.

So our first spin is our first layer of branches.

Our second spin is our second layer of branches, and our third layer is our third spin.

At the end of the tree diagram just like before, we're going to have our sample space where we write down the possible outcomes.

So we could win win win, win win lose, win lose win, win lose lose, lose win win, lose win lose, lose lose win, and lose lose lose.

Jun points out that each outcome in the sample space is made of three letters.

One for each of the spins.

Well noticed jun.

This outcome comes from this first spin landing on W, the second spin landing on W, and the third spin landing on L.

In other words, the first branch we went down was win, and the second branch we went down was win.

But the third branch we went down was for lose.

Let's do a quick check.

Lucas is going to spin spinner 1 once and then spinner 2 twice.

Which outcomes should be written at the end of the branches labelled A and B.

Pause while you work this out.

Welcome back.

What did you put? Well for A, you should have put Y, and for B, you should have put X.

Remember, this was the second and third layer of our branches, and so was referring to spinner two.

And since for A, you could see X was already listed of the top branch.

We knew that A had to be replaced with a Y.

And likewise for B, because Y had already been put on the branch just below it, we knew X had to be on the branch just above.

Which of these statements is true for the highlighted set of branches.

So you can see there, I've gone from the dot to the V and then from the V to the Y.

So which of the three statements is true? Pause now while you make your selection.

Welcome back.

Did you pick B? In order to get to here spinner 1 had to land on V, not W, and then spinner 2 had to land on Y.

What about now? Which of these statements is true for this highlighted set of branches? Pause the video and make your choice.

Welcome back.

Did you pick C? In this situation, spinner 1 landed on W and spinner 2 landed on X, and then for the next spin landed on X again.

Now quick check with our sample space, which outcomes would go in place at the letter C and D in the sample space? Pause the video while you write this down.

Welcome back.

So for C, you should have put V, X and X.

And then for D, you should have put W, Y, and X.

Well done if you've got these right.

How many outcomes are there in the event where two letters repeat themselves? So how many outcomes have that where two letters have repeated themselves? Pause the video while you work this out.

Welcome back.

Did you find them all? There were four.

We had V X X, so the X was repeated.

V Y Y, so the Y is repeated.

And then W X X and W Y Y.

It's time for your first task.

For question one, Sofia will spin, spinner 1, spinner 2, and spinner 3 once each.

For part A, complete this outcome tree and then its sample space.

For part B the event that at least one spinner lands on a vowel has how many outcomes? Pause the video while you complete question one now.

Welcome back.

Let's look at question two.

In question two Aisha plays a game.

Aisha will spin spinner 1, 2 and 3 once each.

If spinner 3 lands on W, then Aisha wins the number of sweets equal to the sum of the numbers on spinner 1 and spinner 2.

Complete this outcome tree and sample space for Aisha's game.

Pause the video and do this now.

Welcome back.

Let's look at question three now.

And question three, Sam will spin spinners 1 and 2 once each, and then flip a coin.

Part A, draw the outcome tree for this three-stage trial.

Stage one is spin spinner 1, stage two is spinner 2, and stage three is the coin.

In part B, modify the outcome tree to satisfy this four-stage trial.

If two spinners land on the same letter and the coin lands on heads, then flip another coin.

Pause and do this now.

Welcome back.

Time to go through the answers.

Question one, you had to complete the outcome tree and the sample space, so feel free to pause the video right now so that you can check your answers against the ones you see on the screen.

And then part B, the event that has at least one spinner landing on a vowel has how many outcomes? Well, there are six outcomes with at least one vowel.

And they are A C E, A C F, A D E, A D F, and B C E, and B D E.

For question two, you had to complete the outcome tree and sample space for Aisha's game.

So again, feel free to pause the video now so that you can check your answers against the ones you see on the screen.

And now question three.

So you had to draw the outcome tree for this three stage trial, and we did say that stage one should be spinner 1, stage two is spinner 2, and stage three is the coin.

So you can see here on our branches I've put for stage one A on the top branch and B on the bottom branch.

Now you may have switched these round and that's fine if you did.

So just check that you've got them producing the same outcomes as I do.

For part B, you had to modify the outcome tree such that if the spin has land on the same letter, so in other words, if it's A then A or B and then B, you then had to consider if the coin showed heads or not.

If it did, then you had to flip another coin.

And that's why if you follow the path that goes A, A, heads, there's then two more branches for the second coin flip.

And the same if you followed the branches for B, B and heads.

Well done if you got this right and modified your tree diagram correctly.

It's now time for the second part of our lesson and that's on outcome trees from multiple events.

Outcome trees can also categorise all of the outcomes of a single stage trial into three or more events.

Jun will roll at fair 12 sided dice with integers from 0 to 11 on it.

If I roll this dice, I wonder how many outcomes are odd and either square number or a cube number.

Now each event can be shown as a layer of branches on an outcome tree.

Each layer will contain an event and its opposite event.

So for example, layer one could be odd on one branch and even on the other.

Layer two would have square on one of the two branches and non-square on the other one.

And then for layer three would have two branches, one showing the cube numbers and one showing the non-cube numbers.

So let's see what this will look like.

Well, here's the start of our tree diagram with our two branches, one for odd and one for even.

Now the numbers 0 to 11 are on this dice, so we can write these outcomes by the respective correct branch.

So there are our odd numbers and there are our even numbers.

We can now consider layer two.

So again, we're going to have two branches here for square numbers and non-square numbers.

We can then look at the outcomes that proceed this work out which ones go on which of the two branches.

So for 1, 3, 5, 7, 9, 11, the square numbers are 1, 9 and the non-square are 3, 5, 7, 11.

We can then consider the even numbers, 0, 2, 4, 6, 8, and 10 and put the numbers that are square, so 0 and 4 by the square event.

And then the non-square values 2, 6, 8, and 10 go at the bottom.

Let's now look at layer three.

So again, we have pairs of branches here where our outcomes are cube and non-cubes.

So looking at the square numbers are 1 and 9, we know that the cube number there is 1, and the non-cube is 9.

For 3, 5, 7, 11.

None of these are cube.

For the square values at 0 and 4, 0 is cube, but 4 is not.

And then for 2, 6, 8, and 10 only 8 is a cube number.

The others are not.

If the dice lands on a 3, 5, 7, or 11, this will be an outcome of the following events.

It is an odd number, it is not square, and it is not cubed.

You'd wanted to find out how many outcomes are odd and either square or cube.

Well, that means he wants to look at the outcomes that are odd, that are square, or they're cube.

So he had a nice choice here.

And that means the outcomes are 1 and 9.

So there are two in total.

Now for the ones that are non-square and cube, well, there weren't any, which means there are just those two outcomes.

Let's do a quick check.

Andy will roll a 10 sided dice with integers from 10 to 19 on it.

Which outcomes would go in the box labelled A? Pause the video while you work this out.

Welcome back.

What did you put? Well, if you look at the outcomes that come just before, so 11, 13, 15, 17, and 19.

The prime ones are all except 15.

So now, which of these boxes will contain no outcomes? So is that B, C, or D? Pause the video now while you work this out.

Welcome back.

Well, it can't be B because we know B is gonna be 15, so it's got to be either C or D.

And it's C, 10, 12, 14, 16, 18 are all even numbers.

And since two is not in that list, there are no prime ones there.

Now how many outcomes are there for the events even and a multiple of four.

So pause the video now while you work this out.

Welcome back.

So you only had to find two outcomes here, and that's 12 and 16.

There were no even numbers that were prime, so you didn't have to follow those branches, so you only had to consider the even non-prime numbers.

So 10, 12, 14, 16, 18, and only two of those are multiples of four.

It's now time for our final task.

For question one Sam will spin a spinner with the integers from 20 to 30 on it.

Sam considers the events that the spinner lands on an even number, a multiple of three and a factor of 100.

Fill in the blanks for this outcome tree with outcomes from this spinner.

Pause and do this now.

Welcome back.

For question two Lucas will spin a spinner with the integers from 15 to 25 on it.

Lucas considers the events that the spinner lands on a square number, a prime number, and a factor of 300.

Fill in the blanks to complete this partial outcome tree.

Pause and do this now.

Welcome back.

For question three you will spin a spinner with the integers from 0 to 15 on it.

You can consider the events that the spinner lands on a factor of 24, a cube number and a two-digit number.

For part A, complete the partial outcome tree and list the outcomes at the end of each branch.

And then in part B, how many outcomes match the events that Jun is looking for? Remember, Jun is looking for the events that we have a factor of 24 that is a cube number and a two digit number.

Pause the video and work on this question now.

Welcome back time for question four.

In question four, Sofia has spinner with the following outcomes.

1, 3, 4, 5, 8, 10, 12, 20, 25, 27, and 36.

Sofia will spin the spinner and consider the events that the spinner lands on an even number, a square number, a cube number, and a triangular number.

For part A construct and fill in a four layer outcome tree for the outcomes of Sofia's spinner.

And in part B, how many outcomes that are even numbers are also both square and triangular.

Pause the video while you complete this question.

Welcome back.

Time to go through our answers.

For question one you had to complete the outcome tree with outcomes from the spinner.

Now, there are quite a lot here, so please feel free to pause the video while you check your answers against mine.

Let's look at question two.

You had to fill in the blanks and complete this outcome tree.

Again, feel free to pause so you can check what you've got as the outcomes against what I have.

For question three, we had to complete the partial outcome tree and list the outcomes at the end of each branch.

So I finished off the outcome tree and you can see the outcomes at the end of each branch.

For part B, how many outcomes match the events that Jun is looking for? Now remember, you wanted events that the spinner lands on a factory of 24, so that's us going down the top branch at the moment.

And a cube number so again, the top branch, and it had to be a two-digit number.

And if you follow those branches through, you'll see there are zero outcomes that match the events that Jun is looking for.

In other words, there were no factors of 24 that were two-digit cube numbers.

And question four, you had to construct and fill in a four layer outcome tree for the outcomes of Sofia's spinner.

Now I've put these in the order they appeared in the question.

So whether it was even odd first, then square or not square, cube not cube, and triangular not triangular.

Now how many outcomes that are even numbers are also square and triangular and there's only one, and that's 36.

And you reach that by going down the even branch, the square branch, the non-cube branch, and then the triangular branch.

There wasn't any other outcome that fulfilled those three requirements of being even, square, and triangular.

It's now time to sum up what you've learned today.

The branches of an outcome tree show the outcomes of a trial or a group of events.

One layer of branches shows one stage of a trial or one event, and its opposite event.

It's possible for there to be three or more layers of branches.

And we saw that our very last question, for example, had four layers.

A sample space could be made from an outcome tree, which lists all outcomes from an outcome tree showing multiple stages of a trial.

Well done.

It was a lot of work today and you've worked so hard.

I look forward to seeing you for more lessons on probability.