Lesson video

Hello, everybody and welcome to this, our next lesson on probability with me, Mr. Gratton.

In this lesson, we will be using two-layer probability trees to show the probabilities of outcomes and events from a two-stage trial.

Pause here to have a quick look at some of the keywords that we'll be using during today's lesson.

But before looking at the probabilities of a two-layer probability tree, let's first see how we can construct a two-layer probability tree.

As a recap, we can construct an efficient single layer of branches by taking each distinct or different outcome from a trial and representing each outcome once on a probability tree, one distinct outcome per branch with the probability of that outcome happening being shown on that branch.

For example, in this probability tree, the five out of eight represents the probability of the spinner landing on any of the five As on the spinner with a total of eight outcomes, both A and B.

These distinct outcomes can be referred to as events, a group of individual outcomes from a trial.

Furthermore, these events are mutually exclusive and exhaustive and therefore, their probabilities must always sum to one.

So the case of this probability tree, we've got 5/8 plus 3/8 equals one.

And so using all of that recap information, here's your first check.

Sofia is hosting a birthday party.

Each person gets one party hat.

Sofia is about to choose a hat at random.

The probability tree shows the probability she chooses a certain hat.

Pause here to consider which probability goes inside this box.

There are a total of eight party hats with four of them having a rabbit on it.

Therefore, the probability of randomly choosing a rabbit hat is four out of eight.

And now what two pieces of information go inside these two boxes? Pause to compare this branch to the other branches on this tree.

The outcome is starred hats and the probability is three out of eight.

You could have replaced the word describing this hat with any other suitable word, such as blue hat or yellow topped hat.

For this next check, this spinner is spun several times with the probability of the outcomes W and Y shown on this tree.

Pause here to consider what probability goes on this branch.

If five out of the seven spins land on W, then the remaining two spins out of the seven must land on the other outcome Y.

And so the probability of the spinner landing on a Y is two out of seven.

Pause here to choose the statement that justifies why 2/7 is the probability of Y if you know that W has a probability of five over seven.

And the answer is C, all the probabilities of all the distinct outcomes of a trial must sum to one and 5/7 plus 2/7 equals one.

Jun wonders if it is possible to represent two-stages to a trial, such as this trial composed of spinning a spinner and flipping a coin on a probability tree.

It is possible if both stages of the trial are performed at the same time, we can choose one of them to be shown on the first layer of branches.

It does not matter which one.

However, if one stage is performed before the other, it is sensible to show that stage on the first set of branches.

Let this first stage of the trial be the spinning of the spinner.

There is a five out of eight chance of the spinner landing on the letter A.

If the spinner did land on the letter A, how could we represent the outcomes of the coin flip as well? We can draw a new pair of branches from the end of the outcome A.

This pair of branches will represent the outcomes of the coin flip, heads and tails, with their probabilities being 1/2 each.

This pair of branches represent the possible outcomes of the coin flip, assuming that the spinner will also land on A.

But how can we show the outcomes of this coin flip if we assume that the spinner lands on B instead? That's right.

We can draw another pair of branches, but this time, starting from the end of outcome B of the previous stage of the trial.

This pair of branches shows the outcome of a coin flip assuming that the spinner has landed on B.

The first layer of branches represents the stage of the trial spinning the spinner whilst this, the second layer of branches, represents the stage flipping the coin instead.

This stage of the trial is shown twice, once for if the spinner will land on A and once for if the spinner lands on B instead.

Okay, now that we've drawn the tree, let's interpret what the tree means.

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

In order to show all of the outcomes of the entire trial, rather than just showing each individual stage of the trial, we must construct a sample space showing each combination of connected branches.

Typically, the sample space is shown to the right of all of the branches of the main probability tree.

For example, this outcome in the sample space shows the connected branches B and tails.

For this check, going back to Sofia's party, Sofia chooses both one hat and one balloon at random.

This probability tree shows the probability she chooses a certain hat and balloon.

Pause here to answer.

What probability goes inside this box? Three out of the four balloons are spotted.

Notice how we are growing this probability tree by constructing more pairs of branches for the balloons from each outcome of the hats layer of branches.

Pause here to answer, what probability goes inside this box? One out of the four balloons is plain.

Next up, pause here to think or discuss what must the probabilities of these two branches sum to? Remember, all the probabilities of a group of branches that meet at a point sum to one.

Therefore, each of these pairs of branches have probabilities that sum to one.

This pair of branches have probabilities that sum to one.

So do these two and so do these two.

And lastly, pause to answer, what do these two marked branches represent, which we can then write in a sample space at the end of the tree? These two connected branches represent the outcome that Sofia chooses a hat with stars and a spotted balloon.

Okay, onto the practise task for question one.

This spinner is spun twice.

Pause here to write down the outcomes and probabilities of this two-stage trial on this probability tree and then write down a sample space showing the outcomes of the entire trial.

And for question two, pause here to complete the probability tree and sample space for this, the two-stage trial of Lucas playing two games that he can either win, lose or draw.

It might be sensible to convert the percentages into either decimals or fractions for this question.

And for question three, time to construct your own probability tree from scratch.

Pause here to show this two-stage trial of spinning a spinner and flipping a coin once each on a probability tree and associated sample space.

Great effort on constructing all of your trees so far.

For the answers, pause here to see if your probability tree and sample space match the correct one that you can see on screen.

And for question two, pause here to see whether your completed tree matches this one.

And for question three, here's one example of a probability tree that you could have constructed.

It was also possible to construct a tree where the first layer of branches showed the coin flip rather than the spinner.

This is absolutely okay as long as the second layer of branches shows the outcomes of the spinner instead.

The sample space should contain the same outcomes regardless of the order in which you constructed your tree.

But you'll see the outcomes written as, for example, heads A rather than A heads.

Okay, we've drawn some two-layer probability trees, that's all good and well.

But for a probability tree, we've not really done anything to do with the probabilities yet.

Let's have a look.

And Jun agrees.

We've done well to list all of the outcomes of a two-stage trial on these trees.

But how do we consider the probabilities of the outcomes of a two-stage trial? For example, what is the probability of spinning an A and flipping A heads? Here we can see the probabilities of each outcome in this, the sample space specifically for stage one of our trial.

And here we can see the probabilities of each outcome of specifically stage two of our trial.

But right at the end of the probability tree is the sample space for the whole entire trial, the spinner and the coin flip combined.

This currently does not have any probabilities represented on it.

To find the probability of both the spinner landing on an A and the coin landing on a heads, the probability of an outcome from the whole trial, we must multiply the probabilities on the branches of the outcomes at each stage of a trial.

Five out of eight is the probability of the spinner landing on an A and one out of two is the probability of the coin landing on heads.

So five over eight times by one over two equals five over 16.

The probability of both the spinner landing on an A and the coin landing on heads is five over 16.

We can represent this probability inside the sample space of the entire trial, that sample space on the far right of the probability tree.

Okay, let's work out the probability of both landing on B and tails with Jun.

The probability of landing on B is three out of eight and the probability of landing on tails is one out of two.

Therefore, three out of eight times by one out of two is three out of 16.

And so the probability three over 16 goes beside the B tails outcome in that sample space.

And so to complete all of the probabilities of this sample space, we have the probability of A and tails, which is 5/8 times by 1/2, which is five over 16.

We also have the probability of B and heads, which is 3/8 times by 1/2, which is three over 16, the probability here.

Remember, the sum of all probabilities of each sample space must sum to one as each sample space is an exhaustive list of all of the outcomes in a trial or stage of a trial.

Adding all of the probabilities at the end of a probability tree is a good way of checking whether your calculations are correct.

So five over 16 plus five over 16 plus three over 16 plus three over 16 equals 16 over 16.

Since 16 over 16 equals one, we could be quite confident that the probability tree has been constructed correctly and the probabilities correctly calculated in that sample space.

For this check, the trial is to spin spinners one and two once each.

This probability tree shows this two-stage trial.

Pause here to answer, what outcome goes in place of the A in that sample space? That first part of the sample space shows the outcome, which is YP.

Spinner one landing on Y and spinner two landing on P.

And next up, pause here to answer what probability goes in place of B in this sample space? In order to calculate the probability of an outcome of an entire trial, an outcome in that sample space at the end of a probability tree, we must the probability of each connected branch that results in the outcome.

In this case, YP.

The probability of Y is 0.

7 and the probability of P is 0.

25.

So 0.

7 times by 0.

25 equals the probability 0.

175, which represents the probability of getting both a Y and a P.

And next up, pause here to complete each remaining part of the sample space with both an outcome and a probability.

For C, the outcome is XQ and the probability of getting XQ is 0.

225.

Whereas for D, the outcome is YQ and the probability is 0.

525.

And to check that our calculations are correct, pause here to answer.

What should all four of these decimals sum two? All the probabilities in that sample space should sum to one.

Notice how the probabilities at each stage of this two-stage trial are given in different forms, percentages, and then fractions.

It is helpful to convert some of the probabilities until all of them are in the same form.

This helps find the probabilities of the whole trial more easily.

Notice how in this question I have probabilities shown as the fractions 1/3.

This isn't easily possible to convert into a decimal without rounding and losing precision so it is sensible to instead convert the percentage into a fraction in order to match the other fractions in the tree.

Next up, two quick checks.

First pause here to consider what is 32% written as a fraction? 32 out of 100 is correct, but simplifying to the equivalent eight out of 25 is also equally correct.

Both are allowed on a probability tree.

Next check, what would go in the part of the sample space currently labelled A? Pause now to give both an outcome and a probability.

We have either 32 out of 100 times by three out of five equals 96 over 500, or if simplified, 24 over 125.

Okay, for question one of this next set of practise tasks, the probabilities of landing on R and G on spinners one and two are shown.

Pause here to complete this probability tree to calculate the probability of each outcome from this trial.

And for question two, currently this probability tree only shows one layer of branches.

Pause here to modify it so it is suitable for a two-stage trial and then complete the probability tree and sample space to show the probabilities of all outcomes of this trial, the two spinners being spun once each.

And finally, question three, time for you to draw three of your own probability trees, one for each person.

Each tree should show the outcomes and probabilities of each type of flower growing that the students take.

Lucas' tree has already been drawn for you, but you still need to put in the outcomes and probabilities.

Pause now to draw and complete three probability trees.

Okay, here are the answers.

Pause here to check if your tree and sample space matches the one on screen.

And for question two, the probability of outcome CQ is two out of 40.

Pause here to compare your tree and completed sample space with the one on screen.

And for question three, pause here to check Lucas' completed tree.

There is a 72% chance of both of his flowers growing.

And pause here to check Izzy's completed tree.

There is a nine out of 16 chance of both of her flowers growing.

And finally, pause here to check Laura's completed tree.

There is a 67.

5% chance of both of her flowers growing.

Now that we've looked at individual outcomes from a two-stage trial, what about events from a two-stage trial instead? Well, let's have a look.

We can still use the same sample space at the end of a probability tree to consider the probabilities of events.

So for the trial of this spinner being spun twice, we have the outcomes and probabilities of the first spin and the second spin, and the sample space showing the outcomes and probabilities of the entire trial.

This outcome, Y on spin one and X on spin two, has a probability of 20 over 81.

But what about the event at least one spin lands on X? How can we consider the probability of this whole event? Well, first, let's see which outcomes satisfy this event.

At least one X means we look at the outcomes that have at least one X mentioned.

X and X satisfies this and has a probability of 16 over 81.

X, then Y has a probability of 20 over 81 and satisfies this criteria.

And Y, then X also satisfies and has a probability of 20 over 81.

Y, then Y does not contain an X, so we do not count it.

To find the probabilities of this event, we add together the probability of each relevant individual outcome.

The probability of at least one spin landing on an X is 56 out of 81 and contains the outcomes X, then X, X then Y and Y then X.

Similarly, we have the event exactly one spin lands on an X.

What is different about this event compared to the previous one? Well, X, then X contains two Xs and so isn't exactly one X.

This means we exclude it as an outcome from this event.

The probability of this event, exactly one spinner landing on X is 20 over 81 plus 20 over 81 equals 40 over 81 for only the outcomes X, then Y and Y then X.

Okay, for this check that uses the same trial, which three outcomes are outcomes for the event at least one spin lands on a Y? Pause now to list all three outcomes from that sample space.

The event at least one Y means that the outcome must contain either one or two Ys and therefore, the outcomes are X, then Y, Y, then X and Y, then Y.

And now pause here to calculate the probability of the event at least one spin lands on Y.

To find the probability of this event, we must add together the probabilities of each individual outcome.

Therefore, 20 over 81 plus 20 over 81 plus 25 over 81 equals 65 over 81.

The probability of the event at least one spin lands on a Y occurring is 65 over 81.

Okay, for this different trial, 70% of trials land on an odd number.

Pause here to consider the probabilities of the outcomes at A and B.

Remember, the probability of the outcomes of a two-stage trial is the product of the probabilities at each stage.

So the probability of odd then even is 0.

7 times by 0.

3, which equals 0.

21.

Similarly, the probability of even, then even is 0.

3 times by 0.

3, which is 0.

09.

And finally, pause here to answer what is the probability of the event the two spins of the spinner will result in at least one even number.

Okay, here are the three outcomes of this event.

Adding the probabilities of these three outcomes gives us 0.

51.

And so the probability of this whole event is 0.

51 or 51%.

Okay, onto the final three practise questions.

For question one, using this probability tree, pause here to find the probabilities of these three events occurring.

And for question two, first, complete this probability tree and sample space and then use your completed diagram to consider which of these two events are more likely to happen.

Pause now to do this question.

Probability plays a massive part in well, so many different video games.

I wish I could talk about all of them right now, but if I did, I'd be here for quite a while.

So instead, let's have a look at this one video game example.

Lucas is playing a video game and is able to control a dragon and a unicorn.

He encounters a cowboy that needs to be scared away.

The cowboy can be scared away if Lucas' character via dragon or unicorn hits the cowboy with an attack or the cowboy misses their attack or both of these things happening.

The probability of an attack hitting its target is shown by the accuracy stat as you can see below each character.

By constructing a probability tree for dragon versus cowboy and a different probability tree for unicorn versus cowboy, pause here to figure out which of Lucas' characters is more likely to scare away the cowboy.

And an amazing job on drawing and analysing all of those probability trees.

Here are the answers for question one.

The probability of getting two heads is 0.

2025.

The probability of at least one heads is 0.

6975, and the probability of exactly one tails is 0.

495.

And for question two, pause here to check if your probability matches the correct one on screen and know that event two was more likely by a total of 21.

73%.

And for question three, pause here to check if the sample space you constructed contains the same outcomes and probabilities as the one on screen, even if the order and layout of your probability tree differs from this one.

Also, the probability of the cowboy being scared away from the dragon is 0.

88 or 88%.

And pause here to check the same for the unicorn, which had a 0.

9 or 90% chance of scaring away the cowboy.

Okay, by comparing the event of the probability that the character hits its attack or the cowboy misses its attack across both the dragon and the unicorn, we can see that the unicorn is 2% more likely to scare away the cowboy.

And a very well done for identifying that if you did, and for the amazing effort in drawing and understanding these complex two-layered probability trees in a lesson where we have looked at one-stage trials on probability trees where the sum of probabilities on all of the branches sum to one.

We've also looked at two-stage trials and probability trees where all groups of branches that meet at a point or create a sample space also must sum to one.

And where we can use these two-layer trees to calculate both the outcomes of a two-stage trial and find the probability of any event by finding the sum of each outcome in that event.

Once again, thank you for joining me today.

I hope to see you again in the future for another round of maths.

But until then, have a great day and goodbye.