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.

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

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 1 of our trial.

And here we can see the probabilities of each outcome of specifically stage 2 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 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.

5/8 is the probability of the spinner landing on an A, and 1/2 is the probability of the coin landing on heads.

So 5/8 times 1/2 equals 5/16.

The probability of both the spinner landing on an A and the coin landing on heads is 5/16.

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 3/8, and the probability of landing on tails is 1/2, therefore 3/8 times 1/2 is 3/16.

And so the probability 3/16 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 1/2 which is 5/16.

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

Remember, the sum of all probabilities of each sample space must sum to 1 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, 5/16 plus 5/16 plus 3/16 plus 3/16 equals 16/16.

Since 16/16 equals 1, 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 1 and 2 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 1 landing on Y and spinner 2 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 multiply 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 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 to? All probabilities in that sample space should sum to 1.

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? 32/100 is correct, but simplifying to the equivalent 8/25 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 32/100 times 3/5 equals 96/500, or if simplified, 24/125.

Okay, for question 1 of this next set of practise tasks, the probabilities of landing on R and G on spinners 1 and 2 are shown.

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

And for question 2.

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 3.

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 2, the probability of outcome CQ is 2/40.

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

And for question 3, 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 9/16 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 1 and X on spin 2, has a probability of 20/81.

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 16/81.

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

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

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 56/81 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 20/81 plus 20/81 equals 40/81 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, 20/81 plus 20/81 plus 25/81 equals 65/81.

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

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 0.

3, which equals 0.

21.

Similarly, the probability of even then even is 0.

3 times 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 1, using this probability tree, pause here to find the probabilities of these three events occurring.

And for question 2, 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 1.

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

73%.

And for question 3, 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 1.

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 1, and where we can use these 2-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.