Lesson video

Hello, everyone, welcome and thank you for joining me, Mr. Gratton, in this our next lesson on probability.

In this lesson we will look at a type of tree diagram, the probability tree, and find out how we can use one to display the outcomes of a trial and how we can present events on a more efficient type of probability tree.

Pause here to check some of the keywords that you'll need to be familiar with for today's lesson.

And of course, our main key phrase, the probability tree, has branches where each one shows a possible outcome from an event or stage of a trial, along with the probability of that outcome occurring.

Let's start off with the most basic of probability trees, ones that show the outcomes of a trial.

So Sofia recalls that an outcome tree can show all of the outcomes of a trial happening.

At the beginning of an outcome tree at the point before any of the branches are drawn, well, this shows the point in time before any trial has taken place.

Each branch of an outcome tree represents one of the possible outcomes from that trial taking place.

For example, on this spinner, there are four possible outcomes represented by the four sectors of that spinner.

Therefore, there are four branches to this outcome tree.

Sofia asks, "Is it also possible to represent probabilities on a different type of tree?" Truthfully, a probability tree looks exactly the same as an outcome tree with one little extra thing.

Each branch still represent an outcome, yes, but the probability of that outcome happening is also labelled on each branch along with the name of the outcome itself.

The arrow on that spinner will land on each of the four sectors approximately once every four spins, and so the probability of the spinner landing on any individual outcome is one over four or one quarter.

So to make this outcome tree a probability tree, we look at each outcome at the end of each branch and label each branch with the probability of that event happening.

For example, the probability of the spinner landing on outcome A is one quarter.

For example, we can interpret this probability tree as saying the probability of the spinner landing on outcome A is one quarter because the probability one over four is labelled on that branch ending with the outcome A.

This is also true for outcome D.

We can see that outcome D also has a probability of one quarter of happening.

It is important to note that every outcome from a trial has to be shown on a probability tree.

It will have less practical use if only some of the outcomes are represented by branches.

A probability tree can still be constructed even if a trial has outcomes that are not unique, meaning that some of the outcomes repeat themselves.

For example, in this spinner, the outcome one appears twice and the outcome two appears three times.

Each individual occurrence of an outcome is represented on a probability tree.

Six equally likely outcomes means six branches as you can see on this outcome tree.

For this to become a probability tree, each branch is labelled with one over six to represent the probability of each individual outcome happening.

For this check, we'll look at a problem in many parts.

For part one, Sofia wants to show each outcome and probability from this spinner on a probability tree.

How many branches should the probability tree have? Pause now to look at the spinner and come up with your answer.

And the answer is three branches, one for each outcome shown by the three equally likely sectors.

For part two, here's the probability tree with two of the three outcomes written on the branches, what would go inside this box? Pause now to look at the spinner and think of your answer.

An outcome goes on the end of a branch and since there are two individual outcomes for B, there will be two branches with B on them.

Next up, what would go inside this box now directed towards the centre of the branch rather than the end of it where the outcomes go.

Pause now to think what else you need for a probability tree.

The outcome goes at the end of the branch.

However, the probability of that outcome goes onto the branch itself.

Whilst there's no perfect correct location, the probability is usually written close to the middle of the branch.

And finally, here's a different check, one that we'll see again later.

This probability tree shows the probability of a spinner landing on a certain outcome.

Pause here to think about what is the probability of the a landing on the outcome B? So we look at the branch ending in B and identify the probability of that branch.

That probability is one fifth, and so the probability of the spinner landing on B is one over five.

Sofia thinks that probability trees are a lot of effort to draw.

For the ones that we've just looked at, I think I agree, especially when trying to represent the outcomes of a spinner like this where five out of the six outcomes are A.

Isn't having a probability tree with six branches where five of them are identical, a bit of a waste? Well, Jun suggests that there may be a better way of collecting these like branches together.

Here is what our original probability tree looked like.

This one shows each individual outcome of that spinner, but we can also construct a different type of probability tree, one whose branches only show the unique outcomes.

In this case, the unique outcomes are A and B.

It doesn't matter that A is repeated many times, A and B are still the two only unique outcomes.

But because there are more outcomes for A or more accurately, a larger proportion of the spinner shows the outcome A, the probability of each unique outcome will be different.

The probability of A and B will be different on this new type of probability tree.

The probability of the spinner landing on B remains the same at one over six.

This is because there is only one individual outcome of B out of the six total outcomes.

But there are five individual outcomes for A.

So we can add together these probabilities at each of these branches.

One six added five times gives an answer of five sixths.

And so the probability that goes on the branch ending in A is five sixths.

The probability of the spinner landing on any of the A's is five outta six.

Here's the same check from earlier.

This probability tree represents a spinner, but this time, what is the probability that the spinner lands on C? Pause now to have a look at this tree and think of your answer.

Because there are three outcomes where the spinner will land on a C, the total probability of landing on any of the Cs is three out of five.

We can collect all three individual branches representing C into one branch with the sum of these probabilities shown on it.

So one fifth plus one fifth plus one fifth equals a three fifths that you see on this modified probability tree.

For this first practise task, we've got two probability trees that show the same trial.

The first tree shows each individual outcome whilst the second shows the unique outcomes where the probabilities of repeated outcomes are collected together.

Pause here to complete tree 2 using the information from tree 1 and then complete the sentence in part B.

For question two, complete the probability tree by filling in the boxes showing probabilities and outcomes.

And for part B, use your completed tree for part A to complete this second probability tree showing the probabilities of unique outcomes instead.

Pause now to complete both of these probability trees.

So far we've used a larger probability tree and collected like branches together to construct a more efficient tree that only shows the unique outcomes.

For question three, let's do the opposite.

Pause here to use the more efficient tree 1 to complete tree 2.

Whilst there are multiple correct answers to this question, each correct one has branches that are simply rearrangements of each other.

And finally, question four, time for you to construct your own tree from scratch to show a more efficient probability tree that shows only the unique outcomes from this spinner.

Once you have constructed your tree, compare the probabilities at each branch and outcome and match the outcome to each of these outcome spinners shown in part B.

Pause now to do parts A and B of this question.

Great job, everyone, on all of your work so far when analysing these probability trees.

Here are the answers.

The probability of the spinner landing on a T is three quarters or 0.

75 or 75%.

The probability of outcome V happening is one quarter instead.

And for question two, pause here to check if your trees match the correct ones shown on screen.

And for question three, each branch should be labelled with the probabilities one over seven.

And as long as you've labelled one W, one X and three Ys, you could have placed them in any order.

However, the order on screen is systematic, which we like because the outcomes have been placed in alphabetical order.

Here's an example of a tree for question four drawn efficiently by only showing the unique outcomes A, B, and C with the matching of one eighth for A, five eighths for B and two eighths for C respectively.

And the three outcome spinners represent A, then C, then B in that order.

And now that we've looked at probability trees that show each and every outcome from a trial or at the very least each unique outcome from a trial, well what about specific events from a trial instead? Let's have a look.

Jun's at the local fair and plays a game to win a prize.

If he spins that spinner and lands on a double digit number, he wins a prize.

Otherwise he leaves empty handed.

Jun says he can show the outcomes of the game on a five branch tree like so.

Notice how each unique outcome is only shown once.

Therefore, the probability of each outcome will be different, such as the probability of the spinner landing on a four being the greatest at three over nine.

For Jun to win, the outcome has to be a double digit number.

And so we have to sum together the probabilities of the two branches whose outcomes are double digit.

These are 10 and 11, each with probabilities two ninths.

Giving a total of four ninths for Jun will win.

This tree is 100% correct, but Sofia again thinks it's a waste of time.

Is there any way of making this tree more efficient and simple to draw, seeing as we only really care about the probabilities of two of the five branches? Jun suggests a tree that doesn't show the outcomes, rather two events from the trial instead.

Event one being represented by one branch and shows the probability of Jun winning and event two represented by the other branch and shows the probability of Jun losing instead.

And so this rightmost tree shows these two opposite events.

I know we haven't put any probabilities on it yet, but doesn't it look much more simple than the abomination that is this five branch tree on the left.

Jun wins if the spinner lands on a double digit number.

And so we sum the probabilities two ninths and two ninths to get four ninths.

So this top branch representing the event that Jun wins has a probability of four ninths shown on it.

Furthermore, Jun loses if the spinner doesn't land on a double digit number, which are the outcomes for seven and eight.

The probability of this happening is five over nine, which we can show on this second branch of this event probability tree.

Remember, a probability tree is most helpful when all outcomes are represented on it.

All nine outcomes are represented by either the events of Jun winning or Jun losing.

And so this probability tree is complete.

Okay, here's a set of checks with a hefty context.

So listen carefully to all of the details.

A spinner is spun to see if Sofia, Laura, or both win some sweets.

If the spinner lands on a seven, Sofia and Laura both draw and win one sweet each.

If this spinner lands on a number greater than seven, then Sofia wins both sweets.

If this spinner lands on a number less than seven, then Laura wins both sweets.

Sofia wants to construct a probability tree.

This tree will show the events of her winning or Laura winning or the game ending a draw and both winning some sweets.

What is the minimum number of branches that this tree could have? Pause now, to answer this question.

This event probability tree could have three branches.

One for Sofia winning, one for Laura winning, and one for them drawing.

Remember, this probability tree only shows events rather than the individual outcomes of that spinner.

So far, this tree has two of the three events already labelled, which remaining event goes in that box? Pause now to look at the events and answer this question.

We've got Sofia winning, them drawing, What's left over is Laura winning.

This is the only one of the three events not already mentioned.

Next, check, what probability goes into this box? What is the probability of Laura winning? Pause here and use the spinner to inform your answer.

And the answer is three ninths.

There are nine outcomes in total, and there are three outcomes four, four, or four that are less than seven, which is the criteria for Laura winning.

Next up, what probability goes into this box instead? Pause to consider the conditions for which Sofia wins.

Five ninths, Sofia has fully constructed the probability tree with these three events shown what is the most likely event from this trial? Pause now to answer this question, Sofia winning is the most likely event at five out of nine compared to the lower three out of nine for Laura winning and one out of nine for them drawing.

Okay, onto the practise task.

For question one, the event Izzy is looking at on this spinner is the event that the spinner lands on a prime number.

Using spinner A and the probability tree of outcomes complete the probability tree of the events of this trial.

Pause now to consider which of those outcomes are prime numbers in order to complete that right hand probability tree showing the two events of this trial.

And for question two, the event that Aisha considers from a spinner with 12 sectors is the event that the arrow will land on a square number.

By looking at the first probability tree, identify the square numbers on the spinner, then use this first tree to complete a probability tree showing the event that if the spinner lands on the square number or doesn't land on the square number.

Pause now to do parts A and B to this question.

Sticking with question two, Aisha also considers a different event from the spinner, that being the event that the spinner lands on an even number.

By looking at the first probability tree, again, identify the even numbers on the spinner, then use this first tree to complete a different probability tree, showing the event that the spinner lands on an even number or doesn't land on an even number.

Pause now to do part C and D to question two.

For question three, Lucas spun the spinner and landed on a five.

Andeep needs and outcome greater than five in order to beat Lucas.

Complete this probability tree to show the events of who could win.

Pause now to consider which outcomes of the spinner would result in a score greater than Lucas's score.

Great work, everyone.

Onto the answers for question one.

The probability of the spinner landing on a prime number is three outta five, whilst landing on a non-prime number is two outta five.

And here are the answers to question two.

Pause here to see if your answers to parts A, B, C, and D match those on screen.

For the tree in part D, the two outcomes could have been placed in either order as long as the probability still matches to the correct outcome.

For question three, the probability that Lucas and Andeep draw is three out of 10, whilst the probability that Andeep alone wins is two out of 10.

Lucas has a much better chance of winning than Andeep does.

And that is some amazing work, everyone, on an introductory lesson to probability trees where we have looked at showing the individual outcomes to a trial on a probability tree, where each and every branch shows one individual outcome.

We've also made some of these trees much more efficient by collecting the probabilities of like outcomes together onto one branch.

And lastly, we've looked at showing the probabilities of events rather than individual outcomes onto a probability tree.

And remember, for every single type of probability tree that we've covered, every outcome to a trial must be represented in some way onto the tree itself.

Thank you so much one and all for joining me for this lesson on probability.

I look forward to seeing you once again as we delve deeper into probability trees and other different types of probability representation.

But until then, have an amazing rest of your day and goodbye.