Lesson video

Hello everyone, I'm Mr. Gratton.

And thank you so much for joining me in today's lesson on probability.

Grab a pencil and a ruler as in today's lesson, we will be looking at many different representations for showing and calculating probabilities and see which representations may be more or less useful in different circumstances.

Pause here to check the keywords that we'll be using today and see if you are familiar with the representations we'll be covering such as Venn diagrams and probability trees.

First up, let's look at comparing where and how both outcomes and probabilities are shown across different diagrams. Sofia thinks it is possible to take the outcomes of this spinner and show the probabilities on a list of outcomes, a frequency table and a probability tree.

Is it possible to transfer the outcomes and probabilities from one of these diagrams onto another? Jun thinks so.

Each diagram shows the outcomes and probabilities from this same trial, that spinner that we saw earlier.

So surely the same information is shown across all of those diagrams just may be in a slightly different way from each other.

For example, each of the unique outcomes we can find in the list is shown by a row of the frequency table and a branch of the probability tree.

Three unique outcomes, win, lose and draw.

So three rows of the frequency table and three branches on the probability tree.

Furthermore, we can also spot how the probabilities are represented consistently across all diagrams. For example, there are two win outcomes in the list each at a 1/6 probability of happening, giving a total probability of the spinner landing on a win at 2/6.

This same 2/6 is seen on both the frequency table and probability tree on the same row and same branch as the win outcome on both of these diagrams. For this first check, we have a frequency table and a probability tree.

Sofia will spin the spinner once.

She begins to show the probability that the spinner will land on each unique outcome using both of these diagrams. But they are both incomplete.

Using the information that you do know on one diagram, complete the missing information on the other.

Pause here to think or discuss.

What values would A, B, C, and D take? A and B are the probabilities of lemon and kiwi as shown on the frequency table.

C is the probability of strawberry as seen on the tree.

And for D, because 3 of the 11 outcomes are lemon, 2 outta the 11 outcomes are kiwi, and so kiwi has a frequency of 2.

And similarly for this check, we have a frequency table and a list of individual sweets in a box of 11 sweets.

Jun will randomly choose one sweet from the box and begins to create a list and table to show the probability he chooses a certain sweet.

Use the information that you do know to complete the missing information from both diagrams. Pause again here to think or discuss.

What values would A, B, C, D and E take? The answer for A requires no interpretation of either diagram.

The question itself says that there are 11 sweets, so the total frequency of sweets is 11.

For B, the probability of Jun selecting a fudge sweet is 2 out of 11 and there are 2 fudge sweets in total.

You can see this from both the table and the list.

So by that same logic, if the probability of selecting a toffee sweet is 6 out of 11, then there must be six toffee sweets in frequency.

And for C, C is a subtraction of frequencies.

11 sweets in total subtract 6 toffees subtract 2 fudge leaves 3 earwax remaining.

And for D, if 3 of the 11 sweets are earwax then the probability of selecting one of them is 3 out of 11.

And for E, there is one missing sweet.

Using the frequency table, we can identify it was an earwax sweet.

Okay, onto the first set of practise questions.

For the first question, use the list of outcomes to complete both the frequency table and probability tree.

Pause now to look at the whole list of outcomes and see how you can make probabilities from them and how you would represent those probabilities on both the frequency table and probability tree.

And for question two, complete the probability tree by identifying the corresponding information in the frequency table that shows the same trial.

Pause now to take all of this information and represent it onto the tree.

And for question three, both diagrams the table and the tree show the same trial, but both are incomplete.

Complete both of them and generate a complete list of outcomes.

A list of every party hat that is described by its colour.

Pause now to inspect what information you do know from the table and the tree and see how you can represent it on the opposite diagram.

And here are the answers for question one.

The missing outcome on the table was N or no move and the missing outcome on the tree was forward or F.

On the frequency table, the missing frequencies were forward had three, and no move or N had four for a total frequency of nine outcomes.

And the probabilities on both the table and the tree are forward had a probability of 3/9.

Backwards had a probability of 2/9 whilst no move or N had a probability of 4/9.

Notice how the order of forwards, backwards, and no move is different between the table and the tree.

This is absolutely okay, but make sure that you are paying attention so that you do not assign the wrong probability to each outcome whilst you're transferring information from one diagram to the other.

And here are the answers to question two.

Once again, the part that might have caught you out was that the order of outcomes on the tree were different to those in the table.

Pause here to check that your tree matches this fully correct one.

And finally for question three, a very well done, if you are able to take all of the information from one diagram and correctly transfer it to the other and vice versa.

Pause here to check that your table, tree, and list of outcomes match the correct ones shown on screen.

Whilst the list could have been in any order, it is best to write it in a systematic way, either in an order such as alphabetically or an order that matches either the table or the tree.

All the representations that we've looked at so far have covered trials with at most four unique outcomes, but are these diagrams still practical in the way that we've already seen them if the trial has far more than four unique outcomes? Well, let's have a look.

This spinner has outcomes from one to seven written on it in varying frequency.

Four seems to appear most frequently with four individual outcomes on the spinner.

We want to consider the probability of the event that the spinner lands on a prime number, but Sofia thinks that it's absurd that we would need to draw a tree with seven branches or a table with seven rows simply to show the probabilities of this one event.

Jun however, has a good solution, consider a probability tree with only two branches, one to show the probability of the prime outcomes and the other showing the probability of the non-prime outcomes.

We can take the outcomes of this spinner and display them in an outcome table.

A table showing the outcomes grouped into the prime and non-prime events.

The ones are not prime.

The two is as are the threes.

The fours are not prime, but the fives are.

The six is not prime, but the sevens are.

There are a total of 15 outcomes, and so the probability of getting a prime number or P bracket prime is 8 outta 15, whereas the probability of not getting a prime number is 7 outta 15.

Now that we know the probabilities for these two opposite events, we can represent them on a two branch probability tree like so with their probabilities being the same as we gathered from the outcome table.

Okay, onto a check.

For this check, the outcomes of a trial have been grouped into both prime and non-prime outcomes of an event in this outcome table.

Using this information, what probability goes on the branch currently labelled x? Pause now to look through all of the information in the outcome table and come to a conclusion as to what your probability will be.

There are 9 prime numbers out of a total of 13 cards, so the probability of picking a prime number is 9 outta 13.

An extra card is added to the deck a three as a prime outcome of the prime number event.

Which of these statements are correct? Pause now and consider how this extra card will affect the number of outcomes in and the probability of each of these two opposite events.

And so notice how the number of prime outcomes has increased because we've added one extra prime number outcome to the prime event, but the number of non-prime outcomes has stayed the same because we haven't added any non-prime numbers to our deck.

However, both probabilities have changed as there are now 14 outcomes total rather than 13, and so the probabilities are 10 outta 14 of being prime and 4 outta 14 of being non-prime.

Then diagrams are great for showing which outcomes do and do not belong to a group of events.

For example, for a trial where a 10 sided dice is rolled, the outcomes 1 and 9 compose the events that the dice lands on an odd number that also is not prime.

But as Sofia asks, if we only want to focus on one event, how can we When the two circle Venn diagram inherently shows two different events? We can because every single outcome in the left-hand circle is a prime number.

This includes the middle of the Venn diagram as the middle is still inside that left hand circle.

There are four prime numbers on the dice.

Yes, three of those prime numbers are also odd because they're in the middle, but when we're only looking at the event of landing on a prime number, we do not care if it's the middle or on the left because we only care about the fact it is somewhere inside that left-hand circle.

Therefore, the probability of selecting a prime number is 4 outta the 10 total outcomes on the dice.

Furthermore, to find the probability of the event that the dice lands on a non-prime, we look at every outcome outside the left-hand circle.

This means looking at the outcomes in the right-most region, not in the middle by the way, but also in the space outside the two circles.

There are six outcomes in these two areas, 1 and 9 in that rightmost region and the 4, 6, 8, and 10 in the space outside the two circles, because there are a total of six outcomes that are not prime out of the 10 in total on the dice, the probability of selecting a non-prime number is 6 outta 10.

As with before now that we've explicitly considered the probability of the event that the dice lands on a prime and its opposite event the dice landing on a non-prime we can take those two probabilities of these two events and represent them on a probability tree like so.

Okay, for this check we can see by the probability tree that the events we are considering are the events that the outcome is a multiple of three and it's opposite event, not a multiple of three.

Pause here to use the Venn diagram to find the probabilities of these two events shown by X and Y.

Right, to focus on all of the multiples of three let's have a look at that whole left-hand circle.

There are five outcomes in that circle on the Venn diagram, two in the leftmost part, the 3 and the 9 and three in the middle, the 6, 12, and 18.

All five of those outcomes are multiples of three.

In total there are nine outcomes in this Venn diagram, and so five out of the nine outcomes are multiples of three.

Its opposite event has a probability of four out of nine as shown by the 4, 8 and 16 on the right-most region and the five on the region outside of the two circles.

Next up, the probability tree for this trial is correctly labelled, but the Venn diagram has one outcome that's missing.

Pause here to inspect the Venn diagram and compare its outcomes to the probability tree showing the probabilities of the events multiples of five and not multiples of five and suggest one possible outcome that could be added to the Venn diagram to make it accurate.

The probability tree says that 9 out of the 12 possible outcomes are multiples of five.

However, only 8 out of the current 11 outcomes in this Venn diagram are multiples of 5 and therefore any multiple of 5, odd or not can be added to the Venn diagram to make it correct.

Sofia is visibly distressed at the thought of writing down all of these outcomes again in a Venn diagram, especially some of its regions are too small and the numbers just won't fit in them.

However, we can spare her pain and hopefully some of yours too, as there is a slightly different type of Venn diagram that we can use.

Let's have a look at what I mean.

Rather than showing every single outcome, some Venn diagrams just show the number of outcomes that belong to each event.

For example, there are two even numbers that are square, the zero and the four, so 2 goes into that middle region.

2 representing those two outcomes.

There are also two odd square numbers.

The 1 and the 9, therefore 2 goes in that left most region.

Furthermore, four outcomes are even and non square.

These are the outcomes, 2, 6, 8, and 10.

And so the quantity 4 goes in the rightmost region to represent even numbers that are not square.

And finally, there are three outcomes that are neither even nor square, and so the quantity 3 goes outside the two circles to describe the numbers 4, 5 and 7.

As with before, whilst a Venn diagram categorises outcomes into two events, we only care about one, the event that the spinner lands on a square number.

Therefore, we only want to pay attention to the numbers of the outcomes in that left-hand circle, there are two plus two equals four outcomes in total that are square as we can see from the outcome table.

We know that there are 11 slides to the spinner in total.

We can also prove that by two plus two plus four plus three, the four numbers inside the Venn diagram, for a total of 11 outcomes in the spinner.

Therefore, the probability of getting a square number is going to be 4 over 11.

And for this check pause here to use the correctly completed Venn diagram to figure out how many outcomes are even and how many outcomes are odd.

This is another case of don't let the order of the outcomes on both diagrams catch you out whilst even is the left hand column of the table, it is the right-most circle in that then diagram and therefore we add together the two numbers in that even circle the 4 plus the 9 to give you 13 outcomes.

For odd we look at everything that is outside of that even circle, which is the 6 and the 15, and so 6 plus 15 outcomes is 21 outcomes.

And lastly, using either the Venn diagram or the outcome table, pause here to find out the probability of the event choosing an odd outcome.

We know that there are 21 odd outcomes and 13 plus 21 is 34 total outcomes, and so the probability of getting an odd outcome is 21 out of 34.

Okay, for question one of this practise task, take the individual outcomes of this Venn diagram and place them into the correct event of this outcome table.

When you've completed the outcome table, find the probability of these three events occurring.

Pause now to do all of this question.

And next up for question two, a spinner has all of these outcomes.

We want to consider the event that the spinner lands on a square number.

Complete the outcome table by grouping the outcomes of the spinner into the events square and not square.

Then write down the number of outcomes, not the outcomes themselves in the events square and even in that Venn diagram and use either of these two representations to complete the probability tree.

For part B, circle or highlight the outcomes or branches in each of the three diagrams required to calculate the probability of the event not square.

Pause now to do this question and think about which of the two diagrams, the table or the Venn diagram will be easier to use to complete the probability tree.

And finally, for question three, the Venn diagram is correct.

Some outcomes, however, in the outcome table are missing.

Use the Venn diagram to suggest possible outcomes that make the outcome table correct.

Once the table is correct, use that information to complete the probability tree.

Pause now to do question three.

Alright, let's have a look at some answers.

For question one here is the correct table grouping the outcomes into the events vowels, A, E, I, O, and U and non vowels or consonants, the L, H, F, and C.

The probability of the event vowel is 5 out of 9.

Whilst the probability of the event not vowel is 4 out of 9.

Out of the nine outcomes, six of them, the A, E, I, L, H, and F are made from only straight lines.

So the probability of the event made from only straight lines is 6 out of 9.

And for question two, pause here to check that if your diagram matches those on screen.

And note, 11 out of the 17 outcomes are not square numbers.

And finally, the outcome table is missing one even cube number such as 8, one odd cube number such as 1 or 27, and one odd number that is also not cube, for example, 5.

And for the probability tree, 10 out of 18 is the probability of the event cube whilst 8 out of 18 is the probability of the event not cube.

Great work everyone on your attention to detail when comparing across all of these probability representations.

In this lesson, we've looked at how the outcomes and probabilities of a one stage trial can be shown using different representations including frequency tables and probability trees.

And when considering events from a trial, we can show the probabilities of events and their opposite events on outcome tables, probability trees, and different types of Venn diagram.

Once again, thank you all for your hard work in bringing together and comparing so many different elements of probability we've looked at so far.

I've been Mr. Gratton.

So until our next maths lesson together, enjoy your day, stay safe, and goodbye.