Lesson video

Hello there.

You made a great choice with today's lesson.

It's gonna be a good one.

My name is Dr.

Rollinson, and I'm gonna be supporting you through it.

Let's get started.

Welcome to today's lesson from the unit of conditional probability.

This lesson is called Algebra in Tree and Venn Diagrams, and by the end of today's lesson we'll be able to apply our knowledge of probability equations to tree and Venn diagrams to solve problems. Here are some previous keywords that will be useful during today's lesson, so you may want to pause the video if you need to remind yourself what any of them mean and press play when you're ready to continue.

The lesson is broken into three learning cycles, and we're going to start by looking at algebraic relationships on Venn diagrams. We can identify probability relationships across a range of different Venn diagrams. For example, here we have a probability Venn diagram where each number that is written inside the Venn diagram represents a probability.

This particular probability Venn diagram shows a probability of a random spinner which has spun once landing on either a square number or a triangular number, or it can land in a number that is both square and triangular or a number that is neither square or triangular.

Currently, the probabilities are expressed as unknowns with letters from W to Z, but even though we don't know what any of those probabilities are, we can still think about the relationships between them.

What probability relationships can we find from just the information we provided here? Perhaps pause video while you think about that.

Think about how these probabilities are related to each other and press play when you're ready to continue.

Well, there are quite a few different relationships between these probabilities, but let's take a look at some together.

Jacob says, "All the probabilities from a trial sum to one.

Therefore, W plus X plus Y plus Z would equal one." There we have an equation.

Jun says, "The probability that the spinner lands on a square number would be equal to W plus X." And Laura says, "The probability that the spinner lands on a number that is not a triangular number will be equal to W plus Z." And there are plenty of other relationships we can explore these probabilities as well.

I wonder if you got any ones? Currently, these unknown probabilities are expressed in terms of different letters, but what if these probabilities were all expressed in terms of the same letter? For example, in this probability Venn diagram, we have the four probabilities, all unknown but expressed in terms of X.

The fact that they're all expressed in terms of X means we can see more relationships between them.

We can see that two of probabilities are equal to each other, they're both equal to X.

One probability is twice that, that is 2X, and there's another probability which is twice that again, which is 4X.

We can use these relationships and other ones that we thought about earlier to construct and solve equations which can help us to find probabilities.

For example, we know that all probabilities in this Venn diagram would sum to one.

Therefore, we can construct this equation here.

That is 2X plus X plus X plus 4X equals one.

And now we have this equation, we could simplify it to 8X equals one and solve it by dividing both sides by eight to get X equals one eighth.

And now we know the value of X, we can substitute that into the expressions on the Venn diagram to get these probabilities here, and we can use these to find probabilities of particular events.

For example, the probability that the Venn diagram lands on a square number would be the sum of two eighths and one eighth, which would be three eighths.

So let's check what we've learned.

Here, we've got a probability Venn diagram where each of the probabilities are expressed in terms of X.

You've got an equation on the left hand side, which is only partially complete.

You've got the right hand side which says it's equal to one.

Could you please complete the equation by writing down what should be on the left hand side? Pause the video while you do it and press play for an answer.

The sum of those four probabilities would be equal to one, so we can express it as an equation like this.

We could even simplify it to 10 X plus 0.

4 equals one.

Could you then please find the value of X and express it as a decimal? Pause the video while you do that and press play for an answer.

Well, if 10X plus 0.

4 is equal to one, 10X would equal 0.

6.

We can divide both sides by 10 to get X equals 0.

06.

So now we know the value of X, could you please use that to find the probability that the outcome is an integer? Pause the video while you do that, and press play for an answer.

You can get this probability by substituting 0.

06 in for wherever you can see X and adding together the probability for integers.

You do that, you should get 0.

58.

Let's take a look at another example now.

Here we have a probability Venn diagram where an outcome can either be prime or not prime, and it can either be over 20 or not over 20.

All the probabilities are unknown, and they're expressed in terms of two different letters: X and Y.

Now the fact that the probabilities are not all expressed in terms of the same letter means we cannot create an equation where we just sum all these probabilities to equal one whole.

Therefore, we may need some other information in order to calculate some of these probabilities.

If given the probability of a certain event occurring, we can find the value of an algebraic expression by identifying relevant regions.

For example, here we are given that the probability and outcome belongs to the union of prime and over 20 is 0.

9.

The union of prime and over 20 means the outcome could either be prime or it could be over 20, or it could be both.

Let's calculate the value of the probability that an outcome is not over 20.

If we look at the Venn diagram, we can see there are two outcomes that are not over 20.

One has a probability of 2X, and the other has a probability of Y.

We need to work out the value of each of those unknowns.

Let's start by working out the value of X.

If we look at these three shaded regions here, the union of prime and over 20, it's probability is 0.

9 and is also the sum of all three of these probabilities that are shaded.

Therefore, we can create an equation that looks something a bit like this.

The sum of the three probabilities in those shaded regions is equal to 0.

9.

We can solve this equation by first simplifying it and then rearranging it to get X equals 0.

2.

So we know X is equal to 0.

2.

We can then start to think about the value of Y.

If the sum of these three regions is not 0.

9, that means the remaining probability that is outside that region must be 0.

1 because all the probabilities must sum to one in all.

So Y is equal to 0.

1.

We have everything we need now to work out the probability that an outcome is not over 20.

It's a sum of the two regions that are not inside the right hand circle, the 2X and the Y, and the probability of an outcome not being over 20 will be the sum of these two probabilities.

So now we know the value of X and Y.

We can substitute those in to get two multiply by 0.

2 plus 0.

1, and that would give 0.

5 for the probability.

So let's check what we've learned.

Here you've got a Venn diagram with probabilities, unknown and expressed in terms of X, Y and Z.

Which algebraic expressions sum to make the probability for an outcome is over 10? Pause the video while you write it down, and press play for an answer.

The answer is 2X and 3x.

If the probability of an outcome is over 10 is equal to 0.

35, could you then please find the value of X? Pause the video while you do that, and press play when you're ready for an answer.

You could do that by creating this equation here.

And then you could simplify it and rearrange it to get 0.

07 for the value of X.

So let's work out another value now.

If the probability that an outcome is a square number is equal to 0.

4, please could you find the value of Y? Pause video while you do it, and press play when you're ready for an answer.

Well, the two outcomes that are square numbers have probabilities, 3X and Y.

You know the value of X.

You can substitute that in, and you know that those two probabilities should sum to 0.

4, so you can create this equation here.

You can simplify it to get Y equals 0.

19.

So far, we've been looking at Venn diagrams that show us probabilities, but some Venn diagrams show us frequencies.

It is also possible to calculate probabilities given a frequency Venn diagram rather than one that's shown probabilities.

So for example, here we have a Venn diagram where each the frequencies are expressed as unknowns from W to Z.

The total frequency of this Venn diagram would be the sum of W, X, Y and Z.

Therefore, we can use this expression to calculate some probabilities.

For example, if we were trying to find the probability that an outcome would satisfy the intersection of multiples of five and not cube.

In other words, the outcome is a multiple five, and it's also not a cube number.

We'd be looking at this section of the Venn diagram.

The numerator therefore would be Y 'cause a frequency of outcomes in that part of the Venn diagram is Y, and the denominator would be the sum of W, X, Y and Z because that's a total frequency.

So we can express the probability like this: Y over the sum of W, X, Y, and Z.

Whatever the frequency is for the event we're trying to find over the total frequency in the Venn diagram.

So let's check what we've learned.

Here you've got a frequency Venn diagram where some of the frequencies are expressed in terms of X.

Could you please write an expression for the total frequency in this Venn diagram? Pause the video while you do it, and press play for an answer.

The answer is eight X plus four.

<v ->Could you now please write an expression</v> for the probability that an outcome belongs to the intersection of A and B? Pause while you do that and press play when you're ready for an answer.

The answer is X over 8X plus four.

So how about if I told you the value of that probability? Given that the probability that an outcome belongs to the intersection of A and B is equal to 5 over 42? Could you please write an equation involving X? Pause video while you do that, and press play for an answer? Well, in the previous question, you expressed this probability in terms of X and now you know it's equal to 5 over 42.

So you can write an equation that says exactly that.

So could you please solve this equation to find the value of X? Pause video while do that, and press play when you're ready for an answer.

If you rearrange this equation, and then solve it, you'll get the value of 10 for X.

And once you know that X is 10, you can work out other things.

For example, if X is 10, it means a total frequency in a Venn diagram is 84.

If you substitute 10 into each value of X and add 'em together, you get 84.

And there are 30 outcomes that satisfy event B that comes from doing X plus 2X.

Substitute 10, you get 10 plus 20.

We could also use a frequency Venn diagram to find the conditional probabilities of events.

For example, if we want to define the probability of A given B, let's think how we'd go about doing that.

We know that B has happened, so rather than the denominator be in the sum of all four of these frequencies, the denominator would just be the sum of the frequencies in this shaded region here, because we are choosing an event from this region, the numerator is the value of the frequency that is in region A and also in region B.

So we can express that probability as X over the sum of X and Y.

So let's check what we've learned there.

Here we have a frequency Venn diagram where some of the frequencies are expressed in terms of X.

You are told the probability that D happens, given that you know C has happened, is three-tenths.

Based on that fact, which of these equations is correct to find the value of X? Choose either A, B, C, or D.

Pause video while you do that, and press play for an answer.

Well, each equation includes three-tenths.

So we are looking for is which one has the correct left hand side.

That would be C.

Okay, it's over to you now for task A.

This task contains four questions and here is question one.

Pause video while you do it and press play when you're ready for more questions.

Here are questions two and three.

Pause while you do these, and press play for question four.

And here is question four.

Pause while you do this, and press play for some answers.

Hare are the answers to question one.

Pause while you check these against your own, and press play for more answers.

Here are the answers to question two.

Pause while you check these, and press play for more answers.

Here are the answers to question three.

Pause while you check these, and press play for more answers.

And then when you do question four, you should get that X is 26 and Y is 50, which means when you complete the frequency tree, it would look a bit like this.

Well done so far.

Let's now move on to the next part of this lesson which looks at algebraic calculations on probability trees.

We can represent two-stage trials with unknown values on a probability tree.

For example, let's look at this scenario here.

A bag contains some marbles, but we don't know how many marbles it contains.

There are seven blue marbles and the rest are purple.

Hmm, that means we don't know how many purple marbles there are either.

A marble is randomly taken from the bag, returned and a second marble is then taken.

Let's see how a probability tree can help us visualise what's going on here.

Before we construct a probability tree, it can be helpful to consider some initial conditions algebraically.

We do not know the initial quantity of marbles, so we can call this unknown quantity X or we can use any other letter, but let's use X.

We know that there are seven blue marbles, but we don't know how many purple marbles there are.

However, it would be equal to the total number of marbles subtract the number of blue marbles.

So you can express that as X subtract seven.

We can use our initial algebraic conditions to write an algebraic expression for the probability of each outcome on the probability tree.

So in the first case, when the first marble is taken, the outcome could either be blue or purple.

We don't know what either of those probabilities actually are, but we could express them in terms of X.

The probability it's blue will be seven over X, where seven is the number of blue marbles and X, the number of marbles altogether.

The probability that it's purple will be equal to X subtract seven over X, where X subtract seven is an expression for the number of purple marbles.

And it's worth double checking that those two probabilities sum to one.

If you add them together, your numerator would be X subtract seven plus seven, which simplifies to X, and your denominator will just be X.

So you have X over X which is equal to one.

Good.

Then we can think about the second marble that is chosen, which again could either be blue or purple.

And because the first marble is returned to the bag, the probabilities of the second marble are independent of the outcome of the first marble.

Therefore the probabilities will be the same.

Now, if we know the value of one probability, we can use it to create an equation and calculate the value of X, which is the number of marbles in the bag.

For example, if we knew that the probability of getting a purple and then a blue was two ninths, we could use that fact and the algebraic expressions for purple and blue to create an equation.

It would look something a bit like this.

On the left hand side, we have the product of the probabilities of getting a purple and a blue.

On the right hand side, we have the value for that probability.

We could then use this to find the value of X, which is how many marbles are in the bag by first simplifying the equation and then rearranging it.

And we can see we have a quadratic.

So in order to solve this, we can make it equal to zero by rearranging it some more and then factorised it.

And then when we solve this, we get two values.

Either X is equal to 21 or it's equal to 21 over two.

But can you see a problem with one of those values? Remember, X is a number of marbles in the bag.

Therefore, since it can only be an integer number of marbles in the bag, the only valid answer answer is X equals 21.

Therefore, there are 21 marbles in the bag.

And once you know that, you can work out other probabilities on this probability tree by subsuming 21 in for the value of X.

So let's check what we've learned with a scenario for you to work with.

A bag contains 20 marbles.

Some of the marbles are green and the rest are yellow.

A marble is randomly taken from the bag returned and a second marble is taken.

Could you please find the values or algebraic expressions for A and B where A is a total number of marbles in the bag and B is a number of yellow marbles.

Pause while you do it, and press play for answers.

The total number of marbles is given to you.

That's 20.

So the number of yellow marbles will be 20, subtract X.

Let's now start to construct a probability tree for this scenario.

Please could you find the probabilities that are labelled C and D on that first layer of branches? <v ->Pause while you do it, and press play for an answer.

</v> C will be equal to X over 20, where X number of green marbles, and 20 is a number of marbles altogether.

And D will be equal to 20 subtract X over 20.

And one thing that's always worth double checking is whether or not those fractions will sum to one whole.

If you add together the numerator, you get 20 subtract X plus X, that just gives you 20, and the denominator is 20.

So yes, they do add up to one whole.

You are told that the probability of drawing two yellow marbles is nine over 25.

Could you please now use that fact to fill in the blanks in this algebraic equation? Pause video while you do it, and press play for an answer.

You would fill in those blanks with the probabilities of drawing a yellow each time.

And that would be something a bit like this.

So now we have that.

Could you please write this quadratic equation as simply as possible? Pause the video while you do that, and press play for an answer.

Okay, here's what you'd get.

If you simplify and rearrange this equation, you'd have X square subtract 40 X plus 256 is equal to zero.

So now we have this quadratic equation.

Could you please solve it to find the value of X? Pause the video while you do it, and press play for answers.

If you factorised this equation and then solve it, you'd get X equals eight or X equals 32.

Now, now that these numbers are non integers, so it's not immediately obvious whether or not one of them needs to be discounted.

Let's go back to our initial conditions.

Here we wrote down the number of marbles in total, the number of green marbles and the number of yellow marbles expressed in terms of X.

Which of these values of X that we found is suitable for this scenario here then? Pause video while you write it down and press play when you're ready for an answer.

The answer is eight because if you substitute 32 in for X, you would get that there are more green marbles in the bag than there are marbles altogether 'cause there are only 20 marbles in the bag.

So X must be equal to eight and therefore there are eight green marbles and 12 yellow marbles.

Okay, it's over to you then for task B.

This task contains two questions and here is question one.

Pause the video, while you do it, and press play for question two.

And here is question two.

Pause while you do this, and press play for some answers.

Okay, let's go through some answers.

Here's what you should get for question one.

Pause the video video while you check it, and press play for more answers.

And then question two, here's what you should get for part A.

Pause while you check this and press play for more.

And then for part B, you should get X equals 60.

And here's your working for Y.

Pause while you check it, and press play for the next part of the lesson.

You're doing great.

Let's now move on to the third and final part of this lesson, which is looking at algebra, ratio, and conditional probability trees.

We can also construct probability trees where proportions are given to us to describe frequencies of outcomes in a trial.

Proportions could be expressed as fractions or ratios.

Let's take this scenario here.

A sweet jar has some orange and blueberry sweets in the ratio of five to two, and Sam eats two sweets at once.

Now let's think about the initial conditions we have here.

In this case, all outcomes have unknown frequencies.

We don't know how many sweets from the jar, and we don't know how many orange and how many are blueberry.

So we consider all initial conditions as algebraic expressions.

We do have the relationship between orange sweets and blueberry sweets express as a ratio of five to two.

So we can express those as 5X and 2X and that means we can express the total number of sweets as a sum of those, which is 7X.

Now let's think about what the probability tree might look like for this.

One thing we should bear in mind is that Sam eats the first sweet which means it's not replaced.

Therefore, the probabilities for the second outcome are dependent on the first outcome.

We should bear that in mind later when it comes to writing probabilities on that second layer of branches.

The first layer of branches would look something a bit like this.

The first sweet that Sam eats could either be orange or blue, and we could use the expressions we wrote to get some probabilities.

The probability that it's orange will be 5X over 7X.

The probability that it's blue will be 2X over 7X, and it's always worth double checking.

but they would sum to one whole.

5X plus 2X is 7X, so would have 7X over 7X, which is equivalent to one whole.

Now you might be thinking to yourself, hmm, we could simplify these fractions by dividing the numerate and denominator by X in each case.

So we'd get five sevens and two sevens.

While we could do that, we are not going to just yet.

These algebraic fractions are really important when considering the second sweet, and you'll see why shortly.

This first layer of branches shows the probabilities before the first sweet was eaten.

And before that first sweet was eaten, there were 5X orange sweets, 2X blueberry sweets, and 7X sweets in total.

Let's consider what that actually means.

We don't know how many orange sweets there are, but we do know it's a multiple of five.

We don't know how many blueberry sweets there are, but we do know it's a multiple of two, and we don't know how many sweets there are in total, but we know it's a multiple of seven.

Now the second layer branches will show the probabilities after the first sweet was taken, but before the second sweet was chosen.

Therefore, the total frequency of sweets will be one less than before.

Initially it was 7X.

One less than that will be 7X subtract one.

Now this is why it was really important we did not simplify those fractions earlier by cancelling the X off the numerate end denominator 'cause if we had done that and got five over seven and two over seven, we may have mistakenly thought that there were seven sweets in the bag to begin with and subtract one from that to get six sweets for the second layer.

But that's not necessarily the case.

All we know is that the initial number of sweets was on multiple of seven.

So we are subtracting one from a multiple of seven and that is not necessarily going to give us six sweets, which is why for the time being we can express it as 7X subtract 1.

Let's consider if the first sweet was orange and what the probabilities would be for the second sweets.

We can do that by looking back at the initial conditions we wrote earlier: 7X, 5X and 2X.

After Sam has eaten the first sweet, there would be 7X subtract one sweet in the bag in total and there'll be one fewer orange sweet in the bag.

So that'll be 5X subtract one.

The number of blueberry sweets would remain the same, so that would be 2X.

And then we could use these expressions to write probabilities in terms of X for the second sweet being either orange or blueberry.

Given that, we know the first one was orange and those probabilities would look something a bit like this.

How about the first sweet was a blueberry sweet? The probabilities for getting an orange or a blueberry sweet for the second one would not necessarily be the same as it were above.

Let's go back to our initial conditions.

7X, 5X and 2X.

Given that the first sweet eaten was blueberry, then the total number of sweets would be 7X subtract one.

The number of orange sweets would remain the same, and the number of blueberry sweets would be 2X subtract one.

And then we can use those expressions to create these probabilities here.

So let's check what we've learned.

In a fruit bowl, three tenths of the fruits are apples and the rest are bananas.

Write down the ratio of apples to bananas.

Pause while you do it, and press play for an answer.

The ratio of apples to bananas will be three to seven.

So two pieces of fruit are randomly chosen and eaten.

Please could you complete this table of initial conditions expressing each frequency algebraically? Can use any letter you like, but maybe go for X.

Pause video while you do it, and press play when you're ready for answers.

It would be 10X for total number of fruits, 3X for apples and 7X for bananas.

So let's now start to construct a probability tree for this.

Could you please find the two probabilities in the first layer of branches on this tree labelled A and B? Pause while you do it, and press play for answers.

Here are your answers.

3X over 10 and 7X over 10.

How about if the first fruit eaten was a banana? Could you please find the probabilities on this pair of branches which are now appeared? They're labelled A and B.

Pause while you do it, and press play when you're ready for answers.

There'll be these probabilities here.

So if we are told that the probability of eating two bananas will be 22 or 45, which of these equations are correct? Pause while you you choose, and press play for an answer.

The equations which are correct are A and B.

Now with B, you might be thinking, hang on, Dr.

Rollinson, you said don't simplify that fraction.

So why have we put seven tens? Well, when we're at these stage of constructing and solving equation, we can simplify the fraction for the first layer of branches.

That's absolutely fine.

The reason why we didn't simplify it to begin with is so that we didn't accidentally subtract one from 10 rather than 10X and subtract one from seven rather than 7X.

But once you've constructed your probability tree, it's fine to then start simplifying fractions.

So it's on to then for task C.

This task contains three questions, and here is question one.

Pause video while you do this, and press play when you are ready for question two.

And here is question two.

Pause while you do this, and press play for question three.

And here is question three.

Pause while you do this, and press play for some answers.

Okay, let's take a look at some answers.

Here's what you should have for question one.

Pause while check this, and press play for more answers.

And then for question two, this is what you should have for part A.

Pause while you check this, and press play for more.

And then for part B, you should have X equals 16.

And here's your working for how to get it.

Pause while you check it, and press play for more.

And finally, here are the answers to question three.

Pause while you check this and press play to summarise today's lesson.

Fantastic work today.

Now let's summarise what we've learned.

Relationships between different regions of a Venn diagram can be used to construct and solve algebraic equations involved in probabilities.

Algebraic probabilities of a two stage trial can be shown on a probability tree when the frequency of each outcome or event is not known.

The algebraic conditional probabilities of a stage of a trial can be written given a particular outcome of a previous stage of a trial has occurred.

And algebra probabilities from proportional frequencies of outcomes can also be written.

Well done today.

Have a great day.