Lesson video

Hello there! You made a great choice with today's lesson.

It's gonna be a good'un.

My name is Dr.

Rowlandson, 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 "Advanced Problem Solving with Conditional Probability", and by the end of today's lesson, we will be able to use our knowledge of conditional probability 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 these words mean, and press play when you're ready to continue.

The lesson is broken into two learning cycles.

I'm going to start by selecting an appropriate representation for the problem we have at hand.

There are many different representations that can be used to organise information when solving problems involving probabilities.

For example, you may use a table for some problems, a Venn diagram, or a tree diagram of some kind.

The type of representation you choose may depend on the information that you are working with.

Let's take this scenario as an example.

A pack of party hats contains hats with the following features.

Some have circles with a topper, some have circles without a topper, some have rings with a topper, and some have rings without a topper.

In a pack of 100 party hats, there are 45 hats with circles, 60 hats with a topper, and 26 hats with circles and a topper.

So what representation could you use to organise this information? Perhaps pause the video and think about what representation you might use, and then press play when you're ready to see how we can do this together.

Let's take a look at a couple of different representations together now.

Here we have Alex, who chooses to represent this information with a Venn diagram.

In the Venn diagram we can see there's one region which is for hats with circles, and there's another region for hats with toppers, and those regions overlap because you can have hats with circles and a topper.

So let's put this information into the Venn diagram now.

We can see that there are 26 hats with circles and a topper, so they would go in the intersection of those two regions.

There are 45 hats with circles, but of those, 26 them are already accounted for in the Venn diagram.

So we can do 45 subtract 26 to get that there are 19 hats with circles that do not have a topper.

There are 60 hats with a topper.

26 of them are accounted for because they have circles.

So there must be 34 in the pack that do not have circles.

And then, the number of hats that do not have circles or a topper would be 100 subtract the sum of the answers we've got already, which would be 21.

Let's now take a look at a different representation.

Here we have Izzy, who chooses to represent this information with a two-way table.

So let's transfer these frequencies into the table.

And to begin with, this is quite straightforward, because we can just write the numbers we can see in the information given straight into the table.

For example, there are 100 party hats in total, 45 of them in total have circles, and 60 of them in total have toppers, and there are 26 hats that have a circle and a topper, so that can go there as well.

Now we've got all of these in the table, we will need to start using calculations to work out missing frequencies.

For example, if we look at the hats with circles, we can see that there are 45 in total.

26 of 'em have a topper, which means 19 of them must have no topper.

If we look at the hats that have a topper, there is 60 in total.

26 of 'em have circles, which means there must be 34 of them that have rings.

And then if there are 100 hats in total and we can see how many have a circle and a topper, and rings and a topper, and a circle and no topper, we could get the number that have rings and no topper by doing 100 subtract the sum of those other three numbers.

That'd be 21.

And then to work out the two remaining totals, we could use addition.

So now they're both filled in.

Each diagram can be used to find probabilities.

For example, a hat is chosen at random from the pack.

What is the probability that the hat satisfies the intersection of rings and topper? In other words, what's the probability it has rings and a topper? We can see the frequency of hats that satisfy this event in each of these representations.

For the Venn diagram, it takes a little bit of interpretation, because the ones that have rings do not have circles.

So you're looking for the frequencies that are outside the region of circles, and then you're looking for the one which has a topper.

In the two-way table, you can see that a bit more clearly.

It's the ones with rings and a topper.

In either case, it's 34, so the probability is 34/100ths.

How about this probability? The probability that the hat chosen has rings? Well, you can see that frequency in both of these two representations.

Once again, the Venn diagram requires a little bit of interpretation, because you're looking for ones that do not have circles, and there are two frequencies that satisfy that, which is 34 and 21, which you need to add together to get 55.

In the two-way table, you can see that 55 more explicitly.

Either way, you get 55 over 100.

So let's check what we learned.

You got the same Venn diagram and two-way table.

A hat is chosen at random from the pack.

Could you please find the probability that it has either a topper or circles? Pause video while you do it, and press play for an answer.

The answer is 79/100ths.

Regardless of which representation you use, you will need to perform some calculations in this case.

You can either do 19 plus 26 plus 34, to get 79, or you could do 100 subtract 21 to get 79.

This time, could you please find the probability that it does not have a topper? Pause the video while you do it, and press play when you're ready for an answer.

The answer is 40/100ths.

This time, could you please find the probability that it does not have a topper, given that it does have circles? Pause the video while you write down your answer, and press play when you're ready to see what it is.

The answer is 19 over 45.

Let's now consider the usefulness of these two representations in a little bit more detail now, by making the scenario just gradually more complex as we go, and seeing in which case one representation is easier to use than another.

So far in this scenario, the party hats can be organised into categories according to two dimensions.

One of those dimensions is whether the hats have circles or rings, and the other dimension is whether the hats have a topper or no topper.

But what would happen if we added a third dimension to this information? For example, what if the third dimension is, they could either have a fringe or no fringe? That means now each hat could either have circles or rings, a topper or no topper, and a fringe or no fringe.

Which of these two representations will be easier to use to organise information this time? Pause the video while you think about which representation might be the easiest to use, or which one might have some problems with this scenario, and think about why as well.

And then press play when you're ready to continue.

I wonder what you thought.

Let's start by looking at the Venn diagram.

Alex says, "I could add a third event for hats with a fringe." So now we can see in the Venn diagram there are three regions that overlap to create lots of intersections.

One of the regions is for the hats with circles, one is for hats with toppers, and one is for hats with a fringe.

And each type of hat, you can see on the right hand side, is represented somewhere in this Venn diagram.

For example, the highlighted section here shows a hat with circles, a topper, and a fringe.

Izzy says, "It would be tricky to show this information with a two-way table," because a two-way table shows information categorised in two ways, one way and another.

So if you have a third way, that'll be quite difficult to show on that table.

You'd need a three-way table and you would have to go 3D in that situation.

So a two-way table will be quite difficult in this situation, where we've got three dimensions of categories.

Let's go back now to that first scenario again, where the hats could only be organised according to two dimensions, circles versus rings as one dimension, and topper versus no topper as the other dimension.

I like to draw your attention this time to the fact that each of those dimensions only has two categories.

We can see in Izzy's table, circles versus rings.

They are two categories in one dimension.

And topper versus no topper, they are two categories in the other dimension.

In Alex's Venn diagram, the circles versus rings is equivalent to circles versus no circles.

So if something can be either inside the region for circles, or outside the region for circles.

But what would happen if we increase the number of categories in at least one of these dimensions? For example, the hats could either have circles, rings, or stars.

In either case, they can still either have a topper versus no topper.

Which of these two representations will be easiest to use for when one of the dimensions has three categories? Perhaps think about which one would you have problems with using as well, and why? Pause the video while you think about this, and press play when you're ready to continue.

I wonder what you thought this time.

Let's take a look at the two-way table, because we still only have two dimensions of information, so we should still be able to use it.

Izzy says, "I could add an extra column to my table." So now we can see the columns have circles, rings, and stars.

Alex says, "Venn diagrams only show categories and their complements," so they show either circles or not circles, rings or not rings, or stars or not stars.

They don't really distinguish between three different choices within the same dimension.

So Venn diagrams can be difficult to use in this situation here.

One thing we could take away is that, when the information can be organised into two dimensions, we can use a two-way table, but if it's three dimensions or more, then the two-way table becomes difficult.

If each of the dimensions only has two categories, either something happens or it doesn't happen, we can use a Venn diagram, so long as it's either two or three dimensions.

After three dimensions, it gets a bit difficult.

But if there are more than two categories within the dimension, the Venn diagram then becomes too difficult to use as well.

So the choice of representation really depends on what the information is you're working with, and what you intend to do with that information.

So let's check what we've learned.

A bag contains four counters numbered one to four.

A counter is chosen at random, and its number is recorded.

It is placed back into the bag, and a counter is chosen at random again, and its number is recorded.

The sum of the two numbers is found.

What is the probability that the sum is five? Please could you work this out, and you may draw a table or diagram first to help you work it out.

Pause the video while you do this, and press play when you are ready for an answer.

In this case, because you are finding the sum of the two outcomes, it may be helpful to use a two-way table like this, and once you've done that, you can work out the probability that the sum is five by doing four over 16, which is one quarter.

How about this scenario? A bag contains 20 counters, nine counters are blue, the rest are green.

Seven counters contain a star, the rest are plain.

And four counters are blue with a star.

A counter is chosen at random.

What is the probability that the counter is green and does not contain a star? Could you work this out please? And you may draw a table or diagram to help you first.

Pause the video while you do this, and press play when you're ready for an answer.

This time, a Venn diagram may be helpful, because the information can be sorted in two different ways, and those categories overlap.

It can be blue and have a star at the same time.

So the Venn diagram is quite helpful for showing that intersection.

If you do that, you can get the probability of it not being blue and not having a star as being 8/20ths.

How about this scenario here? A bag contains 20 counters, nine counters are blue, the rest are green.

Two counters are chosen at random, without replacement.

What is the probability that they are both green? And once again, you may draw a table or a diagram first to help you.

Pause the video while you do this and press play when you're ready for an answer.

A probability tree might be helpful, because you are drawing one counter, leaving it out, and then drawing another one.

If you do that, you can work out the probability that they are both green by multiplying 11/20ths by 10/19ths to get 110 over 380.

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

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

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

Here is question two.

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

Here is question three.

Pause while you do this, and press play to go through some answers.

Okay, let's go through some answers.

In question one, you had the scenario where each spinner is spun once, and you had to work out some probabilities.

It may be helpful to draw a two-way table in this situation.

For the probability that both outcomes are odd, you don't necessarily need to fill in the cells there.

You just need to count all the ones where both of them are odd.

There are eight of them out of 24 cells altogether in that table.

So the probability is 8/24ths.

And then you need to start working out some probabilities based on the sum of the two numbers.

So now it would be helpful to fill out all the possible outcomes.

The probability that the sum is odd would be 12/24ths.

The probability that the sum is greater than four would be 14/24ths.

And the probability that the sum is four, given that spinner B lands on the number three, that would be 3/12ths, or one quarter.

Then for question two, you've got this scenario with the marbles, which can either be blue or green or have a swirl or not have a swirl.

It can be helpful to use a Venn diagram in this situation.

Or you can use a two-way table instead, if you want to.

Or anything else that might help you along the way.

And then you have to work out some probabilities.

The probability that the marble chosen has a swirl will be 30/50ths.

The probability it's from the union of swirl and blue will be 43/50ths.

The probability it's from the intersection of swirl and not blue would be 8/50ths.

And the probability it's swirled, given that it's blue, will be 22/35ths.

Then in question three, you had this scenario we had two box of marbles, where the marbles are either black or white.

And a marble is chosen from box A and placed into box B, and then a marble is chosen from box B.

In this situation, it may be helpful to use a tree diagram, and it would look something a bit like this, with all your probabilities on.

Once you've done that, you can work out the probability that Aisha chooses a black marble from box B by working out this probability as 9/40ths, and this probability as 4/40ths, and adding them together to get 13/40ths.

Great work so far.

Let's now move on to the next part of this lesson, where we're going to look at probability problems that can be solved using algebra.

Let's take a look at this scenario here, where Jacob has a bag containing red and blue counters.

He picks two counters at random from the bag, and the partially completed probability tree helps us see what's going on in this scenario.

We can see that the probability that Jacob gets a red counter on his first pick is one third, and if he does get a red counter on that pick, the probability he gets a red counter on the next pick is one quarter.

And what we are going to do is work out the total number of counters in that bag.

Perhaps pause the video, and think about what steps we might take to try and approach this problem, and then press play when you're ready to work through it with me.

Let's take a look at this together.

We don't know the number of counters in the bag, and we don't know how many are red or how many are blue.

So when we have unknowns, it can be helpful to use some algebra.

And before we do that, it can be even more helpful to define what our variable is representing.

So, let's let's n be the number of counters in the bag at the very start.

If n is the number of counters at the bag at the very start, let's think about what these probabilities tell us about the number of counters.

This third here is the probability of getting a red on the first pick.

That means that one third of those counters are red at the start.

So we can write that as an expression.

The number of red counters at the start is 1/3 n.

And then let's think about what happens after one red counter is removed.

The number of red counters that remain is 1/3 n subtract one.

And the number of counters in the bag altogether will be n subtract one, because each of those is the expressions we previously had for the number of red counters, and the number of counters in total, but we subtracted one from each of them to account for the one red counter we've taken out.

Let's display the expressions we've written so far at the top of the screen, so we can refer back to it as we go through this problem.

Let's imagine now we've drawn one red counter already from the bag.

This probability here means that a quarter of the remaining counters must be red.

Now the remaining counters is n subtract one, so a quarter of n subtract one will be the expression for the number of red counters remaining.

Hang on a minute.

We now have two algebraic expressions for the number of red counters remaining.

One at the top, which is 1/3 n subtract one, and the other one is 1/4 and in brackets n subtract one.

Those two algebraic expressions both represent the same number, the number of red counters remaining after that first pick.

So if they both represent the same number, it means we can create an equation by putting them equal to each other.

And the equation would look something a bit like this.

We can now solve this equation to work out the value of n, which is the number of counters at the start.

We can solve it first by maybe simplifying the equation, and then we could multiply both sides of the equation by four, multiply both sides of the equation by three, add 12 to both sides of the equation, subtract 3n from both sides of the equation, and we get n equals nine.

Therefore there were nine counters in the bag at the start.

So let's check what we've learned.

There are x many cards in a pack.

Five of the cards are red, and the rest are blue.

Please could you write an expression for the probability of drawing a red card, and write it in terms of x.

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

The answer is five over x.

One red card is removed.

Please could you write an expression for the probability of drawing a red card from the remaining deck? Pause the video while you do it, and press play for an answer.

The answer is four over x minus one.

If there are five red cards to begin with, and one is removed, it means there are four red cards remaining.

That's our numerator.

If there are x many cards in a pack to begin with, and one is removed, it means there are x subtract one cards remaining in the pack, and that's our denominator.

We now have a probability tree for this scenario, with the two probabilities that you previously wrote down.

Andeep draws two cards at random from this pack.

Could you please write an expression for the probability that they are both red? Pause while you do it, and press play for an answer.

You'd get that by multiplying your two probabilities together and simplifying it.

And you'd get 20 over x-squared subtract x.

Let's now do something with this answer.

An expression for the probability of drawing two red cards is 20 over x-squared subtract x.

I'm going to tell you now that the actual probability of drawing two red cards is two thirds.

Could you please use those two piece of information to make an equation, and then rearrange it to show that two x-squared subtract two x subtract 60 is equal to zero.

Pause the video while you do it, and press play when you're ready to see how.

Well, if 20 over x-squared subtract x is the probability, and so is two thirds, it means we can put them equal to each other.

And then we can rearrange it by multiplying both sides of the equation by three, multiplying both sides of the equation by x-squared subtract x, and then subtract 60 from each side of the equation.

And we get zero equals two x-squared subtract two x subtract 60.

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

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

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

Here is question two.

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

And here is question three.

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

Okay, let's go through some answers.

In question one, part A, you had to work out how many counters were in the bag at the start, given the information provided.

If you let n, or any letter, equal the number of counters in the bag at the start, then it would lead you to an equation like this, or something equivalent to it.

You could then simplify the equation and rearrange it to get n equals 16.

So there are 16 counters in the bag at the start.

And for part B, you could use the fact that you know that there are 16 counters in the bag at the start to create fractions in the first layer, where the denominator is 16, and fractions in the second layer of your tree, with a denominator of 15.

And then to calculate the probability that Jacob picks two blue counters, you can select the probabilities you want, multiply 'em together, and get 132 over 240.

Then with question two, you had to use the information provided to show that x-squared subtract x subtract 90 is equal to zero.

A good way to start this could be by drawing a probability tree.

You don't have to draw the tree, but it might help you see what's going on.

The probability of drawing a black card at the start is six over x, and once one is drawn, the probability of drawing another black card will be five over x subtract one.

You are told that the actual probability of drawing two black cards is one third.

So you can create an equation like this.

The product of the two probabilities that are expressed in terms of x is equal to one third, and you can simplify your equation and rearrange it to show that zero equals x-squared subtract x subtract 90.

And then using that equation, you have to work out the number of cards in the pack at the very start.

Well, if x is the number of cards in the pack at the very start, you need to solve this equation to find the value of x.

If you do it, you'll get two values.

X equals -9, and x equals 10.

You can't have a negative number of cards in a pack, so there must be 10 cards in the pack at the start.

And then in question three, you had the scenario where two frisbees are flipped in the air.

Now this is quite often what happens at the start of a game or sport that involves frisbees.

Two frisbees are flipped in the air.

Each frisbee could land either heads or tails, but the likelihood of it landing either heads or tails are not equal.

So two events are defined.

One event is called odds, and that is when the two frisbees land in different ways.

And the other event is called evens, and that's when the frisbees land the same way.

And what you had to determine was whether the probability of these two events could be equal to each other, based on the fact that heads and tails do not have the same likelihood.

So, we can visualise this scenario by drawing a probability tree, where we have two layers of branches, one for each frisbee, and each one could land either heads or tails.

We don't know what the probability is that it lands on heads, but we could define it with a letter.

We could let the probability of heads be equal to a, and that means the probability that tails happens is one subtract a.

And we can put all the probabilities on our probability tree like this.

We could then work out the probabilities of combined events, expressing them in terms of a, such as this.

The probability of heads and heads will be a-squared.

The probability of getting heads on one and tails one the other will be a subtract a-squared.

And the probability of getting two tails would be one subtract 2a plus a-squared.

Now we have these probabilities expressed algebraically, we can think about what it would mean if the probability of evens was equal to the probability of odds.

If it was, it would mean that the probability of heads and heads, plus the probability of tails and tails, will be equal to the probability of heads and tails and tails and heads added together.

We can create an equation with that, rearrange it and solve it, to get a is equal to 0.

5.

Now remember, a is the probability that the frisbee lands heads.

Now that would mean that the only way that odds and evens could have the exact same chance of happening would be if the probability that the frisbee lands on heads was 0.

5.

But if the probability that the frisbee lands on heads is 0.

5, it means the probability of tails would also be 0.

5.

And we are told that those two probabilities are not equal.

So that means this is not possible.

It is not possible for the probability of evens to be exactly the same as the probability of odds.

Fantastic work today.

Now let's summarise what we've learned.

There are many different representations that can be used to organise information when solving problems involving probabilities.

Selecting an appropriate and efficient method for solving the problem comes from considering the context and also evaluating methods.

And that comes with practise.

The more probability problems you solve, the more you get used to selecting the most appropriate method that is helpful to you in that particular situation.

Now, probabilities may appear in unfamiliar context, so don't let that put you off.

Always look for the probabilities you need, and the correct representation that would help you in that situation.

And probabilities may require a strong knowledge of converting between fractions, decimals, and percentages.

Well done today.

Have a great day.