Lesson video

Hello, and thank you for choosing this lesson.

My name is Dr.

Rowlandson.

I'm excited to be helping you with your learning today.

Let's get started.

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

This lesson is called Conditional Probability in a Tree Diagram.

And by the end of today's lesson, we will be able to calculate a conditional probability from a tree diagram.

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 two learning cycles and we're going to start with probability trees and conditional probability.

Probability trees are an effective way to represent events from a trial with multiple stages.

For example, spinning a spinner, such as the one you can see here, and flipping a coin.

This coin is biassed, which means its outcomes are not equally likely.

Let's see what happens when we represent this in a probability tree.

The first stage could be spitting the spinner and the outcomes could be getting a prime number or not a prime number.

The probabilities are written on the branches, so the probability of getting a prime number with a spinner is 2/6, and the probability of not getting a prime is 4/6.

The second stage could be flipping the coin.

And we can get either heads or tails in either case, and the probability getting heads is 7/15, and the probability getting tails is 8/15.

Now in this situation, you could have flipping the coin as stage one and spin the spinner as stage two.

Doesn't necessarily matter in this particular case.

But let's stick with the way we got it for now.

The probability of each combined outcome can be calculated by multiplying the probabilities on consecutive branches.

For example, if the spinner first got a prime number and then the coin then showed heads, the probability of that happening could be calculated by doing a multiplication of 2/6 and 7/15, which is 14/19.

I could do the same for the other combined outcomes.

The probability by its prime in the spinner and tails on the coin is 16/90.

The probability of it not being prime on the spinner and being heads on the coin is 28/90.

And the probability of it not being prime on the spinner and being tails on the coin is 32/90.

The probabilities of each set of branches that meet at a point sum to one.

For example, we can see two branches here that meet at a point for prime and not prime and those probabilities, 2/6 and 4/6, sum to one.

Also, we can see it here.

The branches for heads and tails meet at a point and the probabilities, 7/15 and 8/15, sum to one, and the same as well for the two branches above it.

The probabilities of all possible combined outcomes also sum to one.

We can see that here.

These four probabilities all sum to one whole.

Now in this situation, we can see that regardless of whether or not the spinner lands on a prime number, the probabilities for the coin do not change.

We can see that if the spinner lands on a prime number, the probability getting heads on the coin is 7/15.

And if the spinner lands on not a prime number, then the probability getting heads on the coin is still 7/15.

This means that stage two of this trial, flipping the coin, is independent of the outcome of stage one of the trial, which is spinning the spinner.

It doesn't matter what happens with the spinner.

It will not affect the probabilities for the coin.

So let's check what we've learned.

Here we have details for a two-stage trial.

In stage one, a spinner is spun and there are two possible events that could occur.

Either it lands on a square number or not a square number.

In stage two, a coin is flipped.

And there are two outcomes there.

Either it lands heads or tails.

You've got a probability tree which is partially filled out.

Could you please find the values of the probabilities labelled A, B, and C on this probability tree? Pause the video while you do that and press play when you're ready for answers.

Here are the answers.

A is 7/10, which you get from doing 1 - 3/10.

B was 3/5, which you get from doing 1 - 2/5.

And C was 6/50, which you get from multiplying 3/10 by 2/5.

So now all the probabilities are labelled on a tree.

Please, could you explain how you know about the outcomes of flipping the coin are independent of the outcomes of spinning the spinner? Pause the video while you write something down and press play when you're ready to see an answer.

Well, the probabilities of the outcomes heads and tails are the same across both groups of branches, regardless of whether the spinner landed on a square number or a non-square number.

Let's now look at a different scenario.

Here we have Sam, and to get to school each morning, Sam catches a bus and then a train.

The probability of Sam catching the train is dependent on whether or not the bus is on time.

This means that the probability of Sam catching the train will change depending on the outcome of the first stage of the journey, and that is whether or not the bus is late.

Jacob says, "If Sam misses the bus first, then they're also more likely to miss a train too." Hmm, good point, Jacob.

Let's now look at some probabilities by analysing previous data.

On the last 100 school days, Sam's bus was late 17 times.

So let's represent this part of the journey in a probability tree.

We have two possible outcomes: the bus could be on time or the bus could be late.

And let's use the previous data to generate some probabilities.

The probability that the bus is late will be 17/100 because in the last 100 days, the bus was late 17 times.

The probability that the bus is on time would be 83/100.

And then we can think about the next stage of the journey.

When the bus was on time, Sam missed the train 5% of the time and this increased to 80% when the bus was late.

So let's represent this stage of the journey on the tree diagram.

If the bus was on time, two possible outcomes could happen with the train.

Either Sam catches the train or Sam misses the train.

And in this scenario, the probability that Sam misses the train is 5%, which means the probability will be 5/100 and the probability Sam catches the train will be 95/100.

If the bus was late, there are still two possible outcomes for the train.

Either Sam catches the train or Sam misses the train.

And in this situation, the probability that Sam misses the train is 80%, so the probability would be 80/100 and the probability that Sam catches the train would be 20/100.

And then we can calculate some combined probabilities.

The probability that the bus is on time and Sam catches the train could be calculated by multiplying those two probabilities together to get 7885/10000, which simplifies to 1577/2000.

And we can do the same with the other three combined events like so.

Now, if Sam does miss the train, then they are late for school, and the probability that this happens can be calculated by finding the sum of the two combined probabilities that show Sam missing the train.

That is 415/10000 and 1360/10000.

Add those together and simplify it and you get 71/400.

Now in this situation, the probability that Sam catches the train is dependent on whether or not the bus is on time.

We can see that when the bus is on time, the probability that Sam catches the train is 95/100.

However, when the bus is late, the probability that Sam catches the train is 20/100.

The probability of catching the train is different depending on whether or not the bus was late.

This is because the outcome of the first part of the journey affects the probability of the second part of the journey.

Therefore, the probability that Sam catches the train is dependent on whether or not the bus is on time.

So let's check what we've learned.

Each weekend, Sofia plays two games of tennis.

The probability of whether Sofia wins or loses each game is shown on this probability tree.

Explain whether the probability of Sofia winning her second game is dependent or independent of her winning the first game and justify your answer.

Pause the video while you do that and press play when you're ready to see an answer.

Well, it could help to work out the missing probabilities for game 2 or at least one of those missing probabilities because then you would see that the outcomes for game 2 are dependent on the outcomes of game 1, and this is because the probability of Sofia winning game 2 changes depending on the outcome of game 1.

You can see in game 1, if she wins it, then for game 2, the probability she wins that will be 0.

57.

But if she loses game 1, then the probability she wins game 2 is 0.

39.

Those probabilities are different, which means they are affected by the outcome of game 1.

So please find the probability of Sofia winning at least one of her games.

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

You could start this by working out the probability she loses game 1 and then you could calculate the probability that she wins and then wins, wins and then loses, loses and then wins, and finally, add all those together to get the probability she wins at least one game, and that'll be 0.

6462.

Alternatively, you could work out the probability that she loses both of her games and subtract it from 1, and that will give 0.

6462 as well.

Let's now think about conditional probabilities.

We can use a probability tree to identify conditional probabilities.

This would involve inspecting branches based on the outcome for a previous stage in the trial.

For example, this probability tree shows a probability of rain during two consecutive days, day 1 and day 2.

What is the probability that day 2 has rain given that we know day 1 had no rain? Let's think about how we can do that.

Well, we know that day 1 had no rain, so we follow the branch labelled naught rain for day 1, which is written rain and the prime symbol.

Then, follow the branch labelled rain for day 2 and think what is the probability of rain in day 2 for this particular case.

The probability of day 2 having rain given that day 1 did not have any rain would be 0.

2.

Let's generalise this now by looking at a general probability tree that shows the conditional probabilities of the second stage of a trial.

Let's say stage 1 has two possible outcomes, A and B, and the two probabilities are shown on the probability tree here.

And let's say stage 2 has two possible outcomes, C and D, and the probabilities are dependent on the outcomes of stage 1.

If A happens in stage 1, then the probability that C happens in stage 2 would be the probability of C given that A has happened.

The probability that D happens in stage 2 will be the probability of D given that A has happened.

If B happens in stage 1, then the probability that C happens in stage 2 would be the probability of C given that B has already happened.

And the probability of D happening would be the probability of D given that B has happened.

Each of the probabilities for C and D are dependent on what happens in stage 1.

So when we calculate the probabilities for combined outcomes, the probability for A and C both happen would be equal to the probability of A multiplied by the probability of C given that A has happened.

And the same for the other three combined outcomes.

The probability for A and D both happen would be calculated by doing the probability of A multiplied by the probability of D given that A has happened.

The probability that B and C happened will be equal to the probability of B multiplied by the probability of C given that B has happened, and that would not be the same as the probability of C in the top scenario.

That probability would be different, or could be different.

And then the probability that B and D happen will be equal to the probability B multiplied by the probability of D given that B has happened.

So let's check what we've learned.

This probability tree shows the probability of snow during two consecutive days, day 1 and day 2.

Here are three probabilities for you to calculate.

Pause the video while you do that and press play for some answers.

Let's go through some answers.

A useful way to start is to work out the missing probabilities on the tree and write all probabilities in the same form.

For example, write them all as decimals.

Then for the probability in part A, we need to work out the probability that day 2 snows given that day 1 already snowed.

That would be 0.

2.

For B, you have to work out the probability that it snows on both day 1 and day 2.

You get that by multiplying 0.

2 by 0.

03 to get 0.

006.

And in C, you have to work out the probability that day 2 snows given that day 1 did not snow, and that would be 0.

01.

Okay, so to you for task A.

This task contains two questions and here is question 1.

Pause the video while you do it and press play for question 2.

And here is question 2.

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

Here are the answers to question 1.

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

Then question 2.

When you work out the missing probabilities, this is what you should get.

And then you have to justify, using probabilities, why the probability that a pupil completes their assignment is dependent on the location that they started their assignment in.

And you can do that by saying the probability that a pupil completes their assignment changes depending on the location of where they start their assignment in.

And you can see that those probabilities are 0.

6 versus 0.

84.

Well done so far.

Let's now move on to the next part of this lesson, which looks at identifying probabilities of dependent stages.

Here we have a bag that contains six purple marbles and four green marbles.

One marble is chosen at random from the bag, and here are the probabilities of the two outcomes.

The probability that marble is purple is 6/10.

The probability it's green is 4/10.

Let's now consider what could happen if we chose two marbles from this bag instead.

Randomly choosing two marbles from this bag can be done in three different ways.

One way could be choosing one marble from the bag, replacing it, and then choosing a second marble.

Another way could be to choose one marble from the bag, not replacing it, and then choosing a second marble.

And another way could be to choose two marbles from the bag at the same time.

Let's look at these scenarios one at a time.

This first scenario, when we choose a marble from the bag, replace it, and then choose a second marble.

Here we draw a probability tree for that.

In the first stage, we could get either purple or green and the probabilities would be 6/10 and 4/10.

And then we put the marble back in the bag.

So the probability of choosing a purple marble or green marble does not change as the marble is returned to the bag, and that means the outcome of the second stage is independent of which marble is chosen in the first stage.

How about these other two scenarios? One was choosing one marble from the bag, not replacing it, and then choosing a second marble, and the other was choosing two marbles from the bag at the same time.

These may sound like different scenarios, but actually, these two methods are identical and they are still two-stage trials.

They are both taking two marbles out of the bag by the end of the trial.

In one of them, you're doing it one at a time, but in the other, you choose them both at the same time.

But when you choose two marbles at the same time, really, you are choosing one and then another immediately afterwards, so you could think about it as being two stages that happen very quickly after each other instead.

Let's take this scenario here where two marbles are chosen from the bag at the same time.

The first stage of this trial looks at the probabilities before the first marble is chosen.

Can either get a purple or a green.

The probability of purple is 6/10 and the probability of green is 4/10.

The second stage of the trial looks at probabilities after the first marble is chosen but before the second marble is chosen.

Let's say, for example, when I put my hand into the bag to select two marbles out, the first one that I select is a purple marble.

There are still two possible outcomes for the second marble I choose.

Either it's purple or green.

However, once I've chosen that first marble, it means there are only nine marbles left in the bag for me to choose for the second one.

I'm not gonna take the same marble out twice.

So once I've selected that first one, there are only nine left to choose from, and that means in this second stage of the trial, all the probabilities will have a denominator of 9 rather than 10 because there are only nine marbles left to choose from.

And if that first marble I chose was a purple, then it would mean that for the second pick, there are only five purple marbles left.

So the probability of the second choice being a purple marble given that the first one was purple would be 5/9.

Now if the first marble was purple, all four of the green marbles would still remain, so the probability of choosing a green marble for the second one given that the first one was purple would be 4/9.

How about the other scenario then? If the first marble I choose is a green marble, there are still two outcomes for the second choice, either purple or green, but let's consider what the probabilities would be now.

Once I've chosen that first green marble, there are nine marbles left in the bag to choose from, which means the probabilities in the second stage will have a denominator of 9 again.

But this time all six of the purple marbles remain in the bag, so the probability of me choosing a purple marble for the second pick given that the first marble I chose was green would be 6/9, whereas if I've already selected a green marble, it means there are only three other green marbles left to choose for my second pick, which means the probability of me getting a green marble for the second pick given that I've already got one for the first pick would be 3/9.

Each pair of branches on the second stage of the trial should sum to 9/9, which is equal to one whole because each pair of outcomes are mutually exclusive and exhaustive.

For example, we can see here that 5/9 plus 4/9 is equal to one whole, and so are 6/9 and 3/9.

We can calculate the sample space of the entire trial by multiplying the probability at each stage, as usual.

For example, the probability we get a purple and a purple, or two purples, is 30/90, the probability we get a purple and a green is 24/90, the probability we get a green and a purple is 24/90, and the probability we get a green and a green is 12/90.

And all of those probabilities sum to one whole.

So let's check what we've learned.

Here we have a jar with some equally-sized marbles.

12 of them are blue and three of them are green.

Two marbles are randomly chosen at the same time.

Now remember, when you randomly choose two marbles at the same time, what it really means is you're choosing one and immediately afterwards choosing another.

We have a probability tree which sets out this scenario with two unknown probabilities, labelled X and Y.

Please, could you find the values of X and Y? Pause while you do it and press play for an answer.

The answers are 12/15 and 3/15.

So the first marble is chosen and it was blue.

Could you please find the values of C and D, which are the probabilities of blue and green in the next stage of the trial? Pause while you do that and then press play for an answer.

Here are the answers.

The probability of choosing a blue given that we've already got a blue will be 11/14, and the probability of choosing a green given that we've got a blue would be 3/14.

How about this scenario instead? The first marble chosen was green.

Please, could you find the values of E and F, which are the probability of getting blue and green in the next stage of this trial? Pause the video while you do that and press play for some answers.

Here they are.

The probability of getting a blue marble given that we've already got a green one will be 12/14, and the probability of getting a green marble given that we've already got a green marble would be 2/14.

So please, could you find the probability that both marbles were the same colour? Pause while you do it and press play for some answers.

Well, the two different ways that the marbles could be the same colour are blue and blue and green and green.

The probability getting blue and blue would be 132/210 and the probability getting green and green would be 6/210.

And you add those together, you get a probability of 138/210.

Okay, so to you for task B.

This task contains four questions, and here is question 1.

In this scenario, you've got a jar of marbles where one marble is chosen and then returned and a second marble is then chosen.

Could you please answer the questions based on that scenario? Pause the video while you do it and press play for question 2.

And here is question 2.

It is very similar to question 1 but with just one difference, and that is after the first marble is chosen, it is not returned before the second marble is chosen.

Pause the video while you do it and press play when you're ready for question 3.

And here is question 3.

Pause while you do this and press play for question 4.

And here is question 4.

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

Okay, let's go through some answers.

Here are the answers to question 1.

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

Here are the answers to question 2.

Pause while you check these and press play for more.

Here are the answers to question 3.

Pause while you check these and press play for question 4.

And here are the answers to question 4.

Pause while you check all this and press play for a summary of today's lesson.

Fantastic work today.

Now let's summarise what we've learned.

The probability of an outcome occurring may be dependent on whether an outcome from a previous stage of trial has occurred or not.

This means that equivalent groups of branches on the same layer of a probability tree may have different probabilities.

We can consider the conditional probability of randomly choosing two objects from a group on a probability tree, and we can look at how the probability of the second object being chosen may change based on the outcome of the first object being chosen.