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Hello there and welcome to today's lesson.
My name is Dr.
Rowlandson and I'll be guiding you through it.
Let's get started.
Welcome to today's lesson from the unit of conditional probability.
This lesson is called 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, and we're going to start by looking at problems with probabilities expressed in different forms. Probabilities can be expressed in different ways.
For example, here we have a spinner, with letters A to E.
If this spinner is spun once, then the probability that it lands on the letter A could be expressed in different ways.
We could express it as a fraction, as 1/5, could express it as a decimal, as 0.
2, and we can express it as a percentage, as 20%.
Now, while it's great we can express probabilities in different ways, it can create some problems. When dealing with multiple probabilities within the same problem, it can be helpful to ensure that they are all expressed in the same form.
For example, here we have Aisha, Alex, and Sofia, who each attempt to throw a basketball into a hoop.
Based on previously collected data, the experimental probabilities for each person being successful are known.
Aisha says, the probability that I score a basket is 0.
8.
Alex says, the probability that I score a basket is 77%, and Sofia says, the probability that I score a basket is 3/4.
So who is the most likely to score a basket and who is the least likely? Perhaps pause video and think about what makes this problem more difficult than it really needs to be and how we can make it just a little bit easier and then press Play when you're ready to continue.
The thing that makes this problem just a little bit more complex than it really needs to be is that the probabilities are expressed in different forms. It can be easier to compare probabilities when they are all expressed in the same form.
So for example, we could express them all as decimals.
Now, we have all of the probabilities written as their equivalent decimals.
We can see that Aisha is the most likely to score with a probability of 0.
8, and Sofia is the least likely to score with a probability of 0.
75.
Alternatively, we could express them all as percentages, and once again, we can see here Aisha is the most likely with 80% probability and Sofia is the least likely with 75% probability.
Alternatively, could express them at all as fractions, but in this case, you may need to use a common denominator to help you compare the fractions.
So now we've done that.
We can still see that Aisha is the most likely and Sofia is the least likely to score.
So let's check what we've learned.
Here we have the probabilities of three independent events.
They are expressed in different forms at the moment.
Could you please express them all as decimals? Pause video while you do it and press Play when you're ready to see the answer.
Here they are.
The probability For A was already given as a decimal.
That's 0.
3.
The probability for B, which is 24%, is equivalent to 0.
24, and the probability for C, which is 1/5, is equivalent to 0.
2.
Could you please express all these probabilities as percentages now? Pause while I do that and press Play for an answer.
The probability that A happens is 30%.
That's equivalent to 0.
3.
The probability of a B happens was already given as a percentage, 24%, and the probability of that C happens is 20% because that's equivalent to 1/5.
And now could you express all three probabilities as fractions with a common denominator? Pause the video while I do that and press Play when you're ready for answers.
Well, there are different ways you can do this.
Here's one way.
You could express them all as fractions with 100 as your denominator.
That'll be 30/100 for A, 24/100 for B and 20/100 for C, or you can express 'em as fractions with 50 as the denominator, and this is what they'll be in that case, or you can use any equivalent fraction, so long as they have the same denominator and they're equivalent to the original probabilities.
So now we've done that, which event is the most likely to happen? Pause the video while I write the answer down and press Play when you're ready to see what it is.
The answer is Event A.
That one has the greatest probability, which you can see clearly when you express all these in the same form, either as a decimal, percentage, or as a fraction with a common denominator.
So like we said earlier, when dealing with multiple probabilities within the same problem, it can be helpful to ensure that they are all expressed in the same form.
Let's look at another example where this can be helpful.
Here we have the probability of three outcomes to a trial and they're shown in the table.
The outcomes are all exhaustive and mutually exclusive.
So that means the probabilities would sum to one whole, and what we need to do now is calculate the probability that C happens.
The issue here is that one of the probabilities is expressed as a percentage and you want it expressed as a fraction.
Now pause the video and think about what approach you might take to solving this problem here and then press Play when you're ready to look at how we can do it together.
Well, there are a couple different ways you can do this.
One way could be by using percentages.
A is express as a percentage, so we could express B as a percentage and then C as a percentage as well.
If we did that, we'd put the probability of A happens as 21%, that's given to us.
The probability that B happens is 75%.
That's what 3/4 is as a percentage.
So to get the probability of C happens, we can do 100%, which is the total probability altogether, subtract the sum of the probabilities for A and B, and that would give us 4% for a probability of C.
An alternative method could be to use fractions.
We could express A as a fraction.
B is already as expressed as a fraction, and then we could try and get a common denominator and subtract from one whole.
For example, A would be 21/100, B could be written as 75/100, and then we can do one subtract the sum of those two probabilities and that would give us 4/100 for the probability of C.
So let's check what we've learned.
Here we have the probability of three outcomes to a trial which are shown in the table.
The outcomes are exhaustive and they are mutually exclusive, which means the probabilities will sum to one whole.
Please could you calculate the probability that C occurs? Pause the video while I do it and press Play when you're ready for an answer.
Okay, let's take a look.
If you work with decimals, your answer would be 0.
5, whereas if you work with fractions, your answer will be 5/10 or 1/2 or anything that's equivalent to that, or you can express your answer as a percentage.
Okay, it's over to you now for Task A.
This task has three questions and here is question one.
Pause the video while I do it and press Play for Question 2.
And here is Question 2.
Pause video while I do this and press Play for Question 3.
And here is Question 3.
Pause video while I do this and press Play to go through some answers.
Okay, let's go through some answers.
In Question 1, you had the probability of four outcomes to a trial shown the table with one probability missing.
You're told that the outcomes are exhaustive and mutually exclusive, which means that probabilities would sum to one whole and in A, you have to work out the probability that D happens.
Now, there are different ways you can do that.
You can express your probabilities as a fraction, as a percentage, or as a decimal.
For the sake of these answers, let's express our probabilities as decimals and this is what they would be.
Once you do that, you could subtract the sum of those from one whole and you'd get 0.
25 for the probability of D, or you can express that in any equivalent form.
For example, 1/4 or 25%, and now you have all the probability expressed in the same form.
You know that outcome A is the most likely to happen.
Outcome B is the least likely to happen, and then in part D, you have to approximate how many times you'd expect outcome C to occur in 500 trials.
Well, once again, there are different ways you can do this.
One could be to multiply the probability by the number of trials.
For example, 0.
2 multiplied by 500 is 100.
An alternative method could be to think about some of the equivalent forms that that probability could be expressed as.
For example, it could be 20%, which means you could find 20% of 500 because you expect around 20% of those trials to result in outcome C, or you can express their probabilities as a fraction, as 1/5.
That means you expect that around 1/5 of those 500 trials would result in outcome C.
Whichever way you do it, you'll get 100 as your answer.
And then Question 2, Alex has a bag contending only red counters and blue counters.
The number of red counters to blue counters is in the ratio one to four, so now you've got ratios to work with.
Alex picks a counter random from the bag and then replaces it.
In part A, you have to work out the probability that a red counter is drawn, and write it as a fraction.
One way to help you visualise this could be the draw a bar model.
Here we have a bar model that represents the ratio one to four, one red counter to four blue counters, and now we can see it visually.
It's quite easy to see that the fraction would be 1/5 and then part B is express the probability that a blue counter is drawn as a percentage.
That would be 80%.
And then in part C, you are told that there are, in fact, 12 blue counters in the bag and you have to work out how many counters the bag contains in total.
Well, there are plenty of different ways you can work through this.
Let's take a look at a few methods together here.
One method could be to work with the percentages.
You could let the number of counters in the bag be x.
That means 80% of x would be 12 because 80% of the counters in the bag are blue and there are 12 blue counters.
So your job here then is to work out what would 100% of x be.
You can get that in a few different ways.
You could divide both of those numbers by four to get 20% of x is three and then multiply 'em by five to get 100% of x is 15.
Another method could be to visualise this on a bar model.
It looks something a bit like this.
We have our ratio of one to four.
We know there are 12 blue counters in the bag, so we could do 12 divided by four to work out that each box represents three counters and then we could multiply it by five to get the number of counters that all five boxes represents, and that's 15.
Now, with methods one and two, we haven't in fact done the same calculations.
We've done 12 divided by four and then multiplied by five.
It's just been presented in a slightly different way each time.
Another method could be to do 12 divided by 0.
8, which is 15.
That's how you'd work out the original amount before percentage change, or you can do any other method and you should get 15 counts in total regardless of which method you do so long as your calculations are correct.
Then part D, you're told that Alex carries out 400 trials and you had to work out how many more times should Alex expect to pick a blue counter than a red counter? Well, you should expect to draw around 80 red counters and around 320 blue counters, which could be calculated using either fractions, decimals, percentages or ratios, that matter, and then once you've got that, the difference between those would be 240.
Now, with Question 3, Jacob and Sofia each play a tennis match and the partially completed probability tree shows a probability that each person wins their match.
But the issue here is that Jacob's probability is expressed as a fraction while Sofia's is expressed as a decimal.
You have to work out the probability at least one of them wins their match.
Now, there are plenty of different ways you can do this, but a good way is to express your probabilities in the same form.
For example, we could express all our probabilities as decimals and it would look something a bit like this.
If we then work out the probabilities for each of the scenarios where at least one of them wins, we would have these probabilities here, and then we could add them together to get a probability that at least one of them wins as 0.
95, or you can do that with fractions or with percentages if you want to, and you should get whatever's equivalent to 0.
95.
Well done so far.
Let's now move on to the next part of this lesson where we're going to look at using different representations for probability.
Let's begin with this scenario here where we have two spinners.
Each spinner is spun once and the sum of the outcomes is found, and what we're going to do is represent the probability of each total on a probability scale, and Jun's going to help us.
He suggests that a two-way table would be a useful representation as a first step in this problem.
That's a great idea, Jun, and it could look something a bit like this.
The columns show us the outcomes for the bottom spinner and the rows shows the outcomes for the top spinner, and when we add those outcomes together, we get these possible totals.
We can see now that there are four different possible totals.
They are 2, 3, 4, and five, so let's calculate the probabilities.
The probability that the total is two would be 3/25 because there are three outcomes in that sample space where the total is two, out 25 outcomes altogether.
The probability that it is three would be 5/25.
The probability that a total is four would be 11/25, and the probability that the total is five would be 6/25.
So now we've got the probabilities as fractions.
We could leave 'em as fractions and work with those, but because we are going to put these onto a probability scale, it might be easier to express them as decimals and they would be these decimals here.
Let's now put these probabilities onto the scale.
The probability that the total is two is 0.
12, so we can go to 0.
1 the scale and go along one minor interval from there, and that will get us to 0.
12.
The probability that the total is three is 0.
2, and it would go here on the probability scale.
The probability that the total is four is 0.
44, so you could go to 0.
4 on the scale first and then go along two minor intervals, 0.
42, 0.
44, and then the probability that a total is five is 0.
24, so that would go here on the probability scale.
So let's check what we've learned.
Here we've got a two-way table that shows a sample space for a trial where the spinner that you can see on the screen is spun twice and the product of the outcomes is found.
Could you please use a table to write the probability that the product is four and express your answer as a fraction? Pause the video while I do it and press Play when you're ready to see what the answer is.
Well, there are three outcomes at that table where the product is four, out of 25 outcomes altogether, so probability would be 3/25.
Let's now put this probability onto a probability scale.
Which arrow represents this probability? Pause while I do it and press Play when you're ready to see what the answer is.
The answer is B.
It can be helpful to express the probability as a decimal as that's what is on the probability scale, and then once you do that, you can see 0.
12 is where the arrow for B is.
Let's now take a look at a different representation which is used for probability problems. That is a Venn diagram.
A Venn diagram may be a useful representation when people, objects, or elements can be belong to multiple groups at once.
For example, a pack of party hats contains hats with the following features.
They could have spots with toppers, they can have spots without toppers, stripes with toppers, or stripes without toppers, and the Venn diagram shows a frequency of each type in the pack where S represents the hats that have spots, and T represents the hats that have toppers.
I can see that some hats have both spots and toppers and some have neither.
Let's now use this Venn diagram to work out some probabilities.
A party hat is chosen at random from the pack.
Set notation can be used to write the probability of specific events.
For example, this probability here is a probability that then S happens.
In other words, the probability that the hat has spots.
Looking at the Venn diagram, we can see that there are 19 plus 26 hats in the box that do have spots, out of 100 hats altogether, so the probability will be 45/100.
This probability here is for the complement of event S.
In other words, the probability that the hat does not have spots.
We can see from the Venn diagram that there are 34 plus 21 hats in the box that do not have spots, and that means that the probability will be 55/100.
This probability here is for the event that the hat chosen satisfies the intersection of events S and T.
In other words, it's the probability that the hat has both spots and a topper, and that would be 26/100.
This probability here is for the event of the hat satisfies the union of S and T.
In other words, the probability that the hat has either spots or a topper or both.
And from the Venn diagram, we can see that the number of hats that satisfy this is 19 plus 26 plus 34, which means the probability would be 79/100.
And then this probability, this is a probability for the event of S happening given that T has happened.
In other words, the probability that the hat has spots given that we know it has a topper.
Well, this time, it means the hat we have chosen can't be any of the 100 hats in the box because we know it has a topper.
Therefore, it can only be one of the hats that has a topper.
There are 26 plus 34 hats in that box that have a topper, and 26 of those have spots, so probability will be 26/60.
Let's now take a look at four of these probabilities that we found.
These are all expressed as fractions, but they could be expressed in other forms. For example, as decimals or as percentages, and we could even represent these probabilities on a probability scale, a bit like this.
For example, the probability that the hat has spots would go here.
Now, it's worth noting that arrow is pointing in between two of the minor intervals.
That's because the minor intervals in this probability scale go up in 0.
02 each time, and this probability is 0.
45, so it needs to go in the midpoint between 0.
44 and 0.
46 on the probability scale.
And then the probability about the hat has spots and a topper would go here.
The probability of it has either spots or a topper or both would go here and the probability has spots given that it has a topper would go here.
So let's check what we've learned.
Here we've got a Venn diagram with one region shaded, which notation represents that shaded region? Pause the video while I choose and press Play for an answer.
The answer is A.
It's the intersection of A and B.
Or in other words, all the outcomes that satisfy both event A and event B.
What is the probability of this event here happening? It is the event that B happens given that you know A has happened.
Pause while I choose an answer and press Play when you're ready to see what the answer is.
The answer is A.
It's 50/85.
You know A has happened, so it's out of one of the 85 possibilities for A, and there are 50 of those that also satisfy B.
Okay, it's over to you now for task B.
This task contains three questions, and here is Question 1.
Pause video while I do it and press Play for Question 2.
And here is Question 2.
Pause while I do this and press Play for Question 3.
And here is Question 3.
Pause while I do this and press Play for some answers.
Okay, let's go through some answers.
In Question 1, you had two spinners, which are each spun and the total of the outcomes is found.
In part A, you have to complete the two-way table to show the sample space.
It looks something a bit like this.
And then part B, you have to represent the probability of each total on the probability scale.
You've got a few steps to do here.
You first need to work out the probabilities, which you may do as a fraction first, but you'll need it as a decimal to then put it onto the scale.
So for example, the probability that the total is two, you might work out initially as 1/20, but that would be 0.
05 as a decimal, and it would go where the arrow is on the scale.
The probability that the total is three will be 0.
1.
For four, it'll be 0.
15, and for five, it'll be 0.
2.
And then for the remaining probabilities, you may notice that they are the same as the ones you've already worked out.
The probability for six will be 0.
2.
For seven will be 0.
15.
For eight will be 0.
1, and for nine will be 0.
05.
Then in Question 2, you had a Venn diagram, which shows you how many of each type of marble there are in a box.
You had to work out some probabilities and then put 'em onto a probability scale.
For A, the probability that a chosen marble would have a swirl is 0.
4.
The probability that it does not have the swirl will be 0.
6.
The probability that it's blue will be 0.
3.
The probability that it's not blue will be 0.
7.
The probability that it is swirled and blue will be 0.
06.
The probability that it's either swirled or blue or both will be 0.
64.
And the probability that it's swirled, given that it's blue, will be 0.
2.
Then for Question 3, you are working with a scenario about flipping a Frisbee, which can land either heads or tails.
And in part A, you had to complete the probability tree.
It would look something a bit like this.
In part B, you have to represent the probabilities of the following events on the probability scale.
The first event is A, which is both Frisbees land the same way.
That can either happen with heads and heads or tails and tails.
If you add the probability of those together, you would get 0.
505 and it would go just above the halfway mark on that probability scale.
And for B, the probability about the Frisbees land differently.
That could either be heads and then tails or tails and then heads.
You add those probabilities together, you get 0.
495, and that would go just before the halfway point on the probability scale.
So it looks like these two events have very similar probabilities.
They are nearly equally likely, but not quite.
The probability that there's two Frisbees land the same way is just slightly higher than the probability that they land differently.
Fantastic work today.
Now, let's summarise what we've learned.
Problems involving probability may require a strong knowledge of converting between fractions, decimals, and percentages.
Sample spaces can be presented using many different methods.
For example, as a two-way table, or a Venn diagram, or a tree diagram.
Selecting an appropriate and efficient method for solving the problem comes from evaluating the different methods.
The more you practise, the more you get a sense of which representation is more useful in different scenarios, so you get good at choosing the most helpful one for your particular problem.
A clear, logical, structured argument is much easy to follow and understand than providing only the final answer.
So when you solve a problem, make sure you show all the steps you take to get to the answer.
Well done today.
Have a great day.
