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Hi there.
My name is Dr.
Olson and I'm excited to be guiding you through this lesson.
So let's get started.
Welcome to today's lesson from the Unit of Conditional Probability.
This lesson is called "Probabilities Involving Algebra," and by the end of today's lesson we'll be able to work with algebraic statements using the fact for exhaustive events sum to one.
A previous keyword that will be useful during today's lesson, is the word probability.
The probability that an event will occur is the proportion of times that the event is expected to happen in a suitably large experiment.
The lesson is going to be broken into for learning cycles and we're going to start with probabilities from bar models.
Here we have a scenario where we have some marbles in a bag and Jacob informs us that the bag only has purple and green marbles inside.
He says, "The probability of randomly choosing a purple one is 0.
3." So let's think about what else we know based on this information.
Sophia says, "I know that the probability of randomly choosing a green marble is 0.
7 because the two probabilities sum to one." Let's now represent this scenario on a bar model.
The top bar represents the probability of drawing any marble from the back.
That is one whole.
You are certain to get any marble when you draw a marble from the back and underneath it we have the bar representing the probability of drawing a purple marble 0.
3, which takes up part of that whole.
That means the remainder of that whole is the probability of drawing a green marble and that probability would be one subtract 0.
3, which is 0.
7.
We can use bar models to solve more complex probability problems that involve multiple unknown probabilities.
For example, this bag only has purple, green, and blue marbles.
We're told that the probability of drawing a purple marble is 0.
2, and we're told that the probability of choosing a green marble is equal to the probability of choosing a blue marble.
This list is exhaustive as there are no other colours of marbles in the bag.
This list is also mutual exclusive as no marbles are more than one colour.
So let's represent this on a bar model.
Here we have a bar model representing one whole and that is the sum of all three probabilities.
You could even say it's a probability of drawing any marble from the bag.
The probability of drawing a purple marble is 0.
2.
So that means the rest of this bar would be for the probabilities of drawing a green or a blue marble.
Now the probabilities of drawing either or green or blue must be 0.
8.
That is one subtract 0.
2.
And we're told that these probabilities are equal to each other.
So if I want to work out the probability of drawing a green marble, I could do 0.
8 divided by two to get 0.
4, and the probability of drawing a blue marble will be the same as that.
So that would also be 0.
4.
So let's check what we've learned.
A bag contains marbles that are all either purple, green, blue or yellow.
We also have some information on the bottom left side of this screen.
We're told that the probability of drawing a blue marble from the bag is 0.
55, and you're told that the probability of choosing a purple, green or yellow marble are all equal to each other.
So you've got three bar models.
Which of those bar models represents this scenario? Pause while you choose and press play When you're ready for an answer.
The answer is B.
In that bar model, it shows the probability of getting a blue is 0.
55 and 0.
55 is a little bit more than half of the bar.
So what is the total probability of choosing a purple, green or yellow marble from its bag? Pause video while you write it down and press play when you're ready for an answer.
The answer is 0.
45.
We get that from doing one subtract the probability on drawing a blue marble.
So let's now find P of Y.
That is the probability of randomly choosing a yellow marble from this bag.
Pause video while you write it down and press play when you're ready for an answer.
The answer is 0.
15.
We get that from dividing the remaining probability 0.
45 by three.
Bar models can also be used to find multiple unknown probabilities of different value.
For example, Jacob plays a game at a school fair.
The game can either be won, lost or repeated, which means he plays it again and the table shows the probabilities that he knows so far.
He knows that the probability that the game is repeated is 34%, but we don't know the probability of either winning or losing.
However, we are told that the probability of Jacob losing is twice the probability of him winning.
So let's represent this on a bar model.
The top bar represents the probability for any of these three things happen, and that is 100%.
Underneath we have a bar for the probability that he has to repeat the game, which is 34%, which means the rest of that bar is for the probability of either winning or losing.
Now we don't know the probability of Jacob winning, but let's just draw a bar there for now just as a bit of a placeholder.
And we are told that the probability of Jacob losing is twice this amount.
So we could draw two of those bars and label them with lose 'cause that probability is twice the amount of the probability of winning.
Now we've got that we can work out the probability that he either wins or loses by doing 100% subtract 34%, that would give 66%.
That means these three small bars would add up to 66%.
And we know that the value of each of these three small bars are all equal to each other.
So the probability that it wins could be calculated by doing 66% divided by three, which is 22%.
We could also fill that in in the other two smaller bars like this.
Then we can look at the probability that he loses.
That would be 44%, it'd be two lots of those 22%s.
And now we have all the information in the table.
So let's check what we've learned.
Sophia plays a game at a school fair.
The game can either be won, lost or repeated, and the table shows the probabilities you have so far.
You know the probability that the game is repeated is 40%.
You are told that the probability of Sophia winning the game is three times the probability of her losing.
So which of these bar models represents the probability in this game? Pause video while you choose and press play when you're ready for an answer.
The answer is A, You've got the large bar which represents the probability that the game is repeated 40%.
And then you've got four smaller bars that are all of equal size and three of them represent the probability of winning.
So what is the total probability of Sophia winning or losing this game? Pause video while you write it down and press play when you're ready for an answer.
The answer is 60%, which we get from doing 100% subtract 40%.
So with that in mind, could you please complete each sentence on the screen here? The probability of Sophia losing is A and the probability of Sophia winning is B.
What should go in those blanks for A and B? Pause video while writing it down and press play when you're ready for an answer.
The probability of Sophia losing is 15%.
The probability of Sophia winning is 45%.
Okay, it shows you now for Task A, this task contains three questions and here is question one, pause video while you do this and press play for more questions.
And here are questions two and three, pause video while you do these and press play for answers.
Okay, let's take a look at some answers.
Here are the answers to question one, pause while you check these and press play for more answers.
Here are the answers to question two.
Pause while you check these and press play for more answers.
And here are the answers to question three.
Pause while you check these and press play for the next part of this lesson.
Well done so far.
Now let's move on to the next part of this lesson, which is looking at algebraic probabilities from descriptions.
It is possible to create and solve algebraic equations in order to find multiple unknown probabilities, we can write the probability of each outcome as either a number which are fraction or decimal or as an algebraic statement.
Let's take a look at an example.
Lucas, Sophia and Andeep all play a game together.
Exactly one person can win each game.
Andeep has a 28% probability of winning.
Sophia has twice the probability of winning as Lucas.
So what is the probability of Sophia winning this game? Well, we can look at the probability of each outcome.
In other words, each person winning.
So Andeep has a 28% probability of winning.
That means P of A is 0.
28.
Now whilst the probability is given as a percentage, it is helpful to write as a fraction or as a decimal while we perform calculations and solve equations with it.
We don't know the probability of Sophia winning, but we do know that the probability is twice the probability of Lucas winning.
So we could express this algebraically, we could put the probability that Lucas wins is X and then we could put the probability that Sophia wins as two x.
'Cause two X is twice of X.
You could have done it as putting the probability of Sophia wins as X and the probability that Lucas wins as half X, but then you are working with fractions and that's fine, but it does make the calculations a little bit tricky to do.
Using x and two X will be an easier equation to solve later on.
Here comes Jacob, Izzy and Alex who each construct an equation using these probabilities to try and find the value of X.
Jacob says we could write the equation.
Two x plus x equals 0.
28.
Izzy says we can write the equation two x plus 0.
28 equals one.
And Alex says, we can write the equation two X plus X plus 0.
28 equals one.
Now I'm not all of these are correct.
Pause video and think about whose equation is correct here.
And press play when you're ready to continue.
Let's think about these together.
These three outcomes are all exhaustive and they are mutually exclusive.
And so all three probabilities must sum to one whole.
Let's see whose algebra statement does so.
With Jacobs, he's saying that the probability that Sophia wins plus the probability that Lucas wins is equal to the probability that Andeep wins.
That's not necessarily true.
So no, Jacobs is not correct.
Izzy's equation can be interpreted as the probability that Sophia wins plus the probability that Andeep wins is equal to one whole.
Well that can't be correct because it doesn't include the probability that Lucas wins.
So no.
Alex's equation can be interpreted to say that the probability of that either Sophia wins, Lucas wins or Andeep wins equals one whole.
And that is true.
Wonderful.
So let's use that equation to find the value of X.
We can simplify the equation, rearrange it by subtracting 0.
28 from both sides and then divide both sides by three to get 0.
24.
We can then substitute that in for the value of X.
So the probability that Lucas wins is 0.
24.
The probability of Sophia wins will be two lots 0.
24, which is 0.
48.
So let's check what we've learned.
You are given some information about a spinner.
Use this information to match each probability to its algebraic statements.
Pause video while you do this and press play when you're ready for answers.
Okay, let's take a look at some answers.
The probability that the spinner lands on purple is 0.
16.
The probability it lands on green would be 0.
32 'cause it's twice the probability that it lands on purple and the other two probabilities are unknown.
So it's all about now the relationship between those.
We know that the probability of it lands on blue is three times the probability it lands on yellow.
So if we say the probability it lands on yellow is X, then the probability it lands on blue will be three X.
So let's use these now.
The expressions for the probabilities of each outcome are given and they are the ones in the white box on the right side of the screen.
Which equation correctly shows the relationship between all four probabilities.
Your choices from A to E.
Pause while you choose and press play when you're ready for an answer.
The answer is E, because the sum of those four probabilities will be equal to one whole.
So by first solvency equation, find the probability that the spinner lands on and you've got three sub-questions to answer.
Probability lands on yellow, the probability it lands on blue, and the probability it lands on either purple or blue.
Pause video while you do this and press play when you're ready for answers.
Well first we need to solve the equation, which you can do by first simplifying it and then rearranging it.
Get the value of X as 0.
13.
We can then use that to work out some probabilities.
The probability it lands on yellow will be 0.
13.
The probability the spinner lands on blue would be 0.
39.
That's three times 0.
13.
And then the probability that it lands on either purple or blue would be 0.
55.
That's a sum of 0.
16 and 0.
39.
Okay, it's owes you now for Task B, this task contains four questions and here are questions one and two, pause video while you do this and press play for more questions.
And here is question three.
Pause while you do this and press play for question four.
And here is question four.
Pause while you do this and press play for some answers.
Okay, let's take a look at some answers.
Here are the answers to question one.
Pause while you check these against your own and press play when you're ready 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, way check and press play for more.
And finally here he answers to question four, pause, way check and press play for the next part of the lesson.
Well done so far.
Now let's move on to the next part of this lesson where we're going to look at algebra probabilities from ratios.
Here we have a weighted coin.
When a coin is weighted, it means it can be biassed so that its outcomes are not equally likely to happen.
When this coin is flipped, it can land on either heads, tails, or in this particular case on its side edge, it lands on its side edge 1% of the time.
The probability of the coin landing on heads versus tails is in the ratio six to five.
So from that information alone, we can see that it's more likely to land on heads than on tails.
How could we use this information to find the actual probability that the coin lands on heads? Perhaps pause the video while you think about what approach we might take here or what things we could work out before we try and solve the actual problem, and then press play when ready to see how we do it together.
Okay, let's now work through this together.
The probability that the coin does not land on its side is 99%.
That's 100% subtract 1%.
We can divide us 99% in the ratio six to five because the probability that lands on either heads or tails will be 99% and those two outcomes are in the ratio six to five.
We can then write and solve any algae break equation based on the information we work out.
So let's now organise this information into a table.
The probability of it lands in its side is 1%.
We don't have the probability it lands on either heads or tails, but we can write an algebraic expression for each.
We can take each part of the ratio and use them as a coefficient of X.
The probability lands on heads will be six X and the probability lands on tails will be five X.
We could create an equation out of this, but it might be easier to not work with a percentage and work with either a fraction or a decimal.
You might find it easier to work with a decimal, for example.
So we can write it as 0.
01.
Now we know that those three probabilities should sum to one whole, so we can create this equation, six X plus five X plus 0.
01 equals one, and then to find the value of X, we could simplify this equation, rearrange it, and get the value of X as being 0.
09.
Now we know the value of X, we can substitute it in to each expression to find the probabilities.
The probability it lands on heads will be six multiplied by 0.
09, which is 0.
54.
The probability it lands on tails will be five multiplied by 0.
09, which is 0.
45 and it might be worth just double checking, but those numbers all add up to one so you can check you haven't made a multiplication error along the way, and yes they do.
So let's check what we've learned.
In the box you have some information about a scenario.
Could you please match each probability labels from A to E with its algebraic expression? The boxes on the right hand side of the screen, pause video while you do that and press play when you're ready for answers.
Let's go through some answers.
The probability about the dice lands on A is 0.
35 that's given to us.
The probability it lands on B would be three X, C would be four X, and D would be six X.
We can get that from the ratio that is given to us and the probability it lands in either B, C or D.
Well, there are two ways you could write that.
If you have a write it as an algebraic expression as 13 X, that's a sum of three X, four X and six X, or you can write the actual probability, which would be 0.
65, which you get from doing one subtract 0.
35.
Could you please now write down an algebraic equation involving all four of the outcomes? Pause video while I do that and press play when you're ready for an answer.
Well, you know That the probabilities for each of these four outcomes would sum to one whole.
So our equation could be the sum of these four probabilities equals one, and you can simplify it to get 13 X plus 0.
35 equals one.
So could you please now solve this algebraic equation to find the probability of landing on D? Pause video while you do that and press play for an answer.
Well, the value of X will be 0.
05, So the probability of it lands on D will be six.
Lots of that, which would be 0.
3.
Okay, it shows you now for Task C.
This task contains two questions and here they are, pause video while you do and press play for answers.
Okay, let's take a look at some answers.
Here answers to question one, pause while you check and press play for more answers.
And here is your answer for question two.
Pause while you check this against your own and press play for the next part of this lesson.
Excellent work so far.
Now let's move on to the fourth and final part of this lesson where we're going to be using algebra to find frequencies.
We can form and solve an algebraic equation in order to find a missing frequency given other frequencies and at least one probability.
Let's take a look an example of that.
A bag of marbles contains only purple, green, and blue marbles.
The table shows us how many of each marble is in the bag.
However, we don't know how many blue marbles are in there, such represented algebraically as an unknown X in this case.
We're also told that the probability of randomly choosing a blue marble from this bag is one third.
So how can we work out how many blue marbles are in that bag based on this information? Perhaps pause the video while you think about what approach we might take to this and press play when you're ready to continue together.
Well, there are multiple different ways you can do this, but let's do one way together.
To find the number of blue marbles, we should first find an algebraic expression for a total number of marbles in the bag.
Well, the total frequency will be seven plus 11 plus X, which gives 18 plus X.
So we have an expression now for the total number of marbles in the bag.
From this table the theoretical probability of randomly selecting a blue marble is X over 18 plus X, X being how many blue marbles there are and 18 plus X being the total number of marbles altogether.
We are also told that the probability of drawing a blue is one third.
So we have an algebraic expression for the probability of drawing a blue and we have the actual probability of drawing a blue.
That means these two expressions are equal as they both describe the probability of the same outcome.
So if they are equal, we can equate them to create an equation.
We can then start to solve this equation can rearrange it first by multiplying both sides by 18 plus X and multiplying both sides by three.
We can then rearrange it to get two X equals 18, and then we can divide both sides by two to get X equals nine.
Now we know the value of X.
We can say that there are nine blue marbles in the bag, so let's check what we've learned.
A modified deck of playing cards has only hearts, diamonds, clubs, and spade cards in there.
It's a modified deck because it does not contain an equal number of each type of card as it would normally do in a regular deck of playing cards.
The table shows you how many of each card there is, but you don't know how many spade cards there are.
Could you please write down an expression for the total number of cards in the deck? Pause while you do that and press play for an answer.
The answer is 54 plus X.
That's the sum of all four of those frequencies.
So we're told a bit more information now, we're told at the probability of drawing a spade from the deck is 2/11s.
Which of these equations could help you to find the number of spades in the deck? Pause while you choose and press play when you're ready for an answer.
The answer is B.
Each side of that equation sign is the probability of drawing a spade from the deck, on the left hand side that's expressed algebraically.
There are X many spades and the total number of cards a deck is 54 plus X.
So it's X over 54 plus X.
And right on the side it's the actual probability of drawing a spade.
So with that in mind, could you please use this equation to find the total number of spades in the deck.
Pause while you do that and press play for an answer.
Well, you could rearrange this equation by multiplying both sides by 54 plus X and multiplying both sides by 11, that will give you this.
You could then expand the brackets to get this.
You can then rearrange it some more by subtracting two X on both sides, and that will give this.
And then you can divide both sides by nine to get X equals 12.
Okay, it's shows you now for Task D, this task contains two questions and here they are.
Pause video while you do this and press play for some answers.
Okay, here are the answers to question one.
Pause while you check this and press play for more answers.
And here are the answers to question two.
Pause while you check this and press play for the conclusion of today's lesson.
Fantastic work today.
Now let's summarise what we've learned.
Probabilities can be modelled either using a bar model or an algebraic equation.
We can use these models to calculate multiple missing probabilities.
We can also form and solve algebraic equations if given sets of probabilities written as a ratio.
We can form and solve algebraic equations to find a missing frequency if given other frequencies and at least one other probability.
Well done today.
Have a great day.
