Lesson video

Hello there, you made a great choice with today's lesson.

It's gonna be a good one.

My name is Dr.

Ronson and I'm gonna be supporting you through it.

Let's get started.

So welcome to today's lesson from the unit of probability and theoretical probabilities.

This lesson is called Summing Probabilities and by the end of today's lesson we'll be able to show that the probability of all unique outcomes for a trial sum to one.

Here are some previous keywords that we're going to use again during today's lesson.

So if you want to remind yourselves what any of these words mean, you may wanna pause the video while you do that and press play when you're ready to continue.

This lesson will also introduce two new keywords.

One of the keywords is mutually exclusive.

Two or more events are mutually exclusive if they share no common outcome, and we'll see plenty of examples of that during today's lesson.

The second key word is exhaustive.

A set of events are exhaustive if at least one of them has to occur whenever the experiment is carried out.

And we'll see some examples of that too.

The lesson contains two learn cycles.

In the first learn cycle, we're gonna be summing probabilities of events and then we're gonna use what we've learned during the second learn cycle to find some missing probabilities.

But start off with summing probabilities of events.

Here we have Sam and Jun and they have a fair, regular six-sided dice.

The dice is rolled once and the sample space shows the possible outcomes that can occur.

Let's find some probabilities now.

The probability that the dice rolls an even number is three sixth, and the probability the dice rolls an odd number is also three sixths.

Now Sam spot something here.

Sam says, I've spotted something interesting.

The probabilities of these events sum to one whole can see that because we have three sixths plus three sixths, they sum to six sixths, which is equivalent to one whole.

Sam says, I wonder why that's happened.

Why do you think these probabilities sum to one whole? Perhaps pause the video while you think about this and then press play when you're ready to continue.

Jun's gonna explain what he thinks.

Jun says there are six possible outcomes when rolling the dice.

1, 2, 3, 4, 5, and 6.

Jun says three of them are even, that's two, four, and six.

The three remaining outcomes are odd.

That's one, three and five.

So it is certain that you will roll one of these six numbers.

So the probability is one because one represents certainty when it comes to probability.

So let's look at some other probabilities.

The probability of rolling a multiple of three is two sixths.

The probability of rolling a multiple of four is one sixths.

Sam looks at this and thinks, hmm, well these two probabilities, well they sum to three sixths, that's not one whole.

Sam says, these probabilities do not sum to one whole.

So why do you think that's happened? Why do you think these probabilities do not sum to one whole, whereas the last time they did? Pause the video while you think about this and then press play when you're ready to continue.

Jun's gonna explain what he thinks again.

Jun says two of the possible outcomes are multiples of three.

That's three and six.

One of the possible outcomes is a multiple of four, that's four, that accounts only three out of the six possible outcomes.

So that means it isn't certain that one of these events will occur.

So if it's not certain, the probabilities will not sum to one.

Let's look at two other probabilities.

The probability of rolling a factor of 10 is three sixths.

The probability of rolling a factor of 12 is five sixths.

These also do not sum to one whole.

Sam says every number on the dice is accounted for in these two events.

So why do these probabilities not sum to one whole? Why do you think it is this time? Pause the video while you think about it and then press play when you're ready to continue.

Let's hear from Jun again.

Jun says some of the outcomes are accounted for twice with these probabilities.

We can see that one is a factor of 10 and a factor of 12, so is two.

So both of those outcomes are accounted for twice, Jun says this is because there are numbers on the dice that are both factors of 10 and 12.

So they satisfy both events like we can see with the numbers one and two.

That means these probabilities do not necessarily sum to one whole, when a set of events are all unique and account for all possible outcomes of a trial, the probability sum to one, for example, rolling a regular six-sided dice, if your two events are even and odd, well if we represent that in a Venn diagram, what we can see here is that these events are mutually exclusive because there are no numbers on a dice that are both odd and even.

And we can see that from the Venn diagram because the region which is the overlap of sets A and B is empty, these events are also exhaustive.

That's because there are no numbers on a dice that are neither odd or even, and we can see that from the Venn diagram because the space around the two sets is completely empty of outcomes.

So when we think about the events rolling an even number on a dice and rolling an odd number on a dice, every possible outcome of a trial is accounted for exactly once by these two events.

So the probabilities sum to one and the Venn diagram could even look like this because there is no space for the overlap and there is no space around the outside of the two sets as well.

If a set of events are not mutually exclusive or if they are not exhaustive, then the probabilities may not necessarily sum to one.

For example, if we roll a regular six side dice and the two events we're looking at are, rolling a multiple of three and rolling a multiple of four, can you see why these two events do not necessarily sum to one based on the Venn diagram? We can see that these two events are mutually exclusive because there are no outcomes in the intersection between A and B, the region where those two sets overlap, but we can see that there are some numbers on the outside of those two sets.

Therefore these events are not exhaustive because there are numbers that are neither a multiple of three nor are multiple of four, that can be rolled on a dice.

So the probabilities do not sum to one.

Another example is this, rolling a regular six side dice and the events being factors of 10 and factors of 12.

If we look at the Venn diagram for these outcomes, can we see why these two probabilities do not sum to one? It's not that these events are not exhaustive because we can see that the region around the outside of these two sets is empty.

All the numbers on the dice are factors of 10 or 12, but what we can see is these two events are not mutually exclusive.

That's because we can see that there are some numbers that are both factors of 10 and 12.

So the probabilities do not sum to one.

Here we have a spinner with the numbers one to five on it.

A game is being played where players spin the spinner twice and find the sum of the numbers.

And we have an outcome table here that shows the possible totals.

So let's think about some possible events that can occur in this scenario.

One event could be that the total is a prime number and we can see all the possibilities in the table here where that is the case.

Another event could be that the total is not a prime number.

You can see all of the ways that can happen from the table as well.

Now these two events, they are mutually exclusive because there's no way that a number can be both prime and not prime and they are also exhaustive because every single outcome in this table is accounted for by one of those two events.

Therefore, the probabilities of these events would sum to one whole.

And we can see that here with the actual probabilities themselves, the probability of it being prime is 11 25ths.

The probability of not being prime is 14 25ths.

If you add those together you get 25 25ths, which is one whole.

Let's look at some other combinations of events.

If event A is that the total is a prime number and event B is that the total is a square number, would these probabilities sum to one? Well, we can see that they are mutually exclusive.

There are no numbers on that table that are both square and prime, but they are not exhaustive and we can see that from a table because there are some numbers that are not accounted for by either the A or B.

In this case, the numbers six, eight and 10 do not satisfy either of those events.

So these events are not exhaustive.

Therefore these probabilities will not sum to one whole and we can see that by looking at the probabilities themselves.

We have 11 25ths plus five 25ths that gives you 16 25ths, which is not one whole.

Let's look at another combination of events.

If event A is that we get a prime total and event B is that we get a factor of 60, would the probabilities of these events sum to one whole? Well, we can see here that these two events are not mutually exclusive.

There are a few outcomes that satisfy both events.

We can see two, three and five are prime numbers and factors of 60.

So they are not mutually exclusive, they're also not exhaustive.

There are some outcomes that do not satisfy either these events.

They are eight and nine.

So because they are not mutually exclusive and also because they're not exhaustive, these two probabilities will not necessarily sum to one.

We can see that if we look at the probabilities themselves, 11 25ths plus 16 25ths is 27 25ths, which is not equivalent to one whole.

And here's a different scenario.

A bag contains counter labelled with letters A to C, some counters are black and some counters are white.

A counter is drawn at random from the bag and here are some possible events that could happen when we do that.

The counter could contain a letter A, it could contain a letter B, a letter C, it could be black and it could be white.

For which sets of events would the probabilities sum to one? Perhaps pause the video while you think about this and press play when you're ready to go through it together.

Well, we need to find some events that are both mutually exclusive and exhaustive.

One option here could be this.

These three events are exhaustive because you are guaranteed to get one of these three letters.

There are no other letters in the bag and these three events are mutually exclusive because there are no counters in that bag that contain more than one letter.

So these probabilities will sum to one.

Also, this pair of probabilities too, will sum to one.

These two events are mutually exclusive because counters are either black or white, but not both.

And these two events are exhaustive because there are no other counters in there that are not black or white.

Therefore you are guaranteed to get either a black counter or a white counter.

Therefore their probabilities will sum to one whole.

So let's now think about why the probabilities for these two events do not sum to one and let's justify our reasoning with an example.

Our two events are drawing a counter with an A on it and drawing a white counter and we can see that the probabilities sum to nine eighths which is not equivalent to one whole.

So why do you think the probabilities of these two events do not sum to one? Perhaps pause the video while you think about it and press play when you're ready to talk about it together.

The reason why is that these two events are not mutually exclusive.

For example, there are three counters that are white and contain the letter A.

So because they're not mutually exclusive, they don't sum to one whole.

We can also see that these events are not exhaustive because there are some counters that do not satisfy either of these events.

For example, there are two counters that are neither white or containing the letter A.

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

True or false, for this spinner, the probability of it landed on A plus the probability of it landed on B equals one.

Is that true or is it false? Pause video while you choose and then press play when you're ready to continue.

The answer is false.

So what's our justification? Is it A, that the events are not mutually exclusive, or is it B, because the events are not exhaustive? Pause the video while you make a choice and press play when you're ready for an answer.

The answer is B.

These events are not exhaustive.

You can get other outcomes that are not A or B.

The spinner could land in other letters such as C, D, and E.

True or false, the probability of getting a letter from the word ace plus the probability of getting a letter from the word dab equals one.

Is that true or is it false? Pause the video while you make a choice.

Then press play when you're ready to continue with this question.

The answer is false.

So what's our justification? Is it A, that the events are not mutually exclusive or is it B, the events are not exhaustive? Pause the video while you make a choice and press play when you're ready for an answer.

The answer is A, these events are not mutually exclusive.

We can see that the letter A is included in both of those events.

So true or false, the probability of getting a vowel plus the probability of getting a continent equals one.

Is that true or is it false? Pause the video while you make a choice and press play when you're ready for the next part of this question.

The answer is true.

So what's our justification? Here are some options.

Pause the video while you make some choices and press play when you're ready for an answer.

These two probabilities sum to one because the events are mutually exclusive and the events are exhaustive, they have to be both.

Okay, it's over to you now.

For task A, this task contains two questions and here is question one.

We have a spinner here with the numbers one to eight on it.

The spinner is spun once, for each pair of events listed below, state whether the probabilities would sum to one and justify your answers with reasoning.

So to justify your answers, explain whether or not they are mutually exclusive or whether or not they are exhaustive and give some examples in cases where they are not either those things.

Pause the video while you have a go at this and press play when you're ready for question two.

Here is question two.

A game involves the set of cards below, each card shows one of three shapes.

We can see there's a rhombus shape, a lightning shape, and a oval shape.

And also cards are either solid or hollow.

We can see the example of a solid rhombus and example of a hollow rhombus.

So based on how these cards look, from the list of events below, find sets of events where the probabilities would sum to one whole and there are multiple options that sum to one whole.

Pause the video while you have a go at this and then press play when you're ready to go through some answers.

Okay, let's see how we got on with that then.

We wanna decide whether each pair of events would have probabilities that would sum to one and justify our answer.

So in part A, event A is odd, event B is even.

Yes, these probabilities would sum to one because the events are mutually exclusive and exhaustive.

In part B, if event A is odd and event B is multiple of four, these probabilities do not sum to one because the events are not exhaustive.

For example, two and six are neither odd or a multiple of four.

In part C, event A is odd, event B is factor 24.

These probabilities do not sum to one because the events are not mutually exclusive.

For example, one and three are both odd and a factor 24.

Then D, if event A is odd and B is prime, well these probabilities do not sum to one and we could justify in a couple of different ways.

We could say that they are not mutually exclusive.

For example, three, five, and seven are both odd and prime.

Or we can say, well the probabilities do not sum to one because there are events that are not exhaustive.

All we could say it's because the events are not exhaustive.

For example, four, six and eight are neither odd or prime and part E, if event A is integer and event B is non integer, these probabilities sum to one because the events are mutually exclusive and exhaustive.

The probability an integer is one whole.

The probability to get a non integer is zero.

Those two probabilities sum to one.

Then question two, we had to find some events where the probabilities would sum to one whole from this set of cards.

Well the probabilities for events A, D and H would sum to one whole.

That is the probability of getting a rhombus, getting a lightning, and getting an oval.

The probabilities of events C and F sum to one whole, that's the probability of getting a solid card and a hollow card.

The probabilities of events, B, E, G, and I, they sum to one whole and that is the probability of getting one shape on the card, two shapes in the card, three shapes or four shapes and the probabilities of events H and J, they sum to one whole.

That is the probability of getting oval card or not an oval card and the probabilities of events B and K, they sum to one whole.

That's one shape or more than one shape.

You're doing great so far.

Now let's move on to the second learn cycle, which is applying what we've learned to find some missing probabilities.

Here we have Aisha and Jacob, they're about to play each other in a tennis match.

Now the only possible outcome of this match is either Aisha wins or Jacob wins and we can see the probabilities of these two events represented on the probability scale.

We can see that Aisha is more likely to win than Jacob.

What should bear in mind with these probabilities is that the more likely it is that Aisha wins, the less likely it is that Jacob wins.

As the probability that Aisha wins increases, the probability that Jacob wins decreases like this.

Now as these are the only two possible outcomes and it is not possible for both to happen in the same trial, there are probabilities sum to one.

These two events are mutually exclusive.

They can't both win.

And they are exhaustive because at least one of them has to win.

So if these two probabilities sum to one whole, it means that if the probability of one event is known, then the probability for the other then can be calculated by subtracting from one.

For example, if the probability for Aisha wins is seven tenths, then the probability of a Jacob wins is one subtract seven tenths, which is three tenths.

Those two probabilities, seven tenths and three tenths, they sum to one whole.

Calculations with probabilities could also be performed using the equivalent decimals.

For example, if the probability of Aisha wins is naught 0.

7, then the probability that Jacob wins is one subtract naught 0.

7, which is naught 0.

3.

The calculations could also be performed with the equivalent percentages.

If the probability that Aisha wins is 70%, one whole as a percentage is 100%, therefore the probability that Jacob wins is 100% subtract 70%, which is 30%.

Now, as these outcomes are mutually exclusive and exhaustive, the scenario could also be modelled with an outcome tree.

For example, we could use fractions, decimals, or percentages, but in each case there are only outcomes.

Aisha wins or Jacob wins, and we could subtract the probability that we know from one whole to find the missing probability in each of these outcome trees.

So in the one on the left using fractions, we could do one subtract seven tenths to get three tenths for the probability that Jacob wins.

In the middle one we're using decimals to get the probability Jacob wins, we'll do one subtract naught 0.

7, which is naught 0.

3.

And on the outcome tree on the right hand side, using percentages to get the probability that Jacob wins, we could do 100%, subtract 70% to get 30%.

Also, as these are the only two possible outcomes, the two sample spaces below are equivalent.

The top sample space shows us that the possible outcomes are either Aisha wins or Jacob wins.

And the bottom sample space shows us that the possible outcomes are either Aisha wins or Aisha does not win.

Aisha does not win, is equivalent to Jacob winning because if Jacob wins then Aisha does not win.

This could be represented on an outcome tree like this, one outcome is Aisha wins.

Another outcome is that Aisha does not win or on a probability scale like this.

So let's think about this scenario more when we're trying to find the probability that an event does not happen.

If the probability of an event is known, then the probability that the event does not happen could be calculated by subtracting from one.

For example, if the probability that it rains tomorrow is naught 0.

35, well the opposite of that is it not raining.

The probability of it not raining could be calculated by doing one subtract naught 0.

35, which is naught 0.

65.

Now you might be thinking, what about other possible outcomes such as snow and sunny and cloudy, but dry and so on.

Well the event not rain includes all other possible outcomes aside from rain such as sunny, snow and so on.

Here we have Alex and Izzy, they play against each other in a chess match.

The table shows the three possible outcomes of the match and the probabilities for two of these outcomes are shown in the table as well.

And we need to calculate the missing probability.

Well, we know that these three outcomes are mutually exclusive because there are no situations where two of these outcomes can happen at the same time.

And we know that these three events are exhaustive because there are no other options.

Therefore these three probabilities would sum to one.

So to find the missing probability we could do one, subtract the sum of the two probabilities we know.

So the probability of getting a draw is one.

Subtract the sum of naught 0.

4 and naught 0.

45.

That gives us naught 0.

15.

Laura, Andeep and Sofia compete against each other in a race.

The table shows the three possible outcomes of the race.

So either Laura wins, Andeep wins or Sofia wins.

Now for this scenario, we don't know any of the actual probabilities, but we're gonna get some information about how some of these probabilities compare to each other.

Andeep is twice as likely to win as Laura, Sofia is as likely to win as Andeep.

So based on that information, can we find the probability that Laura wins? So let's think about this together now, all three probabilities are unknown, but they are related to each other.

And the unknown that we want to try and find is the probability that Laura wins.

So let's call that probability X.

If the probability that Laura wins is X, well we know that Andeep is twice as likely to win as Laura.

So the probability that Andeep wins is two X and we know that Sofia is as likely to win as Andeep so she has the same probability, which is also two X.

So now we have an expression for each probability and we know that these three probabilities sum to one whole.

So we can make an equation out of this now.

We can write that x plus two x plus two x equals one whole.

That's the sum of the three probabilities is one whole.

And now we've got an equation we can simplify it and rearrange to solve it.

The value of X is naught 0.

2.

So the probability that Laura wins this is naught 0.

2, one fifth or 20%.

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

Find the value of X for the outcome three below, pause the video while you have a go at this and press play when you're ready for an answer.

The answer is three fifths, which you get from subtracting two fifths from one whole.

There is a 55% probability that it rains in town A tomorrow.

What is a probability that it does not rain.

Pause the video while you write down answer and then press play when you're ready to see what the answer is.

The answer is 45%, which we get from subtracting 55% from 100%.

Find the value of X from this outcome tree below.

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

This answer is naught 0.

85, which we get from doing one subtract naught 0.

15.

Find the value of X from this table here, pause the video while you have a go and press play when you're ready for an answer.

The answer is naught 0.

6, which we get from subtracting the sum of naught 0.

3 and naught 0.

1 from one whole.

Right, it's over to you now for task B, this task contains three questions and here is question one.

Find the value of each unknown probability.

Each question presents a complete set of possible outcomes for a trial.

So you can assume with each question that the outcomes are exhaustive and also they are mutually exclusive as well.

Pause the video while you work out each unknown probability and then press play when you're ready for questions two and three.

And here are questions two and three.

In question two, we have Izzy, Laura, and Sofia.

They compete in a swimming race and the outcomes and the probabilities are represented in this table.

You need to find a value of X and in question three you've got Aisha, Jun, Lucas and Sam compete against each other on a video game.

You've got some information about how the probabilities compare with each other and you need to find a probability that Sam wins.

Pause the video, while have a go at these and then press play when you're ready to go through answers.

Let's now go through some answers.

Question one, part A, the probability of rain is 25%.

So the probability does not rain is 75%.

The probability that a train is on time is naught 0.

6.

So the probability that the train is not on time is naught 0.

4.

In part C, the probability of not winning is three sevenths.

So the probability of winning is four sevenths.

In part D, the probability of the draw is 44%.

In part E, the probability of getting red is naught 0.

1, getting green is naught 0.

35.

So the probability getting blue is naught 0.

55.

And then in F, the probability of A, B and D respectively are 1 6ths, 1 quarter and three twelves.

So what's left over? Well, the probability of C is one third or you can have it as the equivalent fraction four twelves or any equivalent fraction.

Question two, so Izzy, Laura and Sofia compete in the swim race, the outcomes are below.

So when to find the value of X based on what we can see, those three probabilities sum to one whole.

So if we make an equation, simplify it, and rearrange it, we get the value of naught 0.

2 for X, or you can have it as any equivalent fraction or percentage.

In question three, let's use this information where we can start by writing an expression for each person's probability.

So if we label the probability that Aisha wins as X, while Jun is as likely as Aisha, that's also X.

Lucas is twice as likely, that's two X, Sam is three times as likely as Lucas.

So that's three times Lucas's probability that gives six X those four probabilities sum to one whole, we've made an equation outta that.

We simplify it, we rearrange and solve it to get the value of X and then you'd need to multiply that by six to get the value of six X 'cause that is Sam's probability.

However you've done it, your answer should be naught 0.

6, 6 tenths, 60% or anything equivalent.

Excellent work today.

Let's now summarise what we've learned in this lesson.

When a set of events are mutually exclusive and exhaustive, the probabilities sum to one and remember, mutually exclusive means that there is no outcome that can satisfy two events.

And exhaustive means there are no outcomes that satisfy neither the events.

So if they are mutually exclusive and exhaustive, the probabilities will sum to one.

If a set of events are not mutually exclusive or if they are not exhaustive, the probabilities may not necessarily sum to one.

And if the probability of all but one outcome is known, then the probability of the remaining outcome can be calculated by subtracting what you know from one.

And if the probability of an event is known, then the probability that the event does not happen can be calculated by subtracting from one as well.

Thank you very much today.

All the best.