Lesson video

Hi, I'm Mrs. Warehouse, and welcome to today's lesson on problem solving with possible outcomes.

This lesson falls within the unit on probability, possible outcomes.

Now, I adore probability, so I'm really looking forward to problem solving with it.

By the end of today's lesson, you'll be able to choose an appropriate diagram to display the outcomes for different events.

Now, here are some keywords we're going to use today in our lesson.

And if you want, feel free to pause to take a moment to read through their definitions.

Now, if any of those words didn't seem familiar to you, I suggest you stop watching this video and go and watch one of the ones from earlier on in this unit.

But if these are familiar, then it seems like you're ready for today.

So let's get going.

Our lesson broken into two parts, and we're going to begin by considering outcome tables and likelihood.

Sofia says, "If I were to spin spinner one twice, I would display the outcomes in this table." So here we have outcomes from the first spin and outcomes from the second spin, and then all the possible outcomes that could occur.

But if I were to spin spinner two twice, would the outcome table also look like this? Ooh, she's got a good point.

Or maybe this.

How would we draw an outcome table in these two different situations? Now, the instance I think actually both are correct.

They just have different uses.

I wonder what he means by that.

Well, the first table is helpful for showing the number of unique outcomes.

Well, that's true.

It didn't matter if I landed on one of the A's or a different A.

It's still an A.

So yeah, seeing unique outcomes really efficiently on that first far smaller table.

Now, if order matters, then there are nine unique outcomes, But if order doesn't matter, for example, CA is the same as AC, then there are only six unique outcomes.

And you can see I've taken three away from the outcome table.

Now, the second outcome table is helpful for showing which outcome is most likely to occur.

Ah, so the second table is representing how many times we saw A on that spinner.

And it's saying, "Well, hang on, there were three A's, only one B and two C's." So it's much more likely you'd have landed on A than on B.

Now, if order matters, then AA is the most likely outcome.

If order doesn't matter, then AC is the most likely outcome because AC and CA would be the same thing and therefore, you'd get 12 of those.

Now, let's do a quick check for this.

Which of these statements are correct for table two? So A, it efficiently shows that there are four unique outcomes; B, it shows the likelihood of the different outcomes; C, it is the quicker table to construct; or D, the two most common outcomes are equally likely? So which of those statements is correct for table two? Pause the video and make your choice now.

You should have picked B and D.

It does show the likelihood of the different outcomes.

So for example, it's showing that there are four ones on the spinner and only two zeros.

And it does show the two most common outcomes that are equally likely.

And that's getting a one on the spinner and a head on the coin or one on the spinner and tail on the coin.

Now, Jun plays a game.

He will spin both spinners once each and multiply the outcomes of each spinner.

If the outcome is a prime number, Jun wins a prize.

If the outcome isn't a prime number, he wins nothing.

Oh, that's a shame.

So Jun wants to construct a table to see whether it's possible for him to even win the game.

And that's what he's done here.

Now, which of these tables is the most appropriate for the purpose of Jun's trial? Is it table one or table two? And remember, you need to select the correct justification for your choice.

So either A, this table shows how likely Jun is to win, or B, this table efficiently shows all unique outcomes.

So Jun, remember, is constructing a table to see whether it's possible for him to win.

So which table is most appropriate for this? Pause and make your choice now.

Welcome back.

Which table did you pick? Well, you should have picked table one because this table efficiently shows all unique outcomes.

Remember, Jun wanted to see whether it was possible to win.

He didn't care how likely it was that he won.

And table one shows all unique outcomes.

None of them are prime, so he can win.

So really, if he'd done table two, he'd have found out the same result, but it would've just taken him longer.

It's now time for our first task.

For question one, Alex will spin this spinner twice.

Complete this outcome table to list all unique outcomes.

And then part B, construct your own outcome table, which shows the most likely outcome.

Then question two, Aisha will spin spinner two and spinner three once each and find the product of the numbers the spinners land on.

Part A, complete this outcome table to list all unique outcomes.

And then in part B, construct a second outcome table.

Starting with the least likely outcome, rank the outcomes in order of likelihood.

So once you've constructed the second outcome table, look at all the outcomes you get and then rank them from least likely to most likely.

Pause and do this now.

Welcome back.

Question three, Sam and Lucas play a game.

Sam will spin spinner four twice and find the sum of the numbers the spinner lands on.

If the score is a double digit number, Sam wins.

If the score is either eight or nine, it's a draw.

And if the score is seven or less, then Lucas wins.

For part A, Construct the most efficient outcome table to show it is impossible for Sam and Lucas to draw.

Then in part B, construct the most efficient outcome table to find where the Sam or Lucas is more likely to win.

And you need to justify your answer.

So you need to explain how you know.

Pause and do this now.

Welcome back.

Let's go through our answers.

Alex will spin this spinner twice.

I asked you to complete this outcome table to list all unique outcomes.

So you should have got AA, BA if I read the first row, and the second row, AB, BB.

Now, it's okay if you are AB and your BA are switched around because you wrote them in different order.

It's still the same outcome table.

And then part B, construct your own outcome table, which shows the most likely outcome.

So you needed to have four rows and four columns, and you needed to see A on three of the rows and three of the columns, and B on the fourth row and fourth column.

Now, when we look at our completed outcome table, we can see that the outcome AA is the most likely, appearing nine times.

Now, even if we said order didn't matter, AB only appears six times in total.

So the AA is still the most likely outcome from spinning twice.

For question two, Aisha will spin spinner two and spinner three twice and find the product of the numbers the spinners land on.

I asked you to first complete this outcome table to list all unique outcomes.

So you needed to fill that in.

So you should have 2, 8, 3, and 12 appearing in your outcome table.

For part B, I asked you to construct a second outcome table.

And then starting with the least likely outcome, rank the outcomes in order of likelihood.

So we needed to do one of the larger tables here.

And across the rows, you either needed to have two, two, three, or you needed to have one, four, four, four.

And then for the columns, you need the other spinner.

So it's okay if your table is transposed, i.

e.

where I have columns, you've got rows and vice versa.

So in other words, you might have one, four, four, four as your row headers as opposed to the column headers like ID.

It doesn't matter.

You'll still get the same outcomes generated in the table.

So we can see here that three only appears once.

So three is the least likely outcome.

After that, it's two, because two appears twice.

The next most likely outcome is 12 because that appears three times.

And then the most likely outcome by quite a way is eight because that appears six times.

Question three, Sam and Lucas play a game and Sam had to spin spinner four twice and find the sum of the numbers the spinner landed on.

Remember, the first part was asking you to construct the most efficient outcome table to show its impossible for Sam and Lucas to draw.

So in other words, I wanted a table showing the unique outcomes.

So if you've been very efficient, you only needed to have this very tiny table.

Just two rows and two columns.

And when we fill it in, remember, by summing the numbers, we know that our table's going to contain the outcomes 4, 7, and 10.

Well, hang on a second.

It's a draw if the score was eight or nine, and there are no outcomes that are either eight or nine.

So it must be impossible for Sam and Lucas to draw.

Part B asked you construct the most efficient outcome table to find whether Sam or Lucas are more likely to win.

And you had to explain how you knew.

Well, in order to do this, we need to consider every single value that the spinner could land on.

So this means it now needs to be a four by four set of outcomes.

In other words, four rows, four columns, and you needed to have two, five, five, and five as your headers.

Now, when we fill this table in, it's really clear to see what's more likely.

Four and seven are coming up a total of seven times, but 10 comes up nine times.

So Sam is more likely to win as there are nine double digit outcomes, but only seven outcomes that are seven or less.

So Sam's only a little bit more likely to win.

It's now time for the second part of our lesson, and that's on outcomes and unequal likelihood.

Sofia will spin spinners one and two once each.

She points out, "If I can use an outcome table to see which outcomes are more likely, then here's my outcome table." The outcome BC is obviously the most likely one to happen.

By looking at spinner one and spinner two, why do you think Sofia's statement is incorrect? Did you spot that our spinners are not quite the same as the spinners we were looking at in the first part of our lesson? Jun says, "The outcomes on these spinners are not all equally likely because some sectors are different in size." Well spotted Jun.

Ah, look what he's done now, he's broken the spinners up so that the outcomes are now equally likely.

In other words, picking one of the A's has the same likelihood as picking one of the B's.

Now, this means each outcome of an outcome table we make is equally likely to happen.

So now our outcome table is going to accurately reflect how likely each outcome is.

My goodness me, this outcome table is large, isn't it? But it had to reflect all the different possible outcomes fairly so that we could see which outcome was the most likely.

So AD is the most likely outcome, not the BC that Sofia suggested because now A is getting the correct weighting as is D.

Let's check our understanding of this.

Laura will spin this spinner twice and find the sum of the numbers the spinner lands on.

Laura says it's pretty obvious that eight is the most likely outcome as there are more fours than any other number on the spinner.

Now, which of these statements explains why Laura is incorrect? A, the outcomes of the spinner aren't all equally likely.

B, there are more fours than negative fours.

C, the negative four sector has the same area as the total area of the four sectors.

And then D, the spinner has too many sectors.

So pause the video and select which of the statements explain why Laura is incorrect.

Welcome back.

You should have chosen statements A and C.

Now, the outcomes of the spinner aren't all equally likely because the sector for negative four is much larger than one of the individual sectors for four.

And then C, the negative four sector has the same area as the total area of the three, four sectors that you can see.

So now we've broken up so that every outcome is equally likely.

Which of these statements is now correct? A, there are four equally likely outcomes.

B, there are four unique outcomes.

C, the outcomes of eight and negative eight are equally likely.

And D, zero is the most likely outcome.

Pause the video and make your selection now.

Welcome back.

Did you pick C and D? Well done if you did.

The outcomes of eight and negative eight are equally likely because we can see eight nine times in the table and we can see negative eight nine times in the table.

So they're equally likely.

And zero is by far the most likely outcome because that appears a total of 18 times in the table.

Sofia points out, "It is important to make sure the area of each sector is equal in size before we can consider likelihood.

But how do you make sure the spinner is equally likely to land on each sector?" For instance, I think you can use a protractor to measure the angle of the smallest sector.

So let's do that.

Okay, I measure that at 72 degrees.

And then split up any larger sector into multiple smaller sectors.

Ah, like this.

Now, I broke up A, but I made sure that each sector of A had an angular 72 degrees just like beaded.

Now, I know all my sectors are the same size.

And Jun says, "I wonder if it's possible to draw an outcome tree to show which outcome is most likely." Hmm.

So Jun's gonna spin spinner one twice.

And Sofia says, "It's definitely possible, but won't it have a lot of branches because there are so many sectors? I think outcome trees are a lot better for showing unique outcomes only that's a lot fewer branches." And she's got a point.

Look at that.

This is our outcome tree for showing each unique outcome.

And it's tiny.

It's only got four listed there.

That looks quite easy to do.

Oh my, that's the outcome tree showing the likelihood of each outcome.

It's massive.

And look how small I've had to write just to get it to fit.

Oh, I absolutely do not wanna be doing that.

That looks like a lot of work when I know my outcome table can show this so much more efficiently.

Outcome trees are useful for listing unique outcomes.

Outcome tables show the likelihood in a far more efficient manner.

Remember, outcome trees, they get super messy when they are looking at likelihood because there are so many branches.

I'm with you.

I'm gonna stick with outcome tables two if I want to show likelihood.

Let's do a quick check now.

The three sectors whose outcomes are B and C are all equally likely.

How many smaller sectors should the sector with outcome A be split up into so that all outcomes on the spinner are equally likely? Pause while you work this out.

Welcome back.

You should have done 225 divided by 45, which is five.

So the sector with outcome A should be split into smaller sectors so that every sector you see has the same size.

It's time now for our final task.

Jacob will spin spinners A and B once each with the number each spinner lands on multiplied together.

Construct an outcome table which lists each unique outcome to this trial.

Then for question two, split sectors on spinners A, B, and C so that all outcomes on the spinner are equally likely.

Pause and do this now.

Welcome back.

Question three, Jacob will spin spinners B and C once each with the number each spinner lands on multiplied together.

Construct an outcome table, which shows how likely each outcome is to occur.

And then question four, Jacob will spin spinners A and B once each, and then he'll spin spinners B and C once each and then spinners A and C once each.

Now these are three separate experiments.

So we're going to spin spinners A and B, work out our outcomes there.

And that is one experiment.

Then for the second experiment, we're going to spin spinners B and C.

So these are not reliant on the outcomes that came from A and B.

This is separate.

So spin spinners B and C ones each and record what our outcomes are.

And then we'll spin spinners A and C once each and consider those outcomes.

So after each trial, so in other words, after spinning A and B, the number each spinner lands on are multiplied together.

And then in our next experiment where we spin B and C, we multiply each number together and then we record those and they're our outcomes, and then for the third experiment, we spin A and C and we take the two values it's landed on and multiply them together.

That's our outcomes for that experiment.

So once we've done that, we're going to look at what happened.

So the outcomes when I span A and B together, we're gonna look at the outcomes from when I spun B and C, and we're gonna look at the outcomes from when I spun A and C.

When I look at the outcomes for each of the three experiments, I'm going to consider which outcome was the most likely for that experiment.

When I've got my three most likely outcomes, one from each experiment, I'm going to consider them and work out which one was the most likely overall.

So across the three experiments.

Pause while you work on this now.

Welcome back.

Let's go through some answers.

So question one, Jacob will spin spinners A, B once each with the number each spinner lands on multiplied and you needed to construct a unique outcome table.

And that's what we have here.

So you should have the values 2, 3, 6, 8, 12, 24 within your table.

And then part C, you had to split the sectors on A, B, and C so that all outcomes on the spinners are equally likely.

And you should see that now.

Spinner A has been broken into six sectors, Spinner B is broken into five sectors, and spinner C is broken into three sectors.

So Jacob is going to spin spinners B and C once each and then multiply the results.

And this produces our outcome.

Construct the outcome table, which shows how likely each outcome is to occur.

So wherever you put B, whether it's your columns or B was your rows, you needed five because you had two, three, six, six, and six.

And then spinner C had to be three rows or three columns, depending where you put it.

You could then generate your set of outcomes and consider which one was most likely.

Well, for this experiment, the outcome 30 appears six times out of 15 total outcomes, and it's the most likely therefore.

So from this experiment, 30 was the most likely outcome.

Question four, you had to spin spinners A and B once each.

Spinners B and C, we'd already done that.

Did you spot? It was in question three.

And then you had to spin A and C once.

So after each trial, remember that trial made up our experiment, the number each spinner lands on are multiplied together.

So I've generated my table of likely outcomes for AB and the table of likely outcomes for AC.

When I consider AB, I can see the outcome 24 appears 12 times out of the 30 possible outcomes.

So it's the most likely for that experiment.

And then for the table, considering A and C being spun together, the outcome 20 appears eight times out of 18 outcomes.

So it's the most likely for that experiment.

Now I've said, when I consider my three outcomes across my three experiments, 20, the result I got from the experiment where I spun A and C is the most likely outcome.

Well, how did I know that? Well, let's consider our likelihood scale.

When we did the experiment spinning A and B, the outcome 24 appeared 12 times out of 30.

Okay, well, if it had been 30 out of 30, that would've been certain.

If it had come up 15 times out of 30, that would've been an even chance.

So I know given it's come out 12 times out of 30, it's got to be unlikely, close to even chance, but not super close.

So I'm gonna put it there.

Remember, this is only a rough scale.

Now, let's look at my next experiment, and that's where I spun B and C.

Now, the outcome 30 appeared six out of 15 times.

Well, hang on a second.

If I imagine scaling that up, six out of 15 is proportionally the same as 12 out of 30.

And that's how often my outcome for AB was appearing.

So actually, it has the same likelihood.

So the outcome of 30, which was the most likely outcome from the experiment, BC, has the same likelihood as the most likely outcome that came from the experiment where I spun A and B.

So they're in the same position.

Hmm.

Let's consider the outcome of 20, which appeared eight times out of 18.

Well, the other ones, remember, were six out of 15.

Or proportionally, we could say that's 12 out of 30.

Now I'm gonna go with the 12 out of 30.

That's quite useful to use.

I know that halfway would've been 15 out of 30, and I have 12 out of 30.

So the difference between those two is only three out of 30, which is a 10th.

So I'm only a 10th away from being an even chance.

That would be my difference.

Here, I've got the outcome 20 appearing eight out of 18.

Well, eight out of 18 is very, very close to being that even chance, an even chance would've been nine out of 18, which means that this outcome of 20 is only, I think about it like a scale, 1/18 away from being an even chance.

Well, 1/18 is a much smaller distance than 1/10.

So I know that this outcome has to be closer to an even chance than the previous ones were.

So the outcome of 20 from the experiment AC is more likely than either of the two outcomes that came from BC and AB.

And this is why 20 is the most likely outcome to occur.

Let's sum up what we've done today.

An outcome table can show the number of unique outcomes of a two-stage trial.

An outcome table can also show the number of individual outcomes of a two-stage trial where some outcomes of at least one stage are repeated.

Well done today.

There were some really tricky problems there, and you did a great job.

I look forward to seeing you for some of our future lessons.