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Hello, I'm Mr. Langton.
And today we're going to look at calculating the probabilities of events.
All you're going to need is something to write with and something to write on.
Try and make sure you're in a quiet space with no distractions.
And when you're ready, we'll begin.
We'll start with the try this activity.
The games below are played by rolling a six-sided die, put the games in order of how likely you are to win them.
Pause the game and have a go.
When you're ready unpause it and we can go through it together.
You pause in three, two, one.
How did you get on? Here's how I'd grouped my answers.
I said that game D was most likely to be won.
And the reason for that is that there are lots of factors of 24.
If I roll the dice, I can score a one, a two, a three, a four, a five or a six.
Now factor the 24, we've got one, two, three, four, and six.
There are five ways that it can happen, which makes this one that most likely won't happen.
Let's compare it with the others.
An even number you could have a two, four or six.
There are three ways in getting an even number.
And a prime number, you could have a two, three, or five.
So in both cases, there are three ways that you can do it.
That makes them both equally likely I'll star them both in green.
And I can see that for game C and game E, I'll put those two together as well.
So game C is a square number, and the square numbers are one and four.
So there's two ways you can do that.
And a number that's in the sequence of 3n - 2.
To the first term in that sequence, 3 X 1 is 3 - 2 3 - 2 is 1.
3 X 2 is 6 - 2 is 4, 3 X 3 is 9 - 2 is 7.
So there are only two numbers in the sequence and it happens that are the two square numbers are in the sequence as well.
So both of these options, you're equally likely to win, but they're the least likely ones out of the five options you were given.
It was easy to compare the probabilities on the previous slide, because we were using the same dice for each game.
So find the most likely winner we just had to see which one had the most chances to win.
This isn't always the case.
When calculating the probability, we need to consider all the possible outcomes and how many ways the outcome that we're investigating can happen.
So here we have two spinners.
I'm going to do an example.
What is the probability of spinning a factor of 12 on each spinner? So before I start, let's just write out the factors of 12.
I can have 1 and 12, I can have two and six, I can have three and four.
So on the first spinner, the pentagon, the factors of 12 that could come up are one, two, three, and four.
So that is four ways we can get a factor of 12.
And there are five outcomes altogether.
Now 4/5 could also be written as 0.
8 or as 80%.
Now over on the other spinner, we could get a one, a two, a three, a four, or a six.
So there are more ways that we can get a factor of 12 there.
There are five ways altogether, that's out of eight.
That's only just a little bit bigger than 1/2, and actually as a decimal that'll be 0.
625 or 62.
5%.
So although there are more ways that we can get a factor of 12 on this spinner, It's more likely that we'll get one on the pentagonal one.
Now which spinner is most likely to land on an odd number? And let's have a close look.
There are three odd numbers on that pentagonal spinner.
With a different colour.
So three of the five possible options.
And on the octagon there are four possible ways out of eight.
Now 4/8 is exactly 1/2, or 0.
5.
3/5 is a little bit bigger than a 1/2.
It's actually 0.
6.
So once again, we're more likely to get an odd number on the pentagonal one.
If we're looking at an even number instead, they'll still be a 50:50 chance on the octagon, but only 2/5 on the pentagon.
So it just depends on what we're looking for each time.
Now it's your turn.
See if you can calculate the probabilities for each option, pause the video and have a go.
Work through the worksheet and when you're ready unpause it and we can go through it together.
Good luck.
Okay, let's go through the answers.
Starting off with question one.
The probability of getting a four on spinner C.
Well there's only one way that we can get a four in the eight options.
The probability of spinning a six on each one.
Well on spinner A, that probability will be zero because it can't happen.
On spinner B, that's 1/6, and on spinner C, it will be 1/8.
Which one is most likely to land on an odd number? On spinner A, there were three ways out of five, on spinner B, there are three out of six, it tends to be less likely, and C is going to be four out of eight, that's also less likely.
So the answer and make sure you answer the question, which one is most likely to land on an odd number? A is most likely to land on an odd number.
And I would always recommend that you make it really clear by writing a statement and not to put a circle around the letter, just make it really clear as well.
Question four, which is most likely to land on a prime number? So I wrapped up my prime numbers, 2, 3, 5, 7, 11, and so on.
That's that's more than I'm going to need it.
So the first one could land on two, three or five.
So A it's once again, 3/5, for B it can land on two, three, or five again, that's 3/6 or 1/2.
And for C it can also land on a seven, so that's 4/8.
So B and C are both 1/2 against, so once again A is most likely.
Question five, which one is most likely to land on a number less than four? So for the first one, it will be 3/8.
So I'm just going to change the colour 'cause I'm running a bit out of space thank you but, sorry 3/5, 3 out of 5.
For B, it will be, number less than four, one, two or three.
So again, 3/6.
And for C, there are 3/8.
So most likely to land on a number less than four, once again, A is most likely.
And for question six, most likely to land on a square number.
Well my square numbers are one, four, nine, and so on.
So are only two options for each one aren't they? So we 2/5 for A, it will be 2/6 for B and for C it's going to be 2/8.
So once again, A is most likely.
For the second part, I'm going to calculate the probabilities for each thing, selecting a prime number.
This time we're going up to 20.
My prime numbers are, 2, 3, 5, 7, 11, 13, 17, and 19.
So that's one, two, three, four, five, six, seven, eight prime numbers that are on that list up to 20.
Which cancels down to 2/5, and you may have written 40% or 0.
4, any of those answers is fine.
Two, selecting a number that is not square.
So our square numbers are 1, 4, 9, 16, 25 is too big.
There are four square numbers.
That means that there are 16 numbers that are not square out of 20, that's 4/5, they're not square numbers or 0.
8 or 80%.
Selecting a factor of 12.
Factors of 12 are 1 and 12, two and six, three and four.
So there are six of them after 20, which is 3/10, which is 0.
3 or 30%.
Selecting a number less than 12, well 11 of the numbers are less than 12.
So it's 11/20, which is 55% or 0.
55.
Selecting a number greater than 20.
Well, that can't happen, that's impossible.
So that is 0, or rather 0, 0.
0, 0% let us stick with 0.
We'll finish with the explorer activity.
The probability scale shows the outcomes of rolling a 12-sided die.
What other events could be placed on the scale in the gaps marked? Pause the video and have a go.
There are lots of possible options here.
So this is really only limited by your imagination.
If you'd like to share your answers to that last problem or anything else that you've come up with, please ask your parent or carer to share your work on Twitter tagging @OakNational and #LearnwithOak.
Thank you.
Goodbye.