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Hi, my name is Mrs. Dennett.
And in today's lesson, we're going to be listing outcomes in a sample space diagram, or a two-way table and calculating probabilities.
sample space diagrams are extremely useful for listing outcomes, where more than one event is happening at the same time.
Here, a third coin is being flipped, and a spinner is being spun.
I could just list all of the outcomes in curly brackets like this, getting a one on the spinner and a head on the coin etcetera, which it will be quite a long list.
Sometimes it is easier and clearer to write the information in a table.
So I could write the outcomes of the coin down the side and the outcomes from the spinner along the top.
We can fill in the table with all of the possibilities.
So a head and one ahead and a two and so on.
And I can do the same for tails.
By writing the sample space as a table is much easier and clearer to see all the possible outcomes.
We can then use this table to work out probabilities.
Here, we're going to work out the probability of getting a prime number and a tail.
I can disregard the top row as this is for heads and just focus on the bottom row.
One is not a prime number, but two is.
So I need two and a tail.
Three is also a prime number.
So I include this one.
Four isn't prime, so I don't need this.
I can see I have two outcomes out of eight in total.
So I write the probability of getting a prime number and a tail is two eighths or one quarter.
I could also write this as not point two, five, or 25%.
Here's a question for you to try.
Pause the video to complete the task and restart when you're finished.
Here are the answers.
Once you have filled in the sample space, we need to look for heads on the top row.
We need to also look for heads with an even number.
There are three of these two H, four H, and six H.
So the probability is three twelves or quator.
If you simplify this fraction.
We're now going to look at finding probabilities from a two-way table.
This table shows some information about some year nine and year 10 students who have entered a prize draw.
We need to find the following probabilities.
Let's begin with finding the probability that the student selected is a girl.
We need to know how many girls there are in total.
There are 238 girls out of a total number of 460 students.
So the probability is 238 out of 460, which we write as a fraction, and we can simplify this.
What's the probability that a girl from year nine is picked? We need to find the number of girls in year nine in the table.
There are 132 girls in year nine.
So this is 132 out of 460.
And again, we can simplify this.
In part c, we're given some information first.
we're told the boy selected so we can ignore the girls now.
What's the probability he's in year 10.
There are 115 boys in year 10 out of 222 boys in total.
So the probability is 115 out of 222.
Be really careful here, it's a common mistake to write the denominators as 460.
But because we're told that the boy is randomly selected, we can ignore the other information and just focus on the boys.
Similarly, in Part d, we're told the year nine was randomly selected, so we can ignore the year 10s.
We want to know the probability that the year nine student is a girl.
There are 132 year nine girls out of 239 students in year nine.
So the probability is 132 out of 239.
Here's a question for you to try.
Pause the video to complete the task and restart when you are finished.
Here are the answers.
For part a, we can see that there are 10 girls who walk to school.
There are a total of 135 students.
So the probability is 10 out of 135.
for part b, we look for buyers who don't get the bus 13 boys walk, and 30 boys the car.
So altogether 13 plus 30 gives us 43 boys who don't take the bus.
So the probability is 43 out of 135.
Finally, we need to look the total number of students who travel by car.
This is 65.
So the probability that a student travels by car is 65 out of 135.
Here is a question for you to try.
Pause the videos complete the task and restart when you are finished.
Here are the answers.
In this question, we've got two dice being rolled, the numbers are multiplied together to get a score.
So when we fill in the table, we need to multiply the numbers together.
So for the first row in the first column, we do one times one.
moving across horizontally, we do two times one, three times one, four times one, etcetra.
Then on the second row, we do one times two, two times two, which is four that was already filled in, and three times two is six.
And you continue in this fushion until you have completed the table.
Then we need to find the probability of getting an even score.
Look at the numbers in red.
There are 27 out of a possible 36 numbers that are even, don't forget to include the four and 24 that were already given in the table.
Then we have to find the probability of getting a square number score.
We look for square numbers.
These are one, four, nine, 16, 25 and 36.
And they're going diagonally from left to right in the table.
But don't forget that we've got one times four, which is four and another four on the other side, too.
So altogether, there are eight square numbers.
So the probability is eight out of 36.
Here's a question for you to try.
Pause the video to complete the task and restart when you are finished.
Here is the answer.
In this question, it is useful to draw a sample space diagram but the score are added together.
There are 36 outcomes, of which the majority are seven.
So seven is the most likely score.
Here's a question for you to try.
Pause the video to complete the task and restart when you're finished.
Here are the answers.
Before I begin this question, it is useful to draw a sample space like this one on the right-hand side.
For part a, you can see that there are four opportunities to get the same letter out of a total of 24 outcomes.
for part b, you needed to find any outcomes that contain a letter B.
There are nine of these.
So the probability is nine out of 24.
That's all for this lesson.
Remember to take the Exit quiz.
Thank you for watching.