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Hi, I'm Mrs. Dennett.

And in this lesson, we're going to be finding the probability of an event not happening.

To begin this lesson, we are going to look at equally likely events.

Read Teddy's statement.

Do you think Teddy is correct? There are two blue sections on the spinner and two yellow sections.

Both colours have a probability of two-tenths.

So yes, Teddy is right.

It is equally likely that the spinner will land on blue or yellow.

Let's recap how to work out probabilities for these events.

Firstly the probability of a spinner lands on red.

Five out of 10 sections are red.

So this is five-tenths or a half.

Now the probability that it will land on blue.

This is two-tenths or one-fifth.

Only one section is green.

So the probability of landing on green is one-tenth.

And finally, the probability of getting yellow.

There are two yellow sections.

So this is two-tenths or one-fifth when simplified.

What do you notice about these two probabilities? That's right.

They are the same.

So they are equally likely.

Do you notice anything about these probabilities? All of these probabilities add up to 10-tenths or one.

They represent all of the possible outcomes of spinning a colour on the spinner.

We say that these events are exhaustive, as they include all the possible outcomes of getting a colour: red, blue, green, or yellow.

This also shows us a very important fact.

The sum of probabilities is equal to one.

Let's look at finding the probability of an event not happening.

We want to find the probability that we do not land on red when we spin the spinner.

We could look at all of the other colours: blue, green, and yellow.

and count colour sections; two blue, one green, and two yellow.

This gives us five-tenths or a half.

But remember we now know that the sum of probabilities is equal to one.

So we could do this calculation instead.

One takeaway the probability of getting a red, which is five-tenths equals five-tenths or a half.

This is much faster if we already know the probability of landing on a red.

So the probability of not landing on a red is a half.

Now let's find the probability that we don't land on a green.

The probability of landing on green is one-tenth.

So we can calculate one takeaway a tenth to get nine-tenths.

Remember that we can also write probabilities as decimals, like this.

Let's use these decimals to find the probability of not landing on blue.

The probability that we get blue is 0.

So the probability that we don't land on blue is one take away 0.

2, which is 0.

We can do the same for the probability that we don't land on green.

We do one take away the probability that we land on green, which is 0.

1 and we get 0.

Notice how the probability that we get green and the probability that we don't get green is equal to one.

Here is a question for you to try.

Pause the video to complete these tasks and restart when you're finished.

Here are the answers.

Remember that the sum of all probabilities is equal to one.

So if we know the probability of Luke winning a race is 0.

6, we can calculate the probability of him not winning as one takeaway 0.

6 which is equal to 0.

Similarly for question two, the sum of probabilities for choosing mint, caramel, or hazelnut is one.

So we calculate one takeaway 0.

3 takeaway 0.

6, or one takeaway 0.

9, leaving us with 0.

Here are two more questions for you to try.

Pause the video to complete these tasks, and restart when you are finished.

Here are the answers.

For question three, we just work out one takeaway two-thirds, which is a third.

For question four, we know that all the probabilities add up to one for each sport.

So adding 0.

15 and 0.

45 gives us 0.

We subtract this from one, leaving us with 0.

4 to share equally between rugby and boxing.

As we are told the probabilities, these two sports are the same.

So the probabilities for rugby and boxing are 0.

Let's look at our numbered spinner again.

Each of these events, such as landing on red, cannot happen at the same time.

We couldn't land on a blue and a red.

We say that these events are mutually exclusive.

Let's consider this question.

Teddy and Malia are playing a game with the spinner.

Teddy wins if it lands on red.

Malia wins if it lands on blue, green, or yellow.

Is this game fair? Explain your answer.

The probability of landing on red is five over 10.

Half of the time, Teddy can win.

The probability of not landing on red, so landing on blue, green, or yellow is also a five-tenths.

So Malia can win half of the time.

This game is fair because the probabilities are the same for each person.

Let's think about Malia's statement.

Malia says "I cannot get an odd number and win." Is she correct? Can she get an odd number with blue, green, or yellow? Four, six, two, two, and eight are all even, so Malia is correct.

We say that these two events are mutually exclusive.

Malia cannot win and get an odd number.

Now it's your turn to have a think about some events.

Which of these events are not mutually exclusive? You can pause the video at this point if you wish.

Remember to restart the video when you've finished.

Let's have a look at the first event.

Pick a card and get a red and a club.

In a pack of cards, the clubs are black.

So this event is mutually exclusive.

The second one.

roll a die and gets an odd number and a four.

Four is even.

So again, these events are mutually exclusive.

The third statement.

There are two red tens in a pack of cards.

So this could definitely happen.

Therefore, it's not mutually exclusive.

What about flipping a coin and getting heads and tails.

You can't flip a coin and get both.

So this is mutually exclusive.

What about winning a football match and scoring three goals.

This is definitely possible.

So not mutually exclusive.

Finally, a student studying maths and English.

It is possible for student to do both of these things.

So these events are not mutually exclusive.

When events are mutually exclusive, like these ones here, it is possible to add the probabilities.

Here we want to work out the probability of getting blue or yellow.

Using decimals, the probability of getting blue is 0.

And the probability of getting yellow is also 0.

So the probability that we get blue or yellow is 0.

2 add 0.

2, which is equal to 0.

Sometimes probabilities can be presented in tables like this.

Let's use the probabilities in this table to work out the probability of getting red or green.

The probability of getting red is five-tenths.

And the probability of getting green is one-tenth.

Add these together to get the probability of getting red or green.

We get six-tenths, which can be simplified to three-fifths if you wish.

Here's a question for you to try.

Pause the video to complete the task, and restart when you've finished.

Here are the answers.

The probability for picking yellow is given in the table; three-twelfths or a quarter.

For green or purple.

We have to calculate five-twelfths add four-twelfths, which is nine-twelfths.

For the probability of not getting a purple crayon, you have to work out one takeaway four-twelfths, which leaves us with eight-twelfths or two-thirds if you simplified.

And finally, there are no blue crayons in the box.

So the probability of picking a blue crayon is zero.

We're now going to have a quick look at a problem involving probability.

As there is quiet a lot of information here, I'm going to give you the opportunity to pause the video now if you wish.

Remember to restart when you've finish reading.

Let's look at part a.

How many counters could be in the bag? It's useful to draw a diagram for this.

I used a bar model divided into five equal parts, as the probability of selecting a red counter is three out of five.

And I've coloured in three out of five equal parts in red.

The probability of selecting a green counter is equally likely as selecting a blue counter.

As there are two-fifth left, the probability of getting green must be one-fifth, and the probability of getting blue must also be a fifth.

I split the two-fifths I had left equally.

So how many counters could be in the bag? Could we have five counters? Well yes; three red, one green, and one blue.

This would give us the right probabilities.

But this isn't the only answer.

There could be 10 counters in the bag: Six red, the six-tenths is equivalent to three-fifths, two green, and two blue as two-tenths is the same as one-fifth.

In fact, we could have 15, 20, or 25 counters in the bag.

Any multiple of five would give us the correct probabilities.

Now let's do part b.

The probability of selecting a blue counter is one-fifth.

But if there are six blue counters in the bag, we need a fraction equivalent to one-fifth with a numerator of six.

I've multiplied the numerator and denominator by six.

So there must be 30 counters in the bag.

I know that the probability of getting a red counter is three-fifths.

I need a fraction equivalent to three-fifths with a denominator of 30.

I multiply the denominator by six, so I do the same to the numerator.

Three times six is 18.

So there are 18 red counters in the bag.

Finally, we have part c.

If there are 15 counters in the bag altogether, how many of each colour are there? Take each probability in turn and divide them with a denominator of 15.

So for red, the probability is three-fifths.

We multiply numerator and denominator by three to get nine-fifteenths.

For green, we have one-fifth.

This is equivalent to three-fifteenths.

And for blue, it is the same.

There are nine red, three green, and three blue counters in the bag.

Here's a question for you to try.

Read the question very carefully, and remember to draw a diagram to help you if you wish.

Pause the video now to complete the task.

And may start when you've finished.

Here are the answers.

For past a, we know that the probability of getting yellow ball is 0.

This is equivalent to one-fifth.

So there could be five, 10, 15 balls in the bag.

Any multiple of five, in fact.

Eight is not a multiple of five, so would not give the correct probabilities.

Therefore, there cannot be eight balls in the bag.

For part b, it is useful to draw a bar model.

We know the probability of picking a yellow ball is 0.

2 or one-fifth.

So my bar model has five equal parts.

One part is shaded yellow because I know the probability of getting a yellow is one-fifth.

It is equally likely that I select a pink or blue ball.

So I share the remaining parts between pink and blue.

There are four parts remaining.

So that is two parts for blue and two parts for pink.

We're told that there are 10 yellow balls.

We know that each part is equal.

So there are 10 balls in each part.

As there are two pink parts, we can work out that there are 20 pink balls.

That's all for this lesson.

Thank you for watching.

I've finished the video