# Video: AQA GCSE Mathematics Higher Tier Pack 1 • Paper 1 • Question 16

AQA GCSE Mathematics Higher Tier Pack 1 • Paper 1 • Question 16

04:56

### Video Transcript

A fair eight-sided dice has two green faces, two blue faces, and the rest of its faces are yellow. A fair spinner has three equal sections. Two sections are yellow and one is blue. The dice is rolled first and then the spinner is spun. Part a) Complete the tree diagram for the dice and the spinner.

A tree diagram is the way of representing all of the possible outcomes for two or more events. We can see that we have the possible outcomes for the dice first: green, blue, or yellow. But the possible outcomes for the spinner have not yet been completed.

In the question, we read that the sections on the spinner are colored yellow and blue. Two of them yellow and one is blue. So the possible outcomes for the color of the spinner are yellow and blue each time, regardless of the color that was shown on the dice.

We now need to complete the tree diagram by adding the relevant probabilities to each branch. We’re told that both the dice and the spinner are fair, which means that the dice has an equal probability of landing on each face and the spinner has an equal probability of landing on each section.

The dice has eight faces in total, two of which are green and two of which are blue, which means that it has four yellow faces. As there is equal chance of the dice landing on each face, the probability that it lands on green is two out of eight as there are two green faces out of a total of eight. And this probability can be simplified to one-quarter.

The same is true for the probability of the dice landing on blue. For yellow, it’s four out of eight, which can be simplified to one-half. Notice that these probabilities must and do sum to one, which we can see if we convert them to fractions with a common denominator of four. We get four over four which is equal to one.

For the spinner, there are two yellow sections out of three equal sections. So the probability that the spinner lands on yellow each time is two-thirds. For blue, it’s one-third as one section is blue out of the total of three.

Again, notice that the probabilities on each branch of the tree diagram sum to one. So we’ve answered part a of the question and completed the tree diagram.

Part b) says work out the probability of getting two colors that are the same.

To work this out, we can first list all of the possible outcomes for the colors of the dice and the spinner at the end of the branches on our tree diagram. So first, we have green on the dice and yellow on the spinner, GY, and then green on the dice and blue on the spinner, GB, and so on.

We want the probability of getting two colors that are the same. So that could be blue on the dice and blue on the spinner or yellow on the dice and yellow on the spinner. We need to work out the probabilities of these two branches.

To work these out, we can remember the rule that if two events A and B are independent — which means they don’t affect each other’s probabilities as the dice and spinner don’t — then to find the probability of both events happening, we multiply the individual probabilities together.

This means that we multiply the probabilities along the branches of the tree diagram. So to find the probability that both the dice and the spinner show blue, we multiply one-quarter by one-third. To multiply fractions, we multiply the numerators and multiply the denominators. So this is equal to one twelfth.

To find the probability that the dice and spinner both show yellow, we multiply the probabilities on these branches of the tree diagram. So we have a half multiplied by two-thirds. We can cross cancel a two in the numerator with a two in the denominator. And the two fractions multiply to one-third.

So we found the individual probabilities of either both dice and spinner being blue or both the dice and spinner being yellow. But how do we combine them? Well, if two events A and B are mutually exclusive, which means there’s no overlap between them, then to find the probability of one or the other happening, we can sum their individual probabilities together.

So the probability of getting either both blue or both yellow is equal to one twelfth plus one-third. We need a common denominator in order to add these two fractions. So we can convert one-third to a fraction with a denominator of 12 by multiplying both the numerator and denominator by four.

We now have one twelfth plus four twelfths which is equal to five twelfths. This fraction can’t be simplified any further as the numerator and denominator have no common factors other than one.

So we’ve answered part a by completing the tree diagram and answered part b by finding that the probability of getting two colors the same is five twelfths.