# Video: Pack 5 • Paper 1 • Question 15

Pack 5 • Paper 1 • Question 15

03:38

### Video Transcript

The probability of event 𝐴 is one-sixth. The probability of event 𝐵 is two-fifths. When can we say that the probability of event 𝐴 and event 𝐵 both occurring is one-sixth multiplied by two-fifths?

This question is asking us under what conditions we can find the probability of two events both occurring by multiplying their individual probabilities together. The answer to this is when events 𝐴 and 𝐵 are independent of each other. By this we mean that the outcome of each event doesn’t affect the outcome of the other event. You only need to give a reference to independence to get this mark, but let’s look at why this is the case.

Suppose we have three sweets in a bag: one red, one blue, and one green. If we select a sweet at random from the bag, the probability that it’s a red sweet is one-third. The probabilities are the same for a green sweet and for a blue sweet.

Let’s think about the scenario when we select one sweet, eat it, and then select another. If we wanted to find the probability of selecting a red sweet and then a blue sweet, do we multiply the two probabilities together? If we did, we’d have one-third multiplied by one-third. And remembering the rules for multiplying fractions; that is, we multiply the numerators together and we multiply the denominators, this would give one-ninth.

However, let’s look at this probability in other way by listing all the possible outcomes for the colours of the two sweets. This is the list of all possible outcomes. We could have a red and then a blue, a red and then a green, a blue and then a red, a blue and then a green, a green and then a red, or a green and then a blue. There’s six possible outcomes in total. And as we had the same number of each colour of sweets to begin with, all of these outcomes are equally likely.

As the sum of the probabilities for all outcomes must be equal to one, the probability of each outcome is one-sixth. However, this shows us that the probability of getting a red and then a blue sweet is one-sixth, not one-ninth.

So multiplying the two probabilities together has given us an incorrect answer. The reason for this is that the two events, which are the colours of the first and second sweet, are not independent. Because we’ve removed one sweet from the bag and not put it back, remember we ate it, then the probabilities for the second sweet are not the same as they were for the first sweet.

Suppose instead that we didn’t eat the sweet, but instead we put it back in the bag. We, therefore, have a different list of possible outcomes. We now have nine outcomes in total. And notice that we have three new possibilities in our list — the outcomes of the two sweets of the same colour: either both red, both blue, or both green. This is because the first sweet has been put back in the bag. So it’s still available to be chosen for the second sweet.

Again, all of these outcomes are equally likely. And this time they are nine of them. So the probability of each outcome is one-ninth. This is the same as the probability we found when we multiply the two individual probabilities together. And this is because this time the colours of the two sweets are independent.

By putting the first sweet back in the bag, after noting its colour, we haven’t changed the setup at all. And so the probabilities for the second sweet are the same as they were originally. This illustrates why two events must be independent in order to find the probability of both occurring by multiplying their individual probabilities together.