### Video Transcript

In this video, we will learn how to calculate probabilities for dependent and independent events and how to check if two events are independent. Remember that in probability, an event is a set of outcomes from an experiment. For example, in the case of rolling a die, an example of an event is rolling an odd number. The outcomes that make up this event are the numbers one, three, and five. In the case of spinning the spinner, an example of an event is the spinner landing on the number one. These two events are an example of independent events.

In this video, weโll learn why that is the case and how we can calculate the probability of both of these events occurring using the formula shown on screen. Weโll also introduce a different formula we can use to calculate the probability of two events occurring if they are dependent. Letโs now consider what it means for two events to be independent. Suppose we have two events ๐ด and ๐ต. These events are independent if the occurrence of one does not affect the probability of the occurrence of the other. In other words, what this means is that if one event happens, it doesnโt make the other event any more or any less likely to occur.

In the same way, if one event doesnโt happen, it doesnโt impact the likelihood of the other. For example, suppose we are rolling a die twice. Successive rolls of the die are independent. The number we get on the first roll doesnโt affect the probabilities for the number we get the second time. So the events of, for example, the first number being odd and the second number being prime are independent of one another. Now we said that two events ๐ด and ๐ต are independent if the occurrence of one does not affect the probability of the occurrence of the other. We can state this more formally as the probability that ๐ต occurs given ๐ด has occurred is equal to the probability of ๐ต. That is, the conditional probability of event ๐ต given event ๐ด is simply the same as the probability of event ๐ต.

Now, this must also be the same as the probability that event ๐ต occurs given that event ๐ด doesnโt occur, and the same must be true the other way around. The probability that event ๐ด occurs given event ๐ต has occurred must be the same as the probability of event ๐ด occurring. If any of these are not true, then we say instead that the events ๐ด and ๐ต are dependent events. Weโll now consider five different scenarios and check whether the pair of events they describe are dependent or independent.

In which of the following scenarios are ๐ด and ๐ต independent events? (A) A student leaves their house on their way to school. Event ๐ด is them arriving at the bus stop in time to catch the bus, and event ๐ต is them getting to school on time. (B) A die is rolled. Event ๐ด is rolling an even number, and event ๐ต is rolling a prime number. (C) A die is rolled and a coin is flipped. Event ๐ด is rolling a six on the die, and event ๐ต is the coin landing with its head side up. (D) A child takes two candies at random from a bag which contains chewy candies and crunchy candies. Event ๐ด is them taking a chewy candy first, and event ๐ต is them taking a crunchy candy second. (E) A teacher select two students at random from a group containing five boys and five girls. Event ๐ด is the teacher selecting a boy first, and event ๐ต is the teacher selecting a girl second.

Remember that two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. So, to establish whether each of these pairs of events are independent, we should examine whether the occurrence of event ๐ด affects the likelihood of event ๐ต occurring. Letโs consider each scenario in turn. In the first scenario, event ๐ด is the student arriving at the bus stop on time to catch their bus, and event ๐ต is them arriving at school on time. It seems reasonable to assume that the probability of the student getting to school on time greatly increases if they catch their bus and, conversely, that the student is more likely to be late to school if they miss their bus. So the occurrence of event ๐ด does affect the probability of the occurrence of event ๐ต, meaning that these two events are dependent.

In scenario (B), this die is only rolled once, and both events referred to the outcome of that single roll. Event ๐ด is rolling an even number, and event ๐ต is rolling a prime number. Assuming this is a six-sided die, then event ๐ด is obtaining any of the outcomes two, four, or six. Event ๐ต, which is rolling a prime number, means obtaining any of the outcomes two, three, or five. Assuming the die is fair, and so all of these outcomes are equally likely, the probability of rolling a prime number is three out of six, which, of course, simplifies to one over two. So we can say that the probability of ๐ต is equal to one-half.

Now if instead we assume that event ๐ด has already taken place, this means we already know that the number we have rolled is even. Itโs either two, four, or six. What now is the conditional probability that the number is prime, the probability of (B) given event ๐ด has occurred? Well, if we know the number was even, so it was either two, four, or six, then in order for it to be prime, it would have to be the number two. In other words, the probability of getting a prime number given that we have an even number is the same as the probability of getting the number two from the set of numbers two, four, and six. This is one out of three equally likely outcomes. So itโs equal to one-third. We found then that the probability of event ๐ต is not the same as the probability of event ๐ต given event ๐ด has occurred. And so these two events are dependent.

Now letโs consider the third scenario. Here, event ๐ด is rolling a six on the die, and event ๐ต is the coin landing with its head side up. Letโs just think about this intuitively. Suppose we have already rolled a six on the die. Does the probability of the coin landing head side up change? No, of course not. The outcome of the die roll does not affect what will happen when we flip the coin. The probability of the coin landing head side up is still one-half, regardless of what happened when we rolled the die. This means that events ๐ด and ๐ต are independent.

We need to check the two remaining scenarios though. In scenario (D) a child is taking candies out of a bag. Event ๐ด is them taking a chewy candy first, and event ๐ต is them taking a crunchy candy second. To determine whether these two events are independent, we need to know whether the probability that event ๐ต occurs is different depending on whether or not event ๐ด has already occurred. We donโt know how many candies are in the bag, but we do know that by taking a candy out, the child changes the proportions of chewy and crunchy candies in the bag. If event ๐ด occurs, so the first candy the child takes is chewy, then thereโs now a greater proportion of crunchy candies in the bag than there was at the start.

On the other hand, if the first candy the child took was crunchy, thereโs now a smaller proportion of crunchy candies in the bag. The child is therefore more likely to take a crunchy candy second if the first candy was chewy than if the first candy was crunchy. So, whether or not event ๐ด occurs affects the probability of event ๐ต occurring, which means that these two events are dependent. The final scenario is similar to the previous one. Event ๐ด is the teacher selecting a boy first, and event ๐ต is the teacher selecting a girl second. We can use a similar argument as we did for the candies. If event ๐ด occurs so the teacher selects a boy first, then there are now nine students left, five of which are girls. This means that the probability of event ๐ต occurring given event ๐ด has occurred is five-ninths.

If, on the other hand, event ๐ด doesnโt happen, this means the teacher must have selected a girl first. So there are now nine students left, but only four of them are girls. The probability that the teacher selects a girl second is now four-ninths. So, the probability of event ๐ต given event ๐ด has occurred is different from the probability of event ๐ต given ๐ด hasnโt occurred. This means that whether or not event ๐ด occurs affects the probability of event ๐ต occurring. And so the events described in scenario (E) are also dependent events. So, only one of the five scenarios describes independent events, scenario (C).

Weโve now looked at several scenarios to help us understand the difference between independent and dependent events. Weโll now consider how to calculate the probability of two events both occurring, and we need to be careful to distinguish between whether the events are independent or dependent. If we illustrate the events ๐ด and ๐ต on a Venn diagram, then the probability of ๐ด and ๐ต both occurring corresponds to the overlap or intersection between the two circles representing event ๐ด and event ๐ต. We should recall the conditional probability formula, which tells us that the conditional probability of event ๐ต occurring given event ๐ด has occurred is equal to the probability of events ๐ด and ๐ต both occurring divided by the probability of event ๐ด. Thatโs the probability of ๐ด intersect ๐ต over the probability of ๐ด.

This is true for all events ๐ด and ๐ต such that the probability of ๐ด is not equal to zero, regardless of whether the events are dependent or independent. We can rearrange this formula by multiplying both sides of the equation by the probability of ๐ด to give the probability of ๐ด multiplied by the probability of ๐ต given ๐ด is equal to the probability of ๐ด intersect ๐ต. And then we can just write the two sides the other way around. This is called the general multiplication rule. And itโs valid for any two events ๐ด and ๐ต whether theyโre dependent or independent. However, we know that for independent events, whether or not event ๐ด occurs has no impact on the probability of event ๐ต occurring. So, the probability of ๐ต given ๐ด is equal to the probability of ๐ต.

This leads to a multiplication rules specifically for independent events. If two events ๐ด and ๐ต are independent, then the probability of ๐ด intersection ๐ต is equal to the probability of ๐ด multiplied by the probability of ๐ต. This rule works both ways. We can use it to calculate the probability of two independent events both occurring if we know their individual probabilities. Or if we know all three of these probabilities, we can test them in this formula to determine whether or not two events are independent. If the probabilities satisfy this rule, then the events are independent. But if not, theyโre dependent. Letโs now consider an example in which we use probabilities presented in a Venn diagram to determine whether two events are independent or dependent.

In a sample space ๐, the probabilities are shown for the combinations of events ๐ด and ๐ต occurring. Are ๐ด and ๐ต independent events?

We will begin by recalling the multiplication rule for independent events. If two events ๐ด and ๐ต are independent, then the probability of the intersection of ๐ด and ๐ต is equal to the probability of ๐ด multiplied by the probability of ๐ต. We can determine each of these probabilities from the Venn diagram and then test whether these probabilities satisfy the multiplication rule for independent events. The probability of the intersection of ๐ด and ๐ต, first of all, is the probability written in the intersection of the two circles. Itโs five nineteenths. The probability of event ๐ด occurring is the total of the probabilities in the circle that represents ๐ด. These are four nineteenths and five nineteenths, so the total probability of event ๐ด occurring is nine nineteenths.

For event ๐ต, the total probability is five nineteenths plus five nineteenths which is ten nineteenths. Now we find the product of the individual probabilities. The probability of ๐ด multiplied by the probability of ๐ต is nine over 19 multiplied by 10 over 19. Thatโs 90 over 361 which does not simplify any further. Now, as this doesnโt simplify, it should be clear that it isnโt equal to five over 19. But if we want to be really explicit about it, we can write the probability of five nineteenths as an equivalent fraction of 95 over 361. We found then that the probability of the intersection of ๐ด and ๐ต is not equal to the probability of ๐ด multiplied by the probability of ๐ต. So, by the converse of the multiplication rule for independent events, this means that ๐ด and ๐ต are not independent events. So, our answer to the question is no.

Weโll now consider one final example in which we calculate the probability of the intersection of two events. Weโll need to be careful to consider whether the two events are dependent or independent.

A bag contains 18 white balls and nine black balls. If two balls are drawn consecutively without replacement, what is the probability that the second ball is black and the first one is white?

Letโs use ๐ด to represent the event that the first ball is white and ๐ต to represent the event that the second ball is black. Weโre asked to determine the probability that both of these events occur. So, weโre looking for the probability of the intersection of events ๐ด and ๐ต. Now, reading the question carefully, we notice that the second ball is drawn without replacing the first ball. This means that the conditions under which we choose the second ball are different from the conditions under which we choose the first. And so the two events are dependent.

We recall the general multiplication rule. The probability of the intersection of events ๐ด and ๐ต is equal to the probability of ๐ด multiplied by the probability of ๐ต given event ๐ด has occurred. Letโs consider the probabilities we need. First, the probability of event ๐ด, thatโs the probability that the first ball is white. Well, at the start, there are 18 white balls in the bag and nine black balls. As weโre equally likely to choose any ball, the probability that the first ball we choose is white is 18 over 18 plus nine. Thatโs the number of white balls over the total number of balls in the bag. That gives 18 over 27 which simplifies to two-thirds.

Now letโs think about the probability for the second event. Weโve already established that these events are dependent, so we need the conditional probability of event ๐ต given event ๐ด has occurred, that is, the probability the second ball is black given that the first one is white. Well, regardless of the color of the first ball, one has been taken out of the bag. This means that the total number of balls left in the bag has reduced by one. So itโs now 26. If event ๐ด occurred, so the first ball we took was white, then the number of black balls in the bag hasnโt changed. So there is still nine black balls.

The probability that the second ball is black given the first was white is therefore nine out of 26. We can now substitute these two probabilities into the general multiplication rule, giving the probability if the intersection of ๐ด and ๐ต is equal to two-thirds multiplied by nine over 26. We can cross cancel a factor of three from the three and the nine and the factor of two from the two and the 26. Weโre left with one multiplied by three over one multiplied by 13, which is three over 13. Using the general multiplication rule, we found that the probability the second ball is black and the first ball is white is three over 13.

Letโs now summarize the key points from this video. Firstly, two events ๐ด and ๐ต are independent if the occurrence of one does not affect the probability of the occurrence of the other. Using probability notation, we can express this as events ๐ด and ๐ต are independent if the conditional probability of event ๐ต given event ๐ด has occurred is equal to the probability of ๐ต. And vice versa, the conditional probability of event ๐ด given event ๐ต has occurred must be equal to the probability of ๐ด.

The general multiplication rule tells us that for any two events ๐ด and ๐ต, regardless of whether theyโre dependent or independent, the probability of the intersection of events ๐ด and ๐ต, so thatโs the probability of ๐ด and ๐ต both occurring, is equal to the probability of ๐ด multiplied by the conditional probability of ๐ต given event ๐ด has occurred. And finally, the multiplication rule for independent events. Events ๐ด and ๐ต are independent if and only if the probability of the intersection of ๐ด and ๐ต is equal to the probability of ๐ด multiplied by the probability of ๐ต. And remember, this rule works both ways. We can use it to calculate the probability of the intersection of two independent events, or, if we know these three probabilities, we can use it to test whether or not two events are independent.