Lesson Video: Dependent and Independent Events | Nagwa Lesson Video: Dependent and Independent Events | Nagwa

# Lesson Video: Dependent and Independent Events Mathematics

In this video, we will learn how to calculate probabilities for dependent and independent events and how to check if two events are independent.

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### 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, we’ll 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. We’ll also introduce a different formula we can use to calculate the probability of two events occurring if they are dependent. Let’s now consider what it means for two events to be independent. Suppose we have two events 𝐴 and 𝐵. 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 doesn’t make the other event any more or any less likely to occur.

In the same way, if one event doesn’t happen, it doesn’t 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 doesn’t 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 𝐴 and 𝐵 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 𝐵 occurs given 𝐴 has occurred is equal to the probability of 𝐵. That is, the conditional probability of event 𝐵 given event 𝐴 is simply the same as the probability of event 𝐵.

Now, this must also be the same as the probability that event 𝐵 occurs given that event 𝐴 doesn’t occur, and the same must be true the other way around. The probability that event 𝐴 occurs given event 𝐵 has occurred must be the same as the probability of event 𝐴 occurring. If any of these are not true, then we say instead that the events 𝐴 and 𝐵 are dependent events. We’ll 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 𝐴 and 𝐵 independent events? (A) A student leaves their house on their way to school. Event 𝐴 is them arriving at the bus stop in time to catch the bus, and event 𝐵 is them getting to school on time. (B) A die is rolled. Event 𝐴 is rolling an even number, and event 𝐵 is rolling a prime number. (C) A die is rolled and a coin is flipped. Event 𝐴 is rolling a six on the die, and event 𝐵 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 𝐴 is them taking a chewy candy first, and event 𝐵 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 𝐴 is the teacher selecting a boy first, and event 𝐵 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 𝐴 affects the likelihood of event 𝐵 occurring. Let’s consider each scenario in turn. In the first scenario, event 𝐴 is the student arriving at the bus stop on time to catch their bus, and event 𝐵 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 𝐴 does affect the probability of the occurrence of event 𝐵, 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 𝐴 is rolling an even number, and event 𝐵 is rolling a prime number. Assuming this is a six-sided die, then event 𝐴 is obtaining any of the outcomes two, four, or six. Event 𝐵, 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 𝐵 is equal to one-half.

Now if instead we assume that event 𝐴 has already taken place, this means we already know that the number we have rolled is even. It’s either two, four, or six. What now is the conditional probability that the number is prime, the probability of (B) given event 𝐴 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 it’s equal to one-third. We found then that the probability of event 𝐵 is not the same as the probability of event 𝐵 given event 𝐴 has occurred. And so these two events are dependent.

Now let’s consider the third scenario. Here, event 𝐴 is rolling a six on the die, and event 𝐵 is the coin landing with its head side up. Let’s 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 𝐴 and 𝐵 are independent.

We need to check the two remaining scenarios though. In scenario (D) a child is taking candies out of a bag. Event 𝐴 is them taking a chewy candy first, and event 𝐵 is them taking a crunchy candy second. To determine whether these two events are independent, we need to know whether the probability that event 𝐵 occurs is different depending on whether or not event 𝐴 has already occurred. We don’t 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 𝐴 occurs, so the first candy the child takes is chewy, then there’s 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, there’s 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 𝐴 occurs affects the probability of event 𝐵 occurring, which means that these two events are dependent. The final scenario is similar to the previous one. Event 𝐴 is the teacher selecting a boy first, and event 𝐵 is the teacher selecting a girl second. We can use a similar argument as we did for the candies. If event 𝐴 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 𝐵 occurring given event 𝐴 has occurred is five-ninths.

If, on the other hand, event 𝐴 doesn’t 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 𝐵 given event 𝐴 has occurred is different from the probability of event 𝐵 given 𝐴 hasn’t occurred. This means that whether or not event 𝐴 occurs affects the probability of event 𝐵 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).

We’ve now looked at several scenarios to help us understand the difference between independent and dependent events. We’ll 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 𝐴 and 𝐵 on a Venn diagram, then the probability of 𝐴 and 𝐵 both occurring corresponds to the overlap or intersection between the two circles representing event 𝐴 and event 𝐵. We should recall the conditional probability formula, which tells us that the conditional probability of event 𝐵 occurring given event 𝐴 has occurred is equal to the probability of events 𝐴 and 𝐵 both occurring divided by the probability of event 𝐴. That’s the probability of 𝐴 intersect 𝐵 over the probability of 𝐴.

This is true for all events 𝐴 and 𝐵 such that the probability of 𝐴 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 𝐴 to give the probability of 𝐴 multiplied by the probability of 𝐵 given 𝐴 is equal to the probability of 𝐴 intersect 𝐵. And then we can just write the two sides the other way around. This is called the general multiplication rule. And it’s valid for any two events 𝐴 and 𝐵 whether they’re dependent or independent. However, we know that for independent events, whether or not event 𝐴 occurs has no impact on the probability of event 𝐵 occurring. So, the probability of 𝐵 given 𝐴 is equal to the probability of 𝐵.

This leads to a multiplication rules specifically for independent events. If two events 𝐴 and 𝐵 are independent, then the probability of 𝐴 intersection 𝐵 is equal to the probability of 𝐴 multiplied by the probability of 𝐵. 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, they’re dependent. Let’s 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 𝑆, the probabilities are shown for the combinations of events 𝐴 and 𝐵 occurring. Are 𝐴 and 𝐵 independent events?

We will begin by recalling the multiplication rule for independent events. If two events 𝐴 and 𝐵 are independent, then the probability of the intersection of 𝐴 and 𝐵 is equal to the probability of 𝐴 multiplied by the probability of 𝐵. 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 𝐴 and 𝐵, first of all, is the probability written in the intersection of the two circles. It’s five nineteenths. The probability of event 𝐴 occurring is the total of the probabilities in the circle that represents 𝐴. These are four nineteenths and five nineteenths, so the total probability of event 𝐴 occurring is nine nineteenths.

For event 𝐵, 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 𝐴 multiplied by the probability of 𝐵 is nine over 19 multiplied by 10 over 19. That’s 90 over 361 which does not simplify any further. Now, as this doesn’t simplify, it should be clear that it isn’t 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 𝐴 and 𝐵 is not equal to the probability of 𝐴 multiplied by the probability of 𝐵. So, by the converse of the multiplication rule for independent events, this means that 𝐴 and 𝐵 are not independent events. So, our answer to the question is no.

We’ll now consider one final example in which we calculate the probability of the intersection of two events. We’ll 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?

Let’s use 𝐴 to represent the event that the first ball is white and 𝐵 to represent the event that the second ball is black. We’re asked to determine the probability that both of these events occur. So, we’re looking for the probability of the intersection of events 𝐴 and 𝐵. 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 𝐴 and 𝐵 is equal to the probability of 𝐴 multiplied by the probability of 𝐵 given event 𝐴 has occurred. Let’s consider the probabilities we need. First, the probability of event 𝐴, that’s 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 we’re equally likely to choose any ball, the probability that the first ball we choose is white is 18 over 18 plus nine. That’s 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 let’s think about the probability for the second event. We’ve already established that these events are dependent, so we need the conditional probability of event 𝐵 given event 𝐴 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 it’s now 26. If event 𝐴 occurred, so the first ball we took was white, then the number of black balls in the bag hasn’t 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 𝐴 and 𝐵 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. We’re 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.

Let’s now summarize the key points from this video. Firstly, two events 𝐴 and 𝐵 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 𝐴 and 𝐵 are independent if the conditional probability of event 𝐵 given event 𝐴 has occurred is equal to the probability of 𝐵. And vice versa, the conditional probability of event 𝐴 given event 𝐵 has occurred must be equal to the probability of 𝐴.

The general multiplication rule tells us that for any two events 𝐴 and 𝐵, regardless of whether they’re dependent or independent, the probability of the intersection of events 𝐴 and 𝐵, so that’s the probability of 𝐴 and 𝐵 both occurring, is equal to the probability of 𝐴 multiplied by the conditional probability of 𝐵 given event 𝐴 has occurred. And finally, the multiplication rule for independent events. Events 𝐴 and 𝐵 are independent if and only if the probability of the intersection of 𝐴 and 𝐵 is equal to the probability of 𝐴 multiplied by the probability of 𝐵. 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.