# Lesson Video: Conditional Probability Mathematics • 10th Grade

In this video, we will learn how to calculate conditional probability using formulas and Venn diagrams.

17:25

### Video Transcript

In this video, we’re going to learn about conditional probability. We’ll recap some basic probability rules, look at mutually exclusive or disjoint events, play with Venn diagrams, and learn how to work out whether two events are independent. Firstly, though, let’s recall some probability rules.

(1) We represent probabilities on the probability scale, with numbers from zero to one. (2) Something’s got to happen; if we sum all of the probabilities of all possible outcomes, they must come to one. Something either happens or it doesn’t happen. (3) We use the notation 𝐴 with a bar over it or 𝐴 prime to represent the complement of 𝐴 or the event that 𝐴 does not occur. If we know the probability that event 𝐴 occurs, if we subtract that from one, we know the probability that event 𝐴 doesn’t occur. We can represent this on a Venn diagram. If circle A represents occasions when event 𝐴 does occur, then the shaded area outside it is all the occasions when it doesn’t occur.

(4) The probability that event 𝐴 occurs or event 𝐵 occurs is known as the probability of 𝐴 union 𝐵. And it can be represented on a Venn diagram like this. (5) The probability that event 𝐴 occurs and event 𝐵 occurs is known as the probability of 𝐴 intersection 𝐵. And it can be represented on a Venn diagram like this. (6) The probability that event 𝐴 occurs but event 𝐵 doesn’t can be represented by 𝐴 intersection the complement of 𝐵. And it looks like this on a Venn diagram. And an alternative way of writing that is the probability of 𝐴 minus the probability of 𝐴 intersection 𝐵.

Next, let’s recap our knowledge of mutually exclusive or disjoint events. For example, an animal might be a cat or it might be a dog. But it can’t be both a cat and a dog. So being a cat and being a dog are mutually exclusive or disjoint events. And when this is the case, the probability of getting an intersection of those two events is zero. And because there’s no overlap, if we want to find the probability that an animal is a cat or a dog, we just simply need to add together the probabilities of it being a cat and the probability of it being a dog. Now, with nonmutually exclusive or nondisjointed events, two events can both occur together. For example, a person may like cats or they may like dogs. They may even like cats and dogs. So likes cats and likes dogs are not mutually exclusive or disjoint events.

So in this case, we need to be careful how we calculate the probability that a person likes cats or dogs, the union of liking cats and liking dogs. This pink-shaded region represents people who like cats, and this green-shaded region represents people who like dogs. So if we just added together the probability that someone likes cats and the probability that someone likes dogs, then we’d have double-counted this region in the middle. So we need to make an adjustment for that in our calculation. So our general formula is that the probability of 𝐴 union 𝐵 is equal to the probability of 𝐴 plus the probability of 𝐵 minus the probability of 𝐴 intersection 𝐵.

Right, now we’ve reviewed what we already know. Let’s talk about conditional probability. If the probability of an event 𝐵 is affected by the occurrence of an event 𝐴, then we say that the probability of 𝐵 is conditional on the occurrence of 𝐴. And we can write this in this way, a 𝐵 with a vertical line followed by 𝐴. And we say the probability of 𝐵 given 𝐴. Okay, let’s look at an example now where the events are not disjoint or mutually exclusive. This leads us to a situation where the probability of one event is conditional on the probability of the other.

On the street, 10 houses have a cat, C, eight houses have a dog, D, three houses have both, and seven houses have neither. Now this question comes in three parts. Let’s look at part one first. Find the total number of houses on the street. Hence, find the probability that a house chosen at random has both a cat and a dog. Give your answer to three decimal places.

Now, a great way to tackle this question is to draw a Venn diagram. In this case, the universal set for a Venn diagram is all of the houses on the street. The left-hand circle represents the houses that have a cat. The right-hand circle represents houses that have a dog. And the intersection of those two circles is the houses which have both. Anything which is outside the circles but inside the rectangle is a house that has neither a cat nor a dog. Now we’re told that 10 houses have a cat, but they’re gonna be distributed across houses which only have a cat and houses which have a cat and a dog. Likewise, the eight houses that have a dog are gonna be distributed between houses that only have a dog and those that have both a cat and a dog.

So it’s gonna be easiest for us to start off by looking at the houses that have both a cat and a dog, and there are three of those. And there are 10 houses that have a cat and three of those are houses which also have a dog. So that leaves 10 minus three, that’s seven houses, which only have a cat. And of the eight houses that have a dog, three of them also have a cat so that leaves eight minus three, that’s five, which only have a dog. And lastly, we’re also told that seven houses have neither a cat nor a dog. So that’s seven out here.

So the total number of houses on the street is made up of the seven houses who just have a cat, the five houses who just have a dog, the three houses who have both a cat and a dog, and the seven houses that neither have a cat nor have a dog. And when you sum those, you get 22. Now we’ve got to find the probability that a house chosen at random has both a cat and a dog. One way of thinking about this probability question is what proportion of houses on the street have both a cat and a dog. Well, we saw that three houses have both a cat and a dog, and there are 22 houses altogether. So the proportion of houses having both a cat and a dog is three over 22. And if you’re picking the houses at random, then the probability of picking a house with a cat and a dog is the same as that proportion. And to three decimal places, that’s 0.316.

Part two of the question asks, find the probability that a house on the street has either a cat or a dog or both. Give your answer to three decimal places.

Okay, so assuming the house is going to be chosen at random. This is just a matter of counting up the cases in which houses have a cat or a dog or both from our Venn diagram. And the probability we’re looking for is just the number of houses with a cat or dog or both as a proportion of the total number of houses in the street. So seven houses just have a cat, five houses just have a dog, and three houses have both. So that’s 15 houses. And we saw earlier that the total number of houses was 22. So the probability we’re looking for is 15 over 22. And correct to three decimal places, that’s 0.682.

Now part three is a conditional probability question. If a house on the street has a cat, find the probability that there is also a dog living there.

So we’ve been given the fact that the house has a cat living there. Given that fact, what’s the probability that there’s also a dog living there? Looking at our Venn diagram then, straightaway, we can disregard all the cases of houses which don’t have cats. So we can think of this question as, of the houses that have cats, what proportion also have dogs? We’re only looking at these seven houses and these three houses. That’s a total of 10 houses that have cats. And of those 10 houses, only these three also have a dog. Three out of the 10 houses that have cats also have a dog. So the probability that a house has got a dog, given that they’ve got a cat, is three-tenths. Now, we gave our other answers as decimals, so let’s do that here as well. The probability that a house has a dog given that it’s got a cat is 0.3.

Now, before we move on to our next example, let’s just generalize that last result. Being given that the house had a cat gave us a subset of cases to consider. We limited our scope to just looking at houses that had cats. And given that we’re only looking at houses that have cats, the houses that have dogs must also have cats. So the probability that a house has a dog given that it has a cat is equal to the probability of dog intersection cat over the probability of cat. Or more generally, the probability of 𝐴 given 𝐵 is the probability of 𝐴 intersection 𝐵 over the probability of 𝐵.

Suppose 𝐴 and 𝐵 are two events. Given that the probability of 𝐴 intersection 𝐵 is two-thirds and the probability of 𝐴 is nine-thirteenths, find the probability of 𝐵 given 𝐴.

Now you may recall the general formula for conditional probability that the probability of 𝐴 given 𝐵 is equal to the probability of 𝐴 intersection 𝐵 over the probability of 𝐵. But we’ve been asked to find the probability of 𝐵 given 𝐴. So let’s reverse 𝐴 and 𝐵 in our formula. Well, that’s great. We’re looking for the probability of 𝐵 given 𝐴. We’ve been given probability 𝐴 in the question, but we were given the probability of 𝐴 intersection 𝐵, not 𝐵 intersection 𝐴. But let’s consider a Venn diagram.

The region 𝐴 intersection 𝐵 is the same as the region 𝐵 intersection 𝐴. So an equivalent formula is the probability of 𝐵 given 𝐴 is equal to the probability of 𝐴 intersection 𝐵 over the probability of 𝐴. We were told that the probability of 𝐴 intersection 𝐵 is two-thirds and the probability of 𝐴 is nine-thirteenths. So the probability of 𝐵 given 𝐴 is two-thirds divided by nine-thirteenths. And an equivalent calculation to dividing by nine-thirteenths is multiplying by its reciprocal, 13 over nine, which gives us our answer. The probability of 𝐵 given 𝐴 is 26 over 27.

So now we’ve seen that we can attack questions of conditional probability either by using Venn diagrams or by using the conditional probability formula. But the conditional probability formula has one extra special use. It can help us to determine whether two events are independent or dependent.

Now, independent events are where the outcome of one event is completely unaffected by the outcome of another event. For example, if I flip a coin and roll a dice, I’ll either get heads or tails on the coin and one or two or three or four or five or six on the dice. Now, regardless of whether the coin lands, heads or tails up, that’s not gonna have any impact on whether I’m going to get a one or a two or three or four or five or six. Those probabilities are completely independent of the probabilities of heads and tails on flipping a coin.

But with dependent events, the outcome of one event is affected by the outcome of another event. For example, let’s say we’ve got a bag with two sweets in it: one is strawberry, and one is orange. If person one comes along and picks a sweet at random and eats it, then out of the two sweets, they’re equally likely to pick strawberry or orange. The probability in each case is a half. Now, if we set up a second event, where person two comes along after person one has eaten their sweet and picks a sweet at random from the bag and eats it, we don’t really know what the probability of getting strawberry or orange is. It all depends what person one ate.

If person one ate the strawberry sweet, then the probability of the second person getting a strawberry sweet is zero and the probability of them getting an orange sweet is one, whereas if the first person ate the orange sweet, now the probability of the second person getting a strawberry is one and the probability of them getting an orange is zero. The probability of the different outcomes of the second event depend on the outcomes of the first event. Now to calculate the probability of both events occurring, the probability of 𝐴 intersection 𝐵 with independent events, we can simply multiply their individual probabilities together.

But we also know from the conditional probability formula that the probability of 𝐴 given 𝐵 is equal to the probability of 𝐴 intersection 𝐵 over the probability of 𝐵. And we can rearrange this to make the probability of 𝐴 intersection 𝐵 the subject. And that’s equal to the probability of 𝐴 given 𝐵 times the probability of 𝐵. Well, that gives us two different expressions for the probability of 𝐴 intersection 𝐵. And it must mean that the probability of 𝐴 is equal to the probability of 𝐴 given 𝐵. And of course, by reversing 𝐴 and 𝐵, we can see that the probability of 𝐵 is equal to the probability of 𝐵 given 𝐴.

Now this gives us a kind of definition of independence, or at least a way to check if events are independent. If the probability of the first event occurring is the same, whether or not the second event occurs, then they’re independent. Again, we can see the probability of 𝐵 is the same whether or not event 𝐴 has occurred. So if both these equations hold, then we can say that the events 𝐴 and 𝐵 are independent. So let’s see how that works in a question.

Suppose the probability of 𝐴 is two-fifths and the probability of 𝐵 is three-sevenths. The probability that event 𝐴 occurs and event 𝐵 also occurs is one-fifth. Calculate the probability of 𝐴 given 𝐵 and then evaluate whether events 𝐴 and 𝐵 are independent.

Well, first, let us recall the conditional probability formula. The probability of 𝐴 given 𝐵 is equal to the probability of 𝐴 intersection 𝐵 over the probability of 𝐵. Well, the probability of 𝐴 intersection 𝐵 is the probability that event 𝐴 occurs and event 𝐵 also occurs, which we were given in the question is a fifth. And we were also told in the question that the probability of event 𝐵 occurring is three-sevenths. So the probability of 𝐴 given 𝐵 is a fifth divided by three-sevenths. And, of course, an equivalent operation to dividing by a fraction is multiplying by the reciprocal of that fraction. So this is equal to a fifth times seven over three. Well, there’s the answer to our first part of the question. The probability of 𝐴 given 𝐵 is equal to seven-fifteenths.

For the second part of the question, let’s recall our test for independence using the conditional probability formulae. If the probability of 𝐴 is equal to the probability of 𝐴 given 𝐵 and the probability of 𝐵 is equal to the probability of 𝐵 given 𝐴, then we can say that events 𝐴 and 𝐵 are independent. Now the question told us that the probability of 𝐴 is two-fifths. So the next step is to work out the probability of 𝐴 given 𝐵. Well, in fact, that’s what we worked out in the first part of the question. So we know that the probability of 𝐴 given 𝐵 is seven-fifteenths.

Now, if two-fifths is the same as seven-fifteenths, then we’d need to go on to check whether the probability of 𝐵 is equal to the probability of 𝐵 given 𝐴. Now to compare two-fifths and seven-fifteenths, we need to get a common denominator. We can do this by multiplying the numerator and denominator by three and two-fifths becomes six-fifteenths. That means that the probability of 𝐴 is not equal to the probability of 𝐴 given 𝐵. And the fact that we need both of these conditions to be true to prove the independence of events 𝐴 and 𝐵, the fact that we’ve shown that the probability of 𝐴 is not equal to the probability of 𝐴 given 𝐵, means that we know that these two events aren’t independent. We don’t even need to go on to check whether the probability of 𝐵 is equal to the probability of 𝐵 given 𝐴.

Now let’s just go over a few key points from our lesson. If the probability of event 𝐵 is affected by the outcome of event 𝐴, then we say that the probability of event 𝐵 is conditional on event 𝐴. We can represent this using the notation, the probability of 𝐵 given 𝐴. That’s a 𝐵 with a vertical line, then an 𝐴. We have a formula for working out conditional probabilities. The probability of event 𝐴 occurring given that event 𝐵 has occurred is equal to the probability of 𝐴 intersection 𝐵 divided by the probability of event 𝐵 occurring.

And we also have a test for the independence of events 𝐴 and 𝐵. They’re independent, if the probability of 𝐴 is the same as the probability of 𝐴 given 𝐵 and the probability of 𝐵 is equal to the probability of 𝐵 given 𝐴. And finally, we’ve also seen how Venn diagrams can be a great way to answer conditional probability questions. And they can also be a help in checking your answers if you do use the formulae.