Lesson Video: Conditional Probability

In this video, we will learn how to calculate conditional probability using the formula and venn diagrams.

17:58

Video Transcript

In this lesson, we will learn how to calculate conditional probability using the formula and Venn diagrams. There are many situations where we might want to apply the rules of conditional probability. So one example is that a person’s eyes are a particular color is affected by or conditional on the color of the parents’ eyes. Or even if we look at a pack of cards, what the value of your second card from a pack if you do not replace the first is also conditional on the value of the first. So these are all situations where conditional probability can be applied.

And we can define conditional probability mathematically as follows. So if we want to look at the definition of a conditional probability, we can say that 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 the notation we use for conditional probability is written here. And it’s 𝑃, and we’ve got parentheses, and then 𝐵 and then a vertical line and then 𝐴. And this is read as the probability of 𝐵 given 𝐴.

So depending on the form of the problem, there are a few different methods we can use to help us calculate conditional probabilities. These include Venn diagrams, tree diagrams, and then also the conditional probability formula. So as well as these, we’ll also need to use some of our rules of probability. So let’s have a quick recap and look at some of those now.

So if we take a look at our notation, we can see that this here means the probability event 𝐴 occurring. So our first rule is that the probability event 𝐴 occurring is greater than or equal to zero but less than or equal to one. And then for our second rule, what we have is that the sum of the probabilities of all possible outcomes is equal to one. So then our third rule states that the probability of the complement of event 𝐴 or not 𝐴 is given by the probability of the complement of an 𝐴 equals one minus the probability of event 𝐴.

So it’s worth noting that, in the notation here, we’ve got 𝐴 bar or 𝐴 with a horizontal line above it. And this means the complement of 𝐴. It’s also worth noting that we could have 𝐴 prime, and it would mean exactly the same thing. So that means not 𝐴. So later on, what we’re also gonna look at is some other rules or formulae that relate to conditional probability.

So now let’s look at some examples. In our first example, we will calculate the probability of the complement of an event. So this question is not actually conditional probability, but it shows us some of the useful skills that we’ll need to use later on.

In an animal rescue shelter, 42 percent of the current inhabitants are cats, and 38 percent are dogs. Find the probability that an animal chosen at random is not a cat. Then find the probability that an animal chosen at random is neither a cat nor a dog.

So the first part of this problem is taking a look at cats because what we want to do is find the probability that an animal chosen at random is not a cat. So first of all, we are told that 42 percent of the current inhabitants are cats. So therefore, the probability of it being a cat that is chosen is 0.42. And we get that because 42 percent means 42 out of 100. So if we divide 42 by 100, this is equal to 0.42. And that’s because if we divide by 100, we move each of our digits two place values to the right.

Well, in this problem, what we’re actually looking for is the probability that an animal chosen at random is not a cat. So what we have to help us with this is a rule or formula. And that is that the probability of an event not occurring or the complement of an event, as it’s also known, is equal to one minus the probability of that event occurring. And it’s also worth reminding ourselves at this point that we’ve got here 𝑃, and then we’ve got inside our parentheses 𝐴 with a horizontal line above it. And we could also write this as 𝑃 and then in parentheses 𝐴 with a prime. And this would give us the same thing because it’s just different notation to mean the complement of 𝐴 or the probability of the complement of 𝐴.

So if we apply this rule to our problem, then we’re gonna have that the probability that the animal chosen at random is not a cat is equal to one minus 0.42, which is equal to 0.58. So therefore, we can say that this is the probability that an animal chosen at random is not a cat.

Okay, so now let’s move on to the second part of the question. Well, for part two, we already know that the probability of a cat being chosen is equal to 0.42. And therefore, we’ve also got the probability of a dog being chosen is equal to 0.38. And we get that because it’s 38 percent of the inhabitants are dogs. So 38 divided by 100 gives us our 0.38.

Now, for the second part of the problem, we could just solve it using a rule we know. But it is always useful to have a look at what different notation can be used. So we’re looking at the probability that an animal chosen at random is neither a cat nor a dog. We can write this as we have done here because we’ve got this U shape in between our complements. And that means all or the union.

Well, we know that all of the probabilities of any of the outcomes that can occur for a certain event must add up to one. So therefore, we can say that the probability that an animal chosen at random is neither a cat nor a dog is gonna be equal to one minus and then if we add together the probability of a cat being chosen and the probability of a dog being chosen. So this is gonna be one minus 0.42 plus 0.38, which is gonna give us the probability that an animal chosen at random is neither a cat nor a dog of 0.2.

So in this example, we actually looked at mutually exclusive events because if we look at an animal from the rescue shelter, we couldn’t have a cat and a dog. However, not all events are mutually exclusive. For example, if you’re looking at the type of animals that people liked, someone might like a cat and a dog. So therefore, they would not be mutually exclusive.

So bearing that in mind, what we’re gonna look at now is some more rules that we can add to the ones that we looked at earlier. So first of all, if we have two events 𝐴 and 𝐵 that cannot occur at the same time, they’re known as mutually exclusive. And in this case, the probability of both 𝐴 and 𝐵 is a good zero. And we can write this as probability of 𝐴 intersection 𝐵 is equal to zero.

Well, following on from this, if events 𝐴 and 𝐵 are mutually exclusive, then the probability that 𝐴 or 𝐵 occurs is the sum of their probabilities. And the way we write that is as follows. So we’d have the probability of 𝐴 union 𝐵 is equal to the probability of 𝐴 plus the probability of 𝐵. However, if events 𝐴 and 𝐵 are not mutually exclusive, we have another rule for either 𝐴 or 𝐵 occurring. And that is that the probability of 𝐴 union 𝐵 equals the probability of 𝐴 plus the probability of 𝐵 minus the probability of 𝐴 intersection 𝐵.

So with the problem up to this point, what we’ve looked at is mutually exclusive events. However, we can have a look at an example now where they’re not mutually exclusive. And what we’re gonna do is find conditional probability using Venn diagrams.

On the street, 10 houses have a cat, eight houses have a dog, three houses have both, and seven houses have neither. 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. Then 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.

And then there’ll be another part that we’ll come on to after that. So to solve this problem, what we’re gonna start by doing is put all our information into a Venn diagram. So we have our Venn diagram with the left circle representing cat and the right circle representing dog. So we’re gonna start with a bit of information that tells us about both.

So we know that three houses have both a cat and a dog. So this is gonna go in the part in the middle, which is our intersection. Then outside our circle, we’re gonna have number seven because seven houses have neither a cat or a dog. And then we’re gonna have seven in the rest of the cat circle. And that’s cause we’re told that 10 houses have a cat. Well, if we got three already in the intersection, then there’s gonna be seven left over. And then, finally, we’re gonna have five in the dog part of our circle, but not in the middle. And that’s because there were eight houses with a dog, and we’ve got three in the intersection. So five add three gives us eight. So we’ve got five in there.

Well, what we’re trying to do is find the probability that the house chosen at random has a cat and a dog. So we’re interested in the intersection. So the probability of 𝐶 intersection 𝐷 is gonna be equal to three. And that says three inside of our intersection oval, and then the total number of houses, which is seven add seven add three add five, so all of our sections added together, which is 22.

Well, we’ve got this as a fraction. But the question wants our answer to three decimal places. So we can calculate this as a decimal. And when we do, we find out that the probability that a house chosen at random has both a cat and a dog is 0.136 to three decimal places. Okay, great, so now let’s move on to part two.

So in part two, what we’re trying to do is find out the probability that a house has either a cat or a dog or both. So to do this, what we do is use one of our probability rules. And that tells us 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. And we use this because it’s not mutually exclusive. So in our example, having a cat or a dog is not mutually exclusive.

So in our example, we know the probability of a cat plus the probability of a dog minus the probability of a cat and a dog. First, we have the probability of a cat, which is 10 over 22. Then we have the probability of dog, which is eight over 22. And then we subtract the probability of cat and dog, which we’ve already worked out to be three over 22. Well, this is gonna give us 15 over 22. But once again, we want this as a decimal to three decimal places. So therefore, we can say that the probability that a house on the street has either a cat or a dog or both is 0.682 to three decimal places. Okay, great, so now let’s move on to part three.

If a house on the street has a cat, find the probability that there is also a dog living there.

So if we look back at the Venn diagram, we’ve got a house on the street that has a cat. So that’s our condition that’s been put in place. So therefore, I’ve drawn a pink circle around cat. So we’ve got seven and three because we’ve got the part which is just houses with cats and then the part which is houses with cats and dogs.

So now what we’re looking at is just a sample of 10 houses because it’s everything within this circle. Well, then we’re told to find what the probability is that there is also a dog living there. Well, this is the intersection because this is the part where we have a dog as well as a cat, so the three in here. The way we’d write this using notation is the probability, and then we’ve got that there’s a dog given, and that’s what the vertical line means, that there is a cat. And this is equal to three over 10 because it’s three over the smaller sample that we’re now looking at.

And then in keeping with the rest of the problem, what we’re gonna do is turn this into a decimal. So we can say that if a house on the street has a cat, then the probability there’s also a dog living there is 0.3.

So what we’ve done here is calculated some conditional probability, and we’ve done that using Venn diagrams. But there’s another method of calculating conditional probability, and that’s using a formula. And that conditional probability formula states that the probability of 𝐴 given 𝐵 is equal to the probability of 𝐴 intersection 𝐵 over the probability of 𝐵. So now what we’re gonna do is we’re gonna have a look at another example that uses this. And once again, it’s gonna involve a cat, a dog, and a house.

On a street, there are 25 houses, of which 12 have a cat and four have both a cat and a dog. If a house has a cat, what is the probability that there is also a dog living there? Give your answer to the three decimal places.

So in order to find what the probability that a house with a cat also has a dog, what we can do is use the conditional probability formula. So what the conditional probability formula tells us is that the probability of 𝐴 given 𝐵 is equal to the probability of 𝐴 intersection 𝐵 or 𝐴 and 𝐵 divided by the probability of 𝐵. So in our scenario, the probability that a dog lives in a house given that a cat also lives there is equal to the probability of a dog and a cat living at the house divided by the probability of a cat living at the house.

So first of all, what we can work out is what the probability of a cat being at house is. Well, if there are 25 houses in total and 12 have a cat, well then therefore the probability that a cat lives at the house is 12 over 25. And the probability of a dog and a cat living at the house is gonna be equal to. Well, we can see that there are four that have both a cat and a dog. And there are still a total of 25 houses. So that’s gonna give us a probability of four over 25.

Okay, great, we now have both bits that we need to use our conditional probability formula. So now let’s get on and use it. Well, if we do, what we’re gonna get is four over 25 divided by 12 over 25 is the probability of a dog being at a house given that there’s a cat there. Well, we know from the rules of dividing fractions that if we’ve got one fraction divided by another fraction, it’s the same as a fraction multiplied by the reciprocal of the second fraction. So we flip the second fraction.

Well, then what we can do is divide both the numerator and denominator by 25 so we can cross cancel. So we’ve got four multiplied by one over one multiplied by 12, which is gonna be four-twelfths. Then if we divide both the numerator and denominator by four, we’re gonna get one-third. Well, we want the answer to three decimal places, not as a fraction. So what we can do is write out what this is going to be. Well, we could use a calculator and do one divided by three. However, we should know that a third is the same as 0.3 recurring. And what this gives us is 0.333, and that’s to three decimal places. So what this tells us is that if a house has a cat, the probability that there’s also a dog living there is 0.333.

So now we’ve taken a look at a few examples, including this one, which include the-the conditional probability formula. What we’re now gonna take a look at is an example where we can determine whether two events are independent using conditional probability.

Suppose the probability of event 𝐴 happening is equal two-fifths and the probability of event 𝐵 equals 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.

So in order to solve the first part of this problem, what we can do is use the conditional probability formula. So in order to use it, what we’re gonna do is substitute in the values we know. So first of all, we know that the probability of 𝐴 and 𝐵 is a fifth. And then this is gonna be divided by the probability of 𝐵, which is three-sevenths, which if we use our rules for fractions is the same as one-fifth multiplied by seven-thirds.

So in order to calculate what this is gonna be, we multiply the numerators and the denominators, which gives us seven-fifteenths. So therefore, we can say the probability of 𝐴 given 𝐵 is seven fifteenths.

Now, for the second part of the question, we’re asked to determine whether they’re independent. Well, if events 𝐴 and 𝐵 are independent, then the probability of 𝐴 given 𝐵 is gonna be equal to the probability of 𝐴. Well, the probability of 𝐴 is equal to two-fifths. Well, to compare this with the probability of 𝐴 given 𝐵, we’re gonna convert this into fifteenths. So to do that, we multiply the numerator and denominator by three. So that gives us six fifteenths. So therefore, as six fifteenths is not equal to seven fifteenths, events 𝐴 and 𝐵 are not independent.

Okay, great, so we’ve now looked at a number of different examples. So now let’s look at the key points of the lesson. So first of all, 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 use the notation such as this, and it means that probability of 𝐵 given 𝐴.

Then we have some general rules of probability. And these are that the probability of an event occurring is between zero and one. The sum of the probabilities of all possible outcomes is equal to one. And then also we have the probability of a complement of event 𝐴, so not 𝐴, is equal to one minus the probability of 𝐴. And then to calculate the conditional probability, we can use Venn diagrams. We can use the formula which we have here, which is the probability of 𝐴 given 𝐵 is equal the probability of 𝐴 intersection 𝐵 over the probability of 𝐵. Or we could use tree diagrams, which we covered in another lesson.

And then, finally, we can say that 𝐴 and 𝐵 are independent if the probability of 𝐴 given 𝐵 is equal to the probability of 𝐴 and the probability of 𝐵 given 𝐴 is equal to the probability of 𝐵.

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