### 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.