### Video Transcript

In this video, we will learn how to
find the probability of the difference of two events. This is written as the probability
of π΄ minus π΅, where π΄ and π΅ are two events such that π΄ minus π΅ is all outcomes
that are in event π΄ but not in event π΅. We will begin this video by
introducing some key notation and formulae.

In the Venn diagram shown
representing the events π΄ and π΅, the probability of π΄ is represented by the
shaded section. The intersection of events π΄ and
π΅ is represented on a Venn diagram by the overlap between the circles. These are the outcomes that occur
in event π΄ and event π΅. Combining these two definitions
leads us to the difference formula, which states that the probability of π΄ minus π΅
is equal to the probability of π΄ minus the probability of π΄ intersection π΅. This can be represented on a Venn
diagram as shown.

We could use the same method to
prove that the probability of π΅ minus π΄ is equal to the probability of π΅ minus
the probability of π΄ intersection π΅. On a Venn diagram, this can be
shown by shading the region that is in circle π΅ but not circle π΄. We will now look at a couple of
examples where we need to use the difference formula.

Suppose π΄ and π΅ are two
events. Given that the probability of π΄ is
0.3 and the probability of π΄ intersection π΅ is 0.03, determine the probability of
π΄ minus π΅.

Letβs begin by recalling the
notation shown in this question. Firstly, we have the intersection
of events π΄ and π΅. This is all the outcomes that occur
in event π΄ and in event π΅. This can be represented on a Venn
diagram as shown. The probability of π΄ minus π΅ is
known as the difference and could be represented on a Venn diagram as shown. It is all of the outcomes in event
π΄ that are not in event π΅.

We recall that the difference
formula states that the probability of π΄ minus π΅ is equal to the probability of π΄
minus the probability of π΄ intersection π΅. Using the values given in the
question, the probability of π΄ minus π΅ is equal to 0.3 minus 0.03. This is equal to 0.27.

In our next example, we will look
at a problem in context.

A ball is drawn at random from a
bag containing 12 balls each with a unique number from one to 12. Suppose π΄ is the event of drawing
an odd number and π΅ is the event of drawing a prime number. Find the probability of π΄ minus
π΅.

We are asked in this question to
find the probability of π΄ minus π΅. And we can do this using the
difference formula. This states that the probability of
π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄
intersection π΅. We are told in the question that
there are 12 balls, each with a unique number from one to 12 as shown. We are told that π΄ is the event of
drawing an odd number. There are six of these in the bag,
the numbers one, three, five, seven, nine, and 11. This means that the probability of
event π΄ selecting an odd-numbered ball is six out of 12 or six twelfths. Dividing the numerator and
denominator by six, we see that this simplifies to one-half.

We are also told that π΅ is the
event of drawing a prime number. We know that a prime number has
exactly two factors, the number one and itself. The prime numbers between one and
12 are two, three, five, seven, and 11. As there are five of these, the
probability of event π΅ is five out of 12 or five twelfths. At this stage, we have the
probability of π΄, but we donβt have the probability of π΄ intersection π΅. The intersection of two events is
all outcomes that occur in both of the events. In this case, the numbers three,
five, seven, and 11 are both odd numbers and prime numbers. The probability of π΄ intersection
π΅ is therefore equal to four twelfths, which in turn simplifies to one-third.

We can now calculate the
probability of π΄ minus π΅ by subtracting one-third from one-half. Using equivalent fractions, this is
the same as three-sixths minus two-sixths. The probability of π΄ minus π΅ is
therefore equal to one-sixth. In context, the probability of
drawing an odd number that is not a prime number is one-sixth.

We could also represent this by
listing all the outcomes on a Venn diagram. We have already mentioned that the
numbers three, five, seven, and 11 are both odd and prime. There are two other odd numbers
between one and 12, the numbers one and nine. And there is one other prime
number, the number two. The numbers four, six, eight, 10,
and 12 are neither odd nor prime. As a result, we write these outside
of the two circles representing event π΄ and event π΅.

Two of the numbers are in the
section represented by the probability of π΄ minus π΅. These are the numbers one and nine,
as they are odd but not prime. This confirms that the probability
of drawing one of these balls is two out of 12, which simplifies to one-sixth.

Before looking at our next example,
we will recall one of our other probability formulae. The probability of the complement
of event π΄, denoted π of π΄ prime or π of π΄ bar, is the probability of event π΄
not occurring. This satisfies the formula the
probability of π΄ prime is equal to one minus the probability of π΄. We will now look at an example
where we need to use this.

Suppose π΄ and π΅ are two
events. Given that π΄ intersection π΅ is
the empty set, the probability of π΄ prime is 0.66, and the probability of π΅ prime
is 0.79, find the probability of π΅ minus π΄.

Before trying to answer this
question, letβs recall some of the notation. We know that π΄ prime and π΅ prime
are the complement of events π΄ and π΅, respectively. And we also know that the
probability of the complement of event π΄ is equal to one minus the probability of
event π΄. Using the information given, we can
therefore calculate the probability of event π΄ along with the probability of event
π΅.

Firstly, we have 0.66 is equal to
one minus the probability of π΄. Rearranging this equation, we have
the probability of π΄ is equal to one minus 0.66. This is equal to 0.34. In the same way, 0.79 is equal to
one minus the probability of event π΅. The probability of π΅ is therefore
equal to one minus 0.79, which is equal to 0.21.

We are also told that π΄
intersection π΅ is equal to the empty set. This means that there are no
elements in event π΄ and event π΅. And we can therefore say that the
two events are mutually exclusive. And the probability of π΄
intersection π΅ is therefore equal to zero. When representing this on a Venn
diagram, there is no overlap as shown. We can fill in the fact that the
probability of event π΄ is 0.34 and the probability of event π΅ is 0.21. We can complete the Venn diagram by
filling in the probability that neither event π΄ nor event π΅ occur. This is equal to 0.45.

We are asked to find the
probability of π΅ minus π΄. And using the difference formula,
we know this is equal to the probability of π΅ minus the probability of π΄
intersection π΅. Substituting in the values we know,
this is equal to 0.21 minus zero, which is just equal to 0.21. This leads us onto an important
rule. If two events π΄ and π΅ are
mutually exclusive, then the probability of π΅ minus π΄ is simply equal to the
probability of π΅. Likewise, the probability of π΄
minus π΅ is equal to the probability of π΄.

Before looking at one final
example, letβs recall the addition rule of probability. The addition rule of probability
states that the probability of π΄ union π΅ is equal to the probability of π΄ plus
the probability of π΅ minus the probability of π΄ intersection π΅. This can be represented using Venn
diagrams as shown. We will now look at one final
example.

Suppose π΄ and π΅ are events in a
sample space which consists of equally likely outcomes. Given that π΄ contains six
outcomes, the probability of π΄ union π΅ is three-quarters, the probability of π΅ is
one-half, and the total number of outcomes is 20, find the probability that only one
of the events π΄ or π΅ occurs.

Letβs begin by looking at how we
can represent the probability that only one of the events π΄ and π΅ occurs on a Venn
diagram. The probability that only event π΄
occurs is shaded in pink. We know that this can be written
using the difference formula. It is the probability of π΄ minus
π΅. And this is equal to the
probability of π΄ minus the probability of π΄ intersection π΅. The probability that only event π΅
occurs is shaded in blue. And this is denoted by the
probability of π΅ minus π΄. This is equal to the probability of
π΅ minus the probability of π΄ intersection π΅. In order to answer this question,
we will need to find the sum of these two values.

Letβs now consider the information
we are given in this question. We are told that there are 20
outcomes in total and that event π΄ contains six of these. The probability of event π΄ is
therefore equal to six out of 20. Dividing the numerator and
denominator by two, this simplifies to three-tenths. We are also told that the
probability of π΄ union π΅ is three-quarters and the probability of π΅ is
one-half. We now have both the probabilities
of event π΄ and π΅ but not the probability of π΄ intersection π΅.

We can calculate this by using the
addition rule of probability, which states that the probability of π΄ union π΅ is
equal to the probability of π΄ plus the probability of π΅ minus the probability of
π΄ intersection π΅. This can be rearranged as
shown. Substituting in our values, we have
the probability of π΄ intersection π΅ is equal to three-tenths plus a half minus
three-quarters. Our three fractions have a common
denominator of 20. So we can rewrite this as six over
20 plus 10 over 20 minus 15 over 20. This is equal to one over 20.

The probability of π΄ intersection
π΅ is one twentieth. We can now use this value together
with the probabilities of π΄ and π΅ to calculate the probability of π΄ minus π΅ and
the probability of π΅ minus π΄. The probability of π΄ minus π΅ is
equal to three-tenths minus one twentieth. This is equal to five
twentieths. The probability of π΅ minus π΄ is
equal to one-half minus one twentieth. And this is equal to nine
twentieths. The probability that only one of
the events π΄ or π΅ occurs is therefore equal to five twentieths plus nine
twentieths. Adding our numerators gives us
fourteen twentieths, which in turn simplifies to seven-tenths or 0.7. The probability that only one of
the events π΄ or π΅ occurs is seven-tenths.

An alternative way to calculate
this would be to list the number of outcomes on our Venn diagram. Since the probability of π΄
intersection π΅ is one twentieth and there are a total of 20 outcomes, there is one
outcome in the intersection of π΄ and π΅. We were told that π΄ contains six
outcomes. Therefore, five of these will occur
in just event π΄. Since the probability of event π΅
is one-half, there are 10 outcomes in event π΅. And nine of these must occur in
just event π΅. As nine plus one plus five equals
15, there must be five outcomes that are not in event π΄ nor event π΅. This confirms that 14 of the
outcomes are in only one of event π΄ or π΅. And 14 out of 20 simplifies to
seven-tenths.

We will now summarize the key
points from this video. In this video, we used the
difference rule of probability together with other probability formulae to solve a
variety of problems. The difference rule of probability
states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus
the probability of π΄ intersection π΅. For any two events π΄ and π΅, it is
also true that the probability of π΅ minus π΄ is equal to the probability of π΅
minus the probability of π΄ intersection π΅.