Lesson Video: Operations on Events: Difference | Nagwa Lesson Video: Operations on Events: Difference | Nagwa

Lesson Video: Operations on Events: Difference Mathematics • Third Year of Preparatory School

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In this video, we will learn how to find the probability of the difference of two events.

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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΅.

Letβs 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 donβt 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, letβs 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, letβs 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.

Letβs 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.

Letβs 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΅.

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