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

# Lesson Explainer: Operations on Events: Difference Mathematics

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

First, recall the operations on events that we have met so far.

### Definition: Complement, Intersection, and Union of Events

The operations on events and are as follows, with the shaded area in the Venn diagram representing each operation respectively.

• The complement of an event is denoted by , which contains elements that are not in .
• The intersection of events and is denoted by , which contains elements that are in both and .
• The union of events and is denoted by , which contains elements that are in either , , or both.

The new operation that we will meet in this explainer is the difference between two events and , as detailed in the definition below.

### Definition: The Difference between Two Events

The difference between two events and is denoted by and is illustrated by the shaded area in the Venn diagram below. This contains elements that are in but not in .

Using our understanding of Venn diagrams, we can derive the formula for the difference between two events.

The area of the shaded region for is equivalent to the area of minus the area of , as seen below.

Therefore, . We can then use this to derive a formula for the probability of the difference between two events.

### Rule of Probability for the Difference between Two Events

The probability of the difference between two events and is

In the following example, we will apply the rule of probability in the definition above in order to find the probability of the difference between two events.

### Example 1: Determining the Probability of the Difference between Two Events

Suppose and are two events. Given that and , determine .

To find the difference between two events and , we use the following formula:

This can be represented using a Venn diagram:

By substituting and into the formula above, we get

Similarly, we can use our Venn diagram to illustrate this.

Therefore, .

In the next example, we will consider how to find the probability of the difference between two events given in a context.

### Example 2: Finding the Probability of a Difference of Two Events

A ball is drawn at random from a bag containing 12 balls each with a unique number from 1 to 12. Suppose is the event of drawing an odd number and is the event of drawing a prime number. Find .

To find , we use the formula for the probability of the difference between two events and , which is

In order to do this, we must find and .

To find , we first identify the set . We know that is the event of drawing an odd number from a bag with balls numbered from 1 to 12. Therefore, is given by the set .

Since the number of outcomes in is 6 and the total number of outcomes is 12 (since there are 12 balls in the bag), then the probability of is given by

In order to find , we start by identifying the sets and , and the set . We know that is given by the set (as stated above). Set is the event of drawing a prime number from a bag with balls numbered from 1 to 12. Therefore, is given by the set .

We can see that , the intersection of and , is the set that contains the elements that occur in both and . In this case, . The probability of is given by

We can now substitute and into the formula in order to find :

Therefore, .

We can use multiple rules of probability for operations on events in order to solve problems. We will next consider two of these rules of probability: the complement and the union of events. Let’s recall what these rules are.

### Definition: Rules of Probability for Complement and Union of Events

• The probability of the complement of an event is or
• The probability of the union of events and is

The following example uses the rules of probability for the union of two events and the difference between two events.

### Example 3: Determining the Probability of the Difference of Two Events Using the Addition Rule

Suppose and are two events with probabilities and . Given that , determine .

As we are required to find , we must use the rule of probability for the difference between two events, which is stated as follows:

Since we do not know , we must use a further rule to determine this. As we know , as well as and , we can use the addition rule for probability:

When we substitute , , and , we get an equation with :

Rearranging to solve for gives us

Since we have found , we can substitute this, along with , into the formula

So, by substituting, we can solve for :

Therefore, .

The next example uses the rules of probability for the complement of an event, the union of an event, and the difference between two events.

### Example 4: Determining the Probability of the Difference of Two Events Using the Addition Rule and the Complement Rule

Suppose that and are events in a random experiment. Given that , , and , determine .

As we are required to find , we must use the rule of probability for the difference between two events, which is stated as follows:

As events and are reversed in this formula, we need to rewrite this as since .

Since we know , , and , but neither nor , we need to use rules of probability for the complement of an event and the probability of the union of two events. First, we will use the rule of probability for a complement of an event to find .

We know that

So, to find , we substitute and rearrange for :

Having found , we can use the addition formula to find . This is

By substituting , , and then rearranging to make the subject, we get

Since we have found , we can now use this, along with , to find . We do this by substituting into the formula and solving for ;

Therefore, .

In the last example, we will consider how to find the probability of exactly one of two events occurring. We will use what we know about the difference between two events to do this.

### Example 5: Using the Addition Rule to Determine the Probability of Exactly One of Two Events Occurring

The probability that Shady passes mathematics is 0.33 and the probability that he fails physics is 0.32. Given that the probability of him passing at least one of them is 0.71, find the probability that he passes EXACTLY ONE of the two subjects.

To find the probability of passing exactly one subject, we find the probability of passing mathematics only and the probability of passing physics only and then add these together.

To find the probability of passing mathematics only, we find the probability of passing mathematics but not physics, which can be done using the difference formula:

If we let the event of passing mathematics be represented by and the event of passing physics be represented by , then by replacing and in the formula above, we can say that the probability of passing mathematics but not physics is

We are told in the question that the probability of passing mathematics is 0.33, so , and the probability of failing physics is 0.32, so . To find the probability of passing physics, we use the complement formula, which states that or, in this case,

So, by substituting and solving for , we get

We are also told that the probability of passing at least one of them is 0.71, which is the same as saying passing mathematics, or physics, or both. We can write this using the union of and , so .

As we need to find in order to find but are told , we use the additional formula for probability which states that

In this case, we are using and to represent the events, so

We know that , , and , so by substituting and solving for , we get

Having found , we can substitute this, along with , into the formula for , which gives us which is the probability of passing mathematics but not physics.

Next, we need to find the probability of passing physics but not passing mathematics. This is done in a very similar way to that above, but by reversing the events and in the formula which is the same as since saying the probability of passing physics and mathematics is the same as passing mathematics and physics.

As we know and , then by substituting, we get which is the probability of passing physics but not mathematics.

Now that we know the probability of passing mathematics but not physics is and the probability of passing physics but not mathematics is , then the probability of passing exactly one subject is the sum of these, which is

Therefore, the probability of passing exactly one subject is 0.41.

In this explainer, we have learned what the difference between two events is, how this is represented on a Venn diagram, and the formula for calculating this.

### Key Points

• The difference between two sets and is denoted as and is represented on the Venn diagram below:
• The rule of probability for the difference between two events and is