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 𝐴 and 𝐵 are as follows, with the shaded area in the Venn diagram representing each operation respectively.

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

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

Definition: The Difference between Two Events

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

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

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

Therefore, 𝐴𝐵=𝐴(𝐴𝐵). 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 𝐴 and 𝐵 is 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

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 𝐴 and 𝐵 are two events. Given that 𝑃(𝐴)=0.3 and 𝑃(𝐴𝐵)=0.03, determine 𝑃(𝐴𝐵).

Answer

To find the difference between two events 𝐴 and 𝐵, we use the following formula: 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

This can be represented using a Venn diagram:

By substituting 𝑃(𝐴)=0.3 and 𝑃(𝐴𝐵)=0.03 into the formula above, we get 𝑃(𝐴𝐵)=0.30.03=0.27.

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

Therefore, 𝑃(𝐴𝐵)=0.27.

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 𝐴 is the event of drawing an odd number and 𝐵 is the event of drawing a prime number. Find 𝑃(𝐴𝐵).

Answer

To find 𝑃(𝐴𝐵), we use the formula for the probability of the difference between two events 𝐴 and 𝐵, which is 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

In order to do this, we must find 𝑃(𝐴) and 𝑃(𝐴𝐵).

To find 𝑃(𝐴), we first identify the set 𝐴. We know that 𝐴 is the event of drawing an odd number from a bag with balls numbered from 1 to 12. Therefore, 𝐴 is given by the set {1,3,5,7,9,11}.

Since the number of outcomes in 𝐴 is 6 and the total number of outcomes is 12 (since there are 12 balls in the bag), then the probability of 𝐴 is given by 𝑃(𝐴)=𝐴=612=12.numberofoutcomesintotalnumberofoutcomes

In order to find 𝑃(𝐴𝐵), we start by identifying the sets 𝐴 and 𝐵, and the set 𝐴𝐵. We know that 𝐴 is given by the set {1,3,5,7,9,11} (as stated above). Set 𝐵 is the event of drawing a prime number from a bag with balls numbered from 1 to 12. Therefore, 𝐵 is given by the set {2,3,5,7,11}.

We can see that 𝐴𝐵, the intersection of 𝐴 and 𝐵, is the set that contains the elements that occur in both 𝐴 and 𝐵. In this case, 𝐴𝐵={3,5,7,11}. The probability of 𝐴𝐵 is given by 𝑃(𝐴𝐵)=𝐴𝐵=412=13.numberofoutcomesinandtotalnumberofoutcomes

We can now substitute 𝑃(𝐴)=12 and 𝑃(𝐴𝐵)=13 into the formula 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵) in order to find 𝑃(𝐴𝐵): 𝑃(𝐴𝐵)=1213=326=16.

Therefore, 𝑃(𝐴𝐵)=16.

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 𝐴 is 𝑃(𝐴)=1𝑃(𝐴) or 𝑃(𝐴)=1𝑃(𝐴).
  • The probability of the union of events 𝐴 and 𝐵 is 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

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 𝐴 and 𝐵 are two events with probabilities 𝑃(𝐴)=57 and 𝑃(𝐵)=47. Given that 𝑃(𝐴𝐵)=67, determine 𝑃(𝐴𝐵).

Answer

As we are required to find 𝑃(𝐴𝐵), we must use the rule of probability for the difference between two events, which is stated as follows: 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

Since we do not know 𝑃(𝐴𝐵), we must use a further rule to determine this. As we know 𝑃(𝐴𝐵), as well as 𝑃(𝐴) and 𝑃(𝐵), we can use the addition rule for probability: 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

When we substitute 𝑃(𝐴)=57, 𝑃(𝐵)=47, and 𝑃(𝐴𝐵)=67, we get an equation with 𝑃(𝐴𝐵): 67=57+47𝑃(𝐴𝐵).

Rearranging to solve for 𝑃(𝐴𝐵) gives us 𝑃(𝐴𝐵)+67=57+47𝑃(𝐴𝐵)=57+4767=37.

Since we have found 𝑃(𝐴𝐵)=37, we can substitute this, along with 𝑃(𝐴)=57, into the formula 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

So, by substituting, we can solve for 𝑃(𝐴𝐵): 𝑃(𝐴𝐵)=5737=27.

Therefore, 𝑃(𝐴𝐵)=27.

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 𝐴 and 𝐵 are events in a random experiment. Given that 𝑃(𝐴)=0.71, 𝑃(𝐵)=0.47, and 𝑃(𝐴𝐵)=0.99, determine 𝑃(𝐵𝐴).

Answer

As we are required to find 𝑃(𝐵𝐴), we must use the rule of probability for the difference between two events, which is stated as follows: 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

As events 𝐴 and 𝐵 are reversed in this formula, we need to rewrite this as 𝑃(𝐵𝐴)=𝑃(𝐵)𝑃(𝐵𝐴)=𝑃(𝐵)𝑃(𝐴𝐵) since 𝑃(𝐵𝐴)=𝑃(𝐴𝐵).

Since we know 𝑃(𝐴), 𝑃(𝐵), and 𝑃(𝐴𝐵), but neither 𝑃(𝐴𝐵) nor 𝑃(𝐵), 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 𝑃(𝐵).

We know that 𝑃(𝐵)=1𝑃(𝐵).

So, to find 𝑃(𝐵), we substitute 𝑃(𝐵)=0.47 and rearrange for 𝑃(𝐵): 0.47=1𝑃(𝐵)0.47+𝑃(𝐵)=1𝑃(𝐵)=10.47𝑃(𝐵)=0.53.

Having found 𝑃(𝐵), we can use the addition formula to find 𝑃(𝐴𝐵). This is 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

By substituting 𝑃(𝐴)=0.71, 𝑃(𝐵)=0.53, and 𝑃(𝐴𝐵)=0.99 then rearranging to make 𝑃(𝐴𝐵) the subject, we get 0.99=0.71+0.53𝑃(𝐴𝐵)0.99+𝑃(𝐴𝐵)=0.71+0.53𝑃(𝐴𝐵)=0.71+0.530.99=0.25.

Since we have found 𝑃(𝐴𝐵)=0.25, we can now use this, along with 𝑃(𝐵)=0.53, to find 𝑃(𝐵𝐴). We do this by substituting into the formula and solving for 𝑃(𝐵𝐴); 𝑃(𝐵𝐴)=𝑃(𝐵)𝑃(𝐴𝐵)𝑃(𝐵𝐴)=0.530.25=0.28.

Therefore, 𝑃(𝐵𝐴)=0.28.

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.

Answer

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: 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

If we let the event of passing mathematics be represented by 𝑀 and the event of passing physics be represented by 𝑃, then by replacing 𝐴 and 𝐵 in the formula above, we can say that the probability of passing mathematics but not physics is 𝑃(𝑀𝑃)=𝑃(𝑀)𝑃(𝑀𝑃).

We are told in the question that the probability of passing mathematics is 0.33, so 𝑃(𝑀)=0.33, and the probability of failing physics is 0.32, so 𝑃(𝑃)=0.32. To find the probability of passing physics, we use the complement formula, which states that 𝑃(𝐴)=1𝑃(𝐴) or, in this case, 𝑃(𝑃)=1𝑃(𝑃).

So, by substituting 𝑃(𝑃)=0.32 and solving for 𝑃(𝑃), we get 0.32=1𝑃(𝑃)𝑃(𝑃)=10.32=0.68.

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 𝑀 and 𝑃, so 𝑃(𝑀𝑃)=0.71.

As we need to find 𝑃(𝑀𝑃) in order to find 𝑃(𝑀𝑃) but are told 𝑃(𝑀𝑃)=0.71, we use the additional formula for probability which states that 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

In this case, we are using 𝑀 and 𝑃 to represent the events, so 𝑃(𝑀𝑃)=𝑃(𝑀)+𝑃(𝑃)𝑃(𝑀𝑃).

We know that 𝑃(𝑀)=0.33, 𝑃(𝑃)=0.68, and 𝑃(𝑀𝑃)=0.71, so by substituting and solving for 𝑃(𝑀𝑃), we get 𝑃(𝑀𝑃)=𝑃(𝑀)+𝑃(𝑃)𝑃(𝑀𝑃)0.71=0.33+0.68𝑃(𝑀𝑃)0.71=1.01𝑃(𝑀𝑃)𝑃(𝑀𝑃)=1.010.71=0.3.

Having found 𝑃(𝑀𝑃)=0.3, we can substitute this, along with 𝑃(𝑀)=0.33, into the formula for 𝑃(𝑀𝑃), which gives us 𝑃(𝑀𝑃)=𝑃(𝑀)𝑃(𝑀𝑃)=0.330.3=0.03, 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 𝑀 and 𝑃 in the formula 𝑃(𝑃𝑀)=𝑃(𝑃)𝑃(𝑃𝑀), which is the same as 𝑃(𝑃𝑀)=𝑃(𝑃)𝑃(𝑀𝑃), since saying the probability of passing physics and mathematics is the same as passing mathematics and physics.

As we know 𝑃(𝑃)=0.68 and 𝑃(𝑀𝑃)=0.3, then by substituting, we get 𝑃(𝑃𝑀)=𝑃(𝑃)𝑃(𝑀𝑃)=0.680.3=0.38, which is the probability of passing physics but not mathematics.

Now that we know the probability of passing mathematics but not physics is 𝑃(𝑀𝑃)=0.03 and the probability of passing physics but not mathematics is 𝑃(𝑃𝑀)=0.38, then the probability of passing exactly one subject is the sum of these, which is 𝑃()=0.03+0.38=0.41.passingexactlyonesubject

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 𝐴 and 𝐵 is denoted as 𝐴𝐵 and is represented on the Venn diagram below:
  • The rule of probability for the difference between two events 𝐴 and 𝐵 is 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

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