Lesson Explainer: Operations on Events Mathematics

In this explainer, we will learn how to find the probabilities of the complement, intersection, and union of events.

We will begin this explainer by introducing some key probability definitions, together with their representation on a Venn diagram.

Definition: The Complement of Events

The complement of an event 𝐴 in a sample space 𝑆, denoted 𝐴 or 𝐴, is the collection of all outcomes in 𝑆 that are not elements of the set 𝐴. In other words, 𝐴 is the event that 𝐴 does not occur.

The probability rule for complements states that 𝑃(𝐴)=1𝑃(𝐴).

The complement of event 𝐴, 𝐴, can be represented on a Venn diagram as shown below.

Definition: The Intersection of Events

The intersection of events 𝐴 and 𝐵, denoted 𝐴𝐵, is the collection of all outcomes that are elements of both of the sets 𝐴 and 𝐵. In other words, 𝐴𝐵 is the event that both 𝐴 and 𝐵 occur.

The intersection of events 𝐴 and 𝐵, 𝐴𝐵, can be represented on a Venn diagram as shown below.

Definition: The Union of Events

The union of events 𝐴 and 𝐵, denoted 𝐴𝐵, is the collection of all outcomes that are elements of one or the other of the sets 𝐴 and 𝐵, or of both of them. In other words, 𝐴𝐵 is the event that 𝐴 or 𝐵 or both 𝐴 and 𝐵 occur.

The union of events 𝐴 and 𝐵, 𝐴𝐵, can be represented on a Venn diagram as shown below.

The definitions of the intersection and union of events lead us to an important rule of probability.

Formula: Additive Rule of Probability

𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵)

This can be demonstrated using the following Venn diagrams.

The above information leads us to six further formulae that we can use to solve problems involving probability and their respective Venn diagrams.

Firstly, we have the complement of the union of events 𝐴 and 𝐵: 𝑃(𝐴𝐵)=1𝑃(𝐴𝐵).

Secondly, we have the complement of the intersection of events 𝐴 and 𝐵: 𝑃(𝐴𝐵)=1𝑃(𝐴𝐵).

Next, we consider the probability of exactly one event occurring.

Firstly, there is the probability of event 𝐴 occurring and event 𝐵 not occurring. This is the intersection of event 𝐴 and the complement of event 𝐵: 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

And we have the corollary to this formula: 𝑃(𝐴𝐵)=𝑃(𝐵)𝑃(𝐴𝐵).

Our last two formulae are the complement of the difference formulae: 𝑃(𝐴𝐵)=1𝑃(𝐴𝐵)

𝑃(𝐴𝐵)=1𝑃(𝐴𝐵)

We will now look at a couple of examples where we can use the rules of probability to calculate probabilities of complements, intersections, and unions of events.

Example 1: Using the Addition Rule to determine the Probability of the Union of Two Events

Denote by 𝐴 and 𝐵 two events with probabilities 𝑃(𝐴)=0.2 and 𝑃(𝐵)=0.47. Given that 𝑃(𝐴𝐵)=0.18, find 𝑃(𝐴𝐵).

Answer

We begin by recalling that the additive rule of probability states that 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

So, 𝑃(𝐴𝐵)=0.2+0.470.18𝑃(𝐴𝐵)=0.49.

We could also demonstrate this on a Venn diagram. To complete all sections of the diagram, we recall that 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

So, 𝑃(𝐴𝐵)=0.20.18=0.02.

Likewise, 𝑃(𝐴𝐵)=𝑃(𝐵)𝑃(𝐴𝐵).

So, 𝑃(𝐴𝐵)=0.470.18=0.29.

We then have that 𝑃(𝐴𝐵)=0.02+0.18+0.29, and therefore, 𝑃(𝐴𝐵)=0.49.

Example 2: Using the Addition Rule to Determine the Probability That Neither of the Events Occur

Suppose 𝐴 and 𝐵 are two events with probability 𝑃(𝐴)=0.6 and 𝑃(𝐵)=0.5. Given that 𝑃(𝐴𝐵)=0.4, what is the probability that neither of the events occur?

Answer

We begin by recalling that the additive rule of probability states that 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

So, 𝑃(𝐴𝐵)=0.6+0.50.4𝑃(𝐴𝐵)=0.7.

In other words, the probability that 𝐴 or 𝐵 or both 𝐴 and 𝐵 occurs is 0.7.

The probability of neither event 𝐴 nor 𝐵 occurring is the complement of the union of events 𝐴 and 𝐵, as shown in the Venn diagram below.

𝑃(𝐴𝐵)=1𝑃(𝐴𝐵)

So, 𝑃(𝐴𝐵)=10.7, and therefore, 𝑃(𝐴𝐵)=0.3.

Example 3: Using the Addition Rule to Determine the Probability of the Union of Two Events

Suppose that 𝑋 and 𝑌 are two events with probabilities 𝑃(𝑌)=13 and 𝑃(𝑋)=𝑃(𝑋).

Given that 𝑃(𝑋𝑌)=18, determine 𝑃(𝑋𝑌).

Answer

We begin by recalling that the sum of the probability of an event occurring together with its complement equals 1: 𝑃(𝑋)=1𝑃(𝑋).

Since 𝑃(𝑋)=𝑃(𝑋), then 1𝑃(𝑋)=𝑃(𝑋)(𝑃(𝑋))1=2𝑃(𝑋)(2)12=𝑃(𝑋).addingtobothsidesdividingbothsidesby

Next, we recall that the additive rule of probability states that 𝑃(𝑋𝑌)=𝑃(𝑋)+𝑃(𝑌)𝑃(𝑋𝑌).

So, 𝑃(𝑋𝑌)=12+1318=1224+824324=1724.

Therefore, 𝑃(𝑋𝑌)=1724.

In our next example, we will use the additive rule for probability to answer a contextual question.

Example 4: Determining the Probability of the Union of Two Events

A group of 68 school children completed a survey asking about their television preferences. The results show that 43 of the children watch channel 𝐴, 26 watch channel 𝐵, and 12 watch both channels. If a child is selected at random from the group, what is the probability that they watch at least one of the two channels?

Answer

Firstly, we are told that 43 out of the 68 school children surveyed watch channel 𝐴; therefore, the probability of selecting a child at random that watches channel 𝐴 is 𝑃(𝐴)=4368.

Secondly, 26 out of the 68 school children surveyed watch channel 𝐵; therefore, the probability of selecting a child at random that watches channel 𝐵 is 𝑃(𝐵)=2668.

Finally, 12 out of the 68 school children surveyed watch both channel 𝐴 and channel 𝐵; therefore, the probability of selecting a child at random that watches both channels is 𝑃(𝐴𝐵)=1268.

Next we recall that the additive rule of probability states that 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

So, 𝑃(𝐴𝐵)=4368+26681268=5768.

We could also demonstrate this on a Venn diagram. To complete all sections of the diagram, we recall that 𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵).

So, 𝑃(𝐴𝐵)=43681268=3168.

Likewise, 𝑃(𝐴𝐵)=𝑃(𝐵)𝑃(𝐴𝐵).

So, 𝑃(𝐴𝐵)=26681268=1468.

We then have that 𝑃(𝐴𝐵)=3168+1268+1468, and therefore, 𝑃(𝐴𝐵)=5768.

In our next example, we will use the additive rule for probability to calculate an unknown variable, 𝑥.

Example 5: Determining the Probability of the Union of Two Events

𝐴 and 𝐵 are two events in a sample space of a random experiment and 𝑃(𝐴)=79𝑃(𝐵). The probability that at most one of 𝐴 or 𝐵 occurs is 0.56. The probability that at least one of 𝐴 or 𝐵 occurs is 0.68. Find the probability that 𝐴 occurs OR 𝐵 does not occur.

Answer

We can begin by letting 𝑃(𝐵)=𝑥, so that 𝑃(𝐴)=79𝑥.

As the probability that at least one of 𝐴 or 𝐵 occurs is 0.68, 𝑃(𝐴𝐵)=0.68.

For at most one of 𝐴 or 𝐵 to occur, either 𝐴 can occur, 𝐵 can occur, or neither. This means that both 𝐴 and 𝐵 cannot occur at the same time, so the probability that at most one of 𝐴 or 𝐵 occurs is given by the complement of the intersection of these, 𝑃(𝐴𝐵). Hence, 𝑃(𝐴𝐵)=0.56.

We know that 𝑃(𝐴𝐵)=1𝑃(𝐴𝐵).

So, 0.56=1𝑃(𝐴𝐵)𝑃(𝐴𝐵)=10.56=0.44.

The additive rule of probability states that 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

So, 0.68=79𝑥+𝑥0.44(0.44)1.12=169𝑥1690.63=𝑥.addingtobothsidesdividingbothsidesby

Therefore, 𝑃(𝐵)=0.63 and 𝑃(𝐴)=79(0.63)=0.49.

We need the probability that 𝐴 occurs OR 𝐵 does not occur, which can be written as 𝑃(𝐴𝐵). This is equal to 1𝑃(𝐴𝐵).

Since 𝑃(𝐴𝐵)=𝑃(𝐵)𝑃(𝐴𝐵), then 𝑃(𝐴𝐵)=0.630.44=0.19.

And as 𝑃(𝐴𝐵)=1𝑃(𝐴𝐵), then 𝑃(𝐴𝐵)=10.19=0.81.

The probability that 𝐴 occurs OR 𝐵 does not occur is 0.81.

We will finish this explainer by recapping some of the key points.

Key Points

  • The complement of an event 𝐴 in a sample space 𝑆, denoted 𝐴 or 𝐴, is the collection of all outcomes in 𝑆 that are not elements of the set 𝐴.
  • The intersection of events 𝐴 and 𝐵, denoted 𝐴𝐵, is the collection of all outcomes that are elements of both of the sets 𝐴 and 𝐵.
  • The union of events 𝐴 and 𝐵, denoted 𝐴𝐵, is the collection of all outcomes that are elements of one or the other of the sets 𝐴 and 𝐵 or of both of them.
  • The additive rule of probability states that 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).
  • Other rules of probability include 𝑃(𝐴𝐵)=1𝑃(𝐴𝐵),𝑃(𝐴𝐵)=1𝑃(𝐴𝐵),𝑃(𝐴𝐵)=𝑃(𝐴)𝑃(𝐴𝐵),𝑃(𝐴𝐵)=𝑃(𝐵)𝑃(𝐴𝐵),𝑃(𝐴𝐵)=1𝑃(𝐴𝐵),𝑃(𝐴𝐵)=1𝑃(𝐴𝐵).
  • We can use these rules together with Venn diagrams to calculate probabilities of complements, intersections, and unions and to solve problems in context.

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