5.2 Combining Events

OBJECTIVES By the end of this section, I will be able to …

  1. Combine events using complement, union, and intersection.
  2. Apply the Addition Rule to events in general and to mutually exclusive events in particular.

1 Complement, Union, and Intersection

In Example 7, if your roommate rolled a 4, then your roommate was to do your laundry for the rest of the semester. Your roommate is keenly interested in not rolling a 4. If is an event, then the collection of outcomes not in event is called the complement of , denoted . The term complement comes from the word “to complete,” meaning that any event and its complement together make up the complete sample space.

EXAMPLE 14 Finding the probability of the complement of an event

If is the event “observing a sum of 4 when the two fair dice are rolled,” then your roommate is interested in the probability of , the event that a sum of 4 is not rolled. Find the probability that your roommate does not roll a sum of 4.

Solution

Which outcomes belong to ? By the definition, is all the outcomes in the sample space that do not belong in . The following outcomes are included in .

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Figure 12 shows all the outcomes, except the outcomes from in the two-dice sample space. There are 33 outcomes in and 36 outcomes in the sample space. The classical probability method then gives the probability of not rolling a sum of 4 to be

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Figure 5.12: FIGURE 12 Outcomes in .

NOW YOU CAN DO

Exercises 7–12.

YOUR TURN #8

Refer to the two-dice sample space on page 247. Define event as rolling a sum of 8.

  1. Find .
  2. Calculate .

(The solutions are shown in Appendix A.)

The probability is high that, on this roll at least, your roommate will not land on Boardwalk.

For event in Example 14, note that

Is this a coincidence, or does the sum of the probabilities of an event and its complement always add to 1? Recall the Law of Total Probability (Section 5.1), which states that the sum of all the outcome probabilities in the sample space must be equal to 1. Because any event and its complement together make up the entire sample space, then it always happens that .

Probabilities for complements

For any event and its complement . Applying a touch of algebra gives the following:

Sometimes we need to find the probability of a combination of events. For example, consider the casino game of craps in which you roll two dice. One way of winning is by rolling the sum 7 or 11. We can find the probability of the following two events: the sum is 7 or the sum is 11. First, we need some tools for finding the probability of a combination of events.

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Union and Intersection of Events

The union of two events and is the event representing all the outcomes that belong to or or both. The union of and is denoted as and is associated with “or.” In other words, .

The intersection of two events and is the event representing all the outcomes that belong to both and . The intersection of and is denoted as and is associated with “and.” In other words, .

If you are asked to find the probability of “ or ,” you should find the probability of . Figure 13 shows the union of two events, with the red dots indicating the outcomes. Note from Figure 13 that the union of the events and refers to all outcomes in or or both. Figure 14 shows that the intersection of the two events is the part where and overlap. Both union and intersection are commutative. That is, . In other words, .

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Figure 5.13: FIGURE 13 Union of events and , .
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Figure 5.14: FIGURE 14 Intersection of events and , .

EXAMPLE 15 Union and intersection

Recall from Section 2.1 that a contingency table (also known as a crosstabulation) is a tabular summary of the relationship between two categorical variables. Table 6 contains a contingency table summarizing the gender and survival status of the passengers and crew of RMS Titanic.

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Suppose our experiment is to select one person at random from the passengers and crew. Define the following events:

  1. Find the intersection of these events, ( and ).
  2. Find the union of these events, ( or ).

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Solution

  1. The intersection of and is the event containing the outcomes that are common to both and . Note in Table 6 that the female outcomes form a column and the survived outcomes form a row. The intersection of and lies at the “intersection” of this column and this row, as illustrated in Table 6. The green cell belongs both to the Female column and the Survived row, and thus belongs to both events. The green cell therefore represents the intersection, , which includes the 344 passengers and crew who were both female and survived.
  2. The union of and is the event containing all the people who were either female or who survived, or both. That is, the union contains the following groups:

Thus, in Table 6, the union is represented by the cells containing 126, 344, and 367.

NOW YOU CAN DO

Exercises 13–18.

YOUR TURN #9

Refer to the experiment in Example 15. Define the following events:

  1. Find ( and ).
  2. Find ( or ).

(The solutions are shown in Appendix A.)

2 Addition rule

We are often interested in finding the probability that either one event or another event may occur. The formula for finding these kinds of probabilities is called the Addition Rule.

Addition Rule

In other words,

What Does the Addition Rule Mean?

We can use Figure 15 to understand the Addition Rule. We are trying to find , the probability of all the outcomes in or or both. The first part of the formula says to add the probabilities of the outcomes in to those of the outcomes in . But what about the overlap between and , which includes outcomes that belong to both events? To avoid counting the outcomes in the overlap (intersection) twice, we have to subtract the probability of the intersection, .

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Figure 5.16: FIGURE 15 How the Addition Rule works.

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EXAMPLE 16 Addition Rule

Continuing with the RMS Titanic data from Example 15, Table 6 is reproduced here. Find the probability that a randomly chosen passenger or crew member had the following characteristics:

  1. Was female
  2. Survived
  3. Was female and survived
  4. Was female or survived
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Solution

  1. Here, we seek . There were 470 female passengers and crew among the 2201 people on board the Titanic. Therefore, .
  2. We are looking for . Of the 2201 people on board the Titanic, 711 survived. Thus, .
  3. Those who were female and survived represent the intersection . In Example 15, we found that these were represented by the green cell with 344 people in it, lying in the intersection of the Female column and the Survived row. Therefore, .
  4. Here, we seek ( or ). By the Addition Rule,

NOW YOU CAN DO

Exercises 19–44.

YOUR TURN #10

Refer to the experiment in Example 15. Define the following events:

Find the probability that a randomly chosen passenger or crew member has the following characteristics:

  1. Was male
  2. Did not survive
  3. Was a male who did not survive
  4. Was a male or did not survive

(The solutions are shown in Appendix A.)

Mutually Exclusive Events

When drawing a card at random from a deck of 52 cards, the events “a heart is drawn” and “a diamond is drawn” have no outcomes in common. That is, no card is both a heart and a diamond. We say that these two events are mutually exclusive.

Two events are said to be mutually exclusive, or disjoint, if they have no outcomes in common.

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Note that any event and its complement are always mutually exclusive. Other examples of mutually exclusive events are given in Table 7.

Table 5.34: Table 7 Examples of mutually exclusive events
Experiment Mutually exclusive events
Draw a single card from a deck of 52 cards Card is red; card is a spade.
Buy a stock Stock will increase in value; stock will not
change in value.
Select a student at random Student is 30 years old or older; student is
under 18 years old.
Select a college course at random Course has 3 credits; course has 4 credits.

Figure 16 shows how mutually exclusive events are represented graphically. It shows the events

with sample space .

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Figure 5.18: FIGURE 16 Even and odd digits are mutually exclusive. Note that the overlap is empty, so that .

Note that no overlap exists between the two events. When two events are mutually exclusive, they share no outcomes, and therefore the intersection of mutually exclusive events is empty. Because the intersection is empty, then for mutually exclusive events, . Therefore, we can formulate a special case of the Addition Rule for Mutually Exclusive Events and :

Addition Rule for Mutually Exclusive Events

If and are mutually exclusive events, .

EXAMPLE 17 Addition Rule for mutually exclusive events

The National Center for Education Statistics conducted a survey regarding the living arrangements of 19,735 college students. The results are shown in Table 8.

Table 5.35: Table 8 Contingency table of college student living arrangements
On campus Off campus With parents Total
Females 1,368 5,741 4,103 11,212
Males 1,240 4,170 3,113 8,523
Total 2,608 9,911 7,216 19,735

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Calculate the following probabilities for a randomly chosen college student:

  1. Lives on campus, .
  2. Lives with parents, .
  3. Lives on campus or with parents, .

Solution

  1. A total of 2,608 of the 19,735 students live on campus, so .
  2. Of the 19,735 students, 7,216 live with parents, so .
  3. Living on campus and living with parents are mutually exclusive. So, by the Addition Rule for Mutually Exclusive Events, .

NOW YOU CAN DO

Exercises 45–48.

YOUR TURN #11

Refer to Example 17. Find the probability that a randomly chosen student lives on campus or lives off campus.

(The solution is shown in Appendix A.)