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3.2 Unions and Intersections

An event can often be viewed as a composition of two or more other events. Such events, which are called compound events, can be formed (composed) in two ways.

The union of two events A and B is the event that occurs if either A or B (or both) occurs on a single performance of the experiment. We denote the union of events A and B by the symbol $A \cup B . A \cup B$ $A \cup B . A \cup B$ consists of all the sample points that belong to A or B or both. (See Figure 3.7a.)

Figure 3.7
Venn diagrams for union and intersection

The intersection of two events A and B is the event that occurs if both A and B occur on a single performance of the experiment. We write $A \cap B$ $A \cap B$ for the intersection of A and B. $A \cap B$ $A \cap B$ consists of all the sample points belonging to both A and B. (See Figure 3.7b.)

Example 3.8 Probabilities of Unions and Intersections— Die-Toss Experiment

Problem

Consider a die-toss experiment in which the following events are defined:
- $\begin{matrix} A : {Toss an even number .} \end{matrix}$ $\begin{matrix} A : {Toss an even number .} \end{matrix}$
- $\begin{matrix} B : {Toss a number less than or equal to 3.} \end{matrix}$ $\begin{matrix} B : {Toss a number less than or equal to 3.} \end{matrix}$
1. Describe $A \cup B$ $A \cup B$ for this experiment.
2. Describe $A \cap B$ $A \cap B$ for this experiment.
3. Calculate $P (A \cup B)$ $P (A \cup B)$ and $P (A \cap B)$ $P (A \cap B)$ , assuming that the die is fair.

Solution

Draw the Venn diagram as shown in Figure 3.8.

Figure 3.8
Venn diagram for die toss
1. The union of A and B is the event that occurs if we observe either an even number, a number less than or equal to 3, or both on a single throw of the die. Consequently, the sample points in the event $A \cup B$ $A \cup B$ are those for which A occurs, B occurs, or both A and B occur. Checking the sample points in the entire sample space, we find that the collection of sample points in the union of A and B is
  
  [&A|Opunion|B|=||cbo|*N*[-1%0]1, 2, 3, 4, 6*N*[-1%0]|cbc| &]
  $A \cup B = {1, 2, 3, 4, 6}$ $A \cup B = {1, 2, 3, 4, 6}$
2. The intersection of A and B is the event that occurs if we observe both an even number and a number less than or equal to 3 on a single throw of the die. Checking the sample points to see which imply the occurrence of both events A and B, we see that the intersection contains only one sample point:
  
  [&A|inter|B |eq||cbo|*N*[-1%0]2*N*[-1%0]|cbc| &]
  $A \cap B = {2}$ $A \cap B = {2}$
  
  In other words, the intersection of A and B is the sample point Observe a 2.
3. Recalling that the probability of an event is the sum of the probabilities of the sample points of which the event is composed,` we have
  
  [&*AS*P|pbo|A|Opunion|B|pbc|*AP*|=|P|pbo|1|pbc||+|P|pbo|2|pbc||+|P|pbo|3|pbc||+|P|pbo|4|pbc||+|P|pbo|6|pbc| &]
  ***
  [&*AS**AP*|=|*frac*{1}{6}|+|*frac*{1}{6}|+|*frac*{1}{6}|+|*frac*{1}{6}|+|*frac*{1}{6}|=|*frac*{5}{6} &]
  $\begin{array}{l} \begin{matrix} P (A \cup B) \end{matrix} & = & P (1) + P (2) + P (3) + P (4) + P (6) \\ = & \begin{matrix} \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{5}{6} \end{matrix} \end{array}$ $\begin{array}{l} \begin{matrix} P (A \cup B) \end{matrix} & = & P (1) + P (2) + P (3) + P (4) + P (6) \\ = & \begin{matrix} \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{5}{6} \end{matrix} \end{array}$
  
  and
  
  [&P|pbo|A|inter|B|pbc||=|P|pbo|2|pbc||=|*frac*{1}{6} &]
  $P (A \cap B) = P (2) = \frac{1}{6}$ $P (A \cap B) = P (2) = \frac{1}{6}$

Look Back

Since the six sample points are equally likely, the probabilities in part c are simply the number of sample points in the event of interest, divided by 6.

Now Work Exercise 3.47a–d

Unions and intersections can be defined for more than two events. For example, the event $A \cup B \cup C$ $A \cup B \cup C$ represents the union of three events: A, B, and C. This event, which includes the set of sample points in A, B, or C, will occur if any one (or more) of the events A, B, and C occurs. Similarly, the intersection $A \cap B \cap C$ $A \cap B \cap C$ is the event that all three of the events A, B, and C occur. Therefore, $A \cap B \cap C$ $A \cap B \cap C$ is the set of sample points that are in all three of the events A, B, and C.

Example 3.9 Finding Probabilities from a Two-Way Table—Mother’s Race versus Maternal Age

Problem

Family Planning Perspectives reported on a study of over 200,000 births in New Jersey over a recent two-year period. The study investigated the link between the mother’s race and the age at which she gave birth (called maternal age). The percentages of the total number of births in New Jersey, by the maternal age and race classifications, are given in Table 3.4.

Table 3.4 Percentage of New Jersey Birth Mothers, by Age and Race

	Race
Maternal Age (years)	White	Black
$\leq 17$ $\leq 17$	2%	2%
18–19	3%	2%
20–29	41%	12%
$\geq 30$ $\geq 30$	33%	5%

This table is called a two-way table because responses are classified according to two variables: maternal age (rows) and race (columns).

Define the following event:

$A : {A New Jersey birth mother is white .}$ $A : {A New Jersey birth mother is white .}$
$B : {A New Jersey mother was a teenager when giving birth .}$ $B : {A New Jersey mother was a teenager when giving birth .}$

Find P(A) and P(B).
Find $P (A \cup B)$ $P (A \cup B)$ .
Find $P (A \cap B)$ $P (A \cap B)$ .

Solution

Following the steps for calculating probabilities of events, we first note that the objective is to characterize the race and maternal age distribution of New Jersey birth mothers. To accomplish this objective, we define the experiment to consist of selecting a birth mother from the collection of all New Jersey birth mothers during the two-year period of the study and observing her race and maternal age class. The sample points are the eight different age–race classifications:

$\begin{array}{l} \begin{array}{r} E_{1} : {\leq 17 yrs ., white} & E_{5} : {\leq 17 yrs ., black} \\ E_{2} : {18 - 19 yrs ., white} & E_{6} : {18 - 19 yrs ., black} \\ E_{3} : {20 - 29 yrs ., white} & E_{7} : {20 - 29 yrs ., black} \\ E_{4} : {\geq 30 yrs ., white} & E_{8} : {\geq 30 yrs ., black} \end{array} \end{array}$ $\begin{array}{l} \begin{array}{r} E_{1} : {\leq 17 yrs ., white} & E_{5} : {\leq 17 yrs ., black} \\ E_{2} : {18 - 19 yrs ., white} & E_{6} : {18 - 19 yrs ., black} \\ E_{3} : {20 - 29 yrs ., white} & E_{7} : {20 - 29 yrs ., black} \\ E_{4} : {\geq 30 yrs ., white} & E_{8} : {\geq 30 yrs ., black} \end{array} \end{array}$

Next, we assign probabilities to the sample points. If we blindly select one of the birth mothers, the probability that she will occupy a particular age–race classification is just the proportion, or relative frequency, of birth mothers in that classification. These proportions (as percentages) are given in Table 3.4. Thus,
- $\begin{array}{l} \begin{matrix} P (E_{1}) = Relative frequency of birth mothers in age - race class {\leq 17 yrs ., \end{matrix} \\ \begin{matrix} white} = .02 \end{matrix} \end{array}$ $\begin{array}{l} \begin{matrix} P (E_{1}) = Relative frequency of birth mothers in age - race class {\leq 17 yrs ., \end{matrix} \\ \begin{matrix} white} = .02 \end{matrix} \end{array}$
- $\begin{aligned} P (E_{2}) = .03 \end{aligned}$ $\begin{aligned} P (E_{2}) = .03 \end{aligned}$
- $\begin{aligned} P (E_{3}) = .41 \end{aligned}$ $\begin{aligned} P (E_{3}) = .41 \end{aligned}$
- $\begin{aligned} P (E_{4}) = .33 \end{aligned}$ $\begin{aligned} P (E_{4}) = .33 \end{aligned}$
- $\begin{aligned} P (E_{5}) = .02 \end{aligned}$ $\begin{aligned} P (E_{5}) = .02 \end{aligned}$
- $\begin{aligned} P (E_{6}) = .02 \end{aligned}$ $\begin{aligned} P (E_{6}) = .02 \end{aligned}$
- $\begin{aligned} P (E_{7}) = .12 \end{aligned}$ $\begin{aligned} P (E_{7}) = .12 \end{aligned}$
- $\begin{aligned} P (E_{8}) = .05 \end{aligned}$ $\begin{aligned} P (E_{8}) = .05 \end{aligned}$
You may verify that the sample point probabilities sum to 1.
1. To find P(A), we first determine the collection of sample points contained in event A. Since A is defined as ${w h i t e},$ ${w h i t e},$ we see from Table 3.4 that A contains the four sample points represented by the first column of the table. In other words, the event A consists of the race classification ${w h i t e}$ ${w h i t e}$ in all four age classifications. The probability of A is the sum of the probabilities of the sample points in A:
  
  [&P|pbo|A|pbc||=|P|pbo|E_{1}|pbc||+|P|pbo|E_{2}|pbc||+|P|pbo|E_{3}|pbc||+|P|pbo|E_{4}|pbc||=|.02|+|.03|+|.41|+|.33|=|.79 &]
  $P (A) = P (E_{1}) + P (E_{2}) + P (E_{3}) + P (E_{4}) = .02 + .03 + .41 + .33 = .79$ $P (A) = P (E_{1}) + P (E_{2}) + P (E_{3}) + P (E_{4}) = .02 + .03 + .41 + .33 = .79$
  
  Similarly, $B = {teenage mother, age \leq 19 years}$ $B = {teenage mother, age \leq 19 years}$ consists of the four sample points in the first and second rows of Table 3.4:
  
  [&P|pbo|B|pbc||=|P|pbo|E_{1}|pbc||+|P|pbo|E_{2}|pbc||+|P|pbo|E_{5}|pbc||+|P|pbo|E_{6}|pbc||=|.02|+|.03|+|.02|+|.02|=|.09 &]
  $P (B) = P (E_{1}) + P (E_{2}) + P (E_{5}) + P (E_{6}) = .02 + .03 + .02 + .02 = .09$ $P (B) = P (E_{1}) + P (E_{2}) + P (E_{5}) + P (E_{6}) = .02 + .03 + .02 + .02 = .09$
2. The union of events A and $B, A \cup B$ $B, A \cup B$ , consists of all sample points in either A or B (or both). That is, the union of A and B consists of all birth mothers who are white or who gave birth as a teenager. In Table 3.4, this is any sample point found in the first column or the first two rows. Thus,
  
  [&P|pbo|A|Opunion|B|pbc||=|.02|+|.03|+|.41|+|.33|+|.02|+|.02|=|.83 &]
  $P (A \cup B) = .02 + .03 + .41 + .33 + .02 + .02 = .83$ $P (A \cup B) = .02 + .03 + .41 + .33 + .02 + .02 = .83$
3. The intersection of events A and $B, A \cap B$ $B, A \cap B$ , consists of all sample points in both A and B. That is, the intersection of A and B consists of all birth mothers who are white and who gave birth as a teenager. In Table 3.4, this is any sample point found in the first column and the first two rows. Thus,
  
  [&P|pbo|A|inter|B|pbc||=|.02|+|.03|=|.05. &]
  $P (A \cap B) = .02 + .03 = .05 .$ $P (A \cap B) = .02 + .03 = .05 .$

Look Back

As in previous problems, the key to finding the probabilities of parts b and c is to identify the sample points that make up the event of interest. In a two-way table such as Table 3.4, the total number of sample points will be equal to the number of rows times the number of columns.

Now Work Exercise 3.51f–g

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Table of Contents for 3.2 Unions and Intersections

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3.2 Unions and Intersections