To solve the problem of identifying the outcomes that are in event [tex]\(A\)[/tex] (places that are cities) or event [tex]\(B\)[/tex] (places that are in North America), we will use the information given in the table and follow a set theory approach.
First, we define the sets for events [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
1. Event [tex]\(A\)[/tex] (places that are cities):
[tex]\[
A = \{\text{Rome}, \text{Tokyo}, \text{Houston}, \text{Miami}, \text{Toronto}\}
\][/tex]
2. Event [tex]\(B\)[/tex] (places that are in North America):
[tex]\[
B = \{\text{Houston}, \text{Peru}, \text{Miami}, \text{Toronto}, \text{Canada}\}
\][/tex]
To find outcomes that are in [tex]\(A\)[/tex] or [tex]\(B\)[/tex], we need the union of sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[
A \cup B
\][/tex]
The union of two sets includes all elements that are in either set, without duplicates. Therefore, we list all unique outcomes from both sets:
Combining the outcomes from set [tex]\(A\)[/tex] and set [tex]\(B\)[/tex]:
[tex]\[
\{\text{Rome}, \text{Tokyo}, \text{Houston}, \text{Miami}, \text{Toronto}\} \cup \{\text{Houston}, \text{Peru}, \text{Miami}, \text{Toronto}, \text{Canada}\}
\][/tex]
Arrange all unique places:
[tex]\[
\{\text{Rome}, \text{Tokyo}, \text{Houston}, \text{Miami}, \text{Toronto}, \text{Peru}, \text{Canada}\}
\][/tex]
So, the combined outcomes that are in event [tex]\(A\)[/tex] or event [tex]\(B\)[/tex] are:
[tex]\[
\{\text{Rome}, \text{Tokyo}, \text{Houston}, \text{Miami}, \text{Toronto}, \text{Canada}, \text{Peru}\}
\][/tex]
Now, let's compare this result with the provided options:
A. \{Rome, Tokyo, Houston, Miami, Toronto\}
B. (Houston, Miami, Toronto)
C. \{Houston, Miami, Toronto, Canada\}
D. \{Rome, Tokyo, Houston, Miami, Toronto, Canada\}
The correct set of outcomes that are in [tex]\(A\)[/tex] or [tex]\(B\)[/tex] is:
[tex]\[
\{\text{Rome}, \text{Tokyo}, \text{Houston}, \text{Miami}, \text{Toronto}, \text{Canada}, \text{Peru}\}
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{\text{D}}
\][/tex]
This is because option D includes all places listed in either event [tex]\(A\)[/tex] (cities) or event [tex]\(B\)[/tex] (places in North America).
