To solve this question, let's first understand the sample space and the event described.
### Sample Space
The sample space [tex]\( S \)[/tex] represents all the possible pairings of the three players (Joe, Keitaro, and Luis) in the first match. Since we are choosing two players out of three, the sample space includes every possible combination of two players:
1. Joe and Keitaro (JK)
2. Joe and Luis (JL)
3. Keitaro and Joe (KJ)
4. Keitaro and Luis (KL)
5. Luis and Joe (LJ)
6. Luis and Keitaro (LK)
Therefore, the sample space [tex]\( S \)[/tex] is:
[tex]\[ S = \{ JK, JL, KJ, KL, LJ, LK \} \][/tex]
### Event of Joe Playing in the First Match
Next, we need to identify the subset of the sample space where Joe is one of the players in the first match:
1. Joe and Keitaro (JK)
2. Joe and Luis (JL)
So the event where Joe plays in the first match, say [tex]\( E \)[/tex], is:
[tex]\[ E = \{ JK, JL \} \][/tex]
### Complement of the Event
The complement of the event [tex]\( E \)[/tex] (denoted as [tex]\( E^c \)[/tex]) includes all outcomes in the sample space where Joe is not one of the players in the first match. To find this, we subtract the event [tex]\( E \)[/tex] from the sample space [tex]\( S \)[/tex]:
[tex]\[
E^c = S - E = \{ JK, JL, KJ, KL, LJ, LK \} - \{ JK, JL \}
\][/tex]
Performing this subtraction, we get:
[tex]\[
E^c = \{ KJ, KL, LJ, LK \}
\][/tex]
### Conclusion
Thus, the complement of the event where Joe plays in the first match (denoted as [tex]\( E^c \)[/tex]) is the set of all possible pairings where Joe does not play in the first match:
[tex]\[
E^c = \{ KJ, KL, LJ, LK \}
\][/tex]
Looking at the provided options:
- [tex]\( A = \{ KL \} \)[/tex]
- [tex]\( A = \{ KJ, KL \} \)[/tex]
- [tex]\( A = \{ KL, LK \} \)[/tex]
None of these options entirely match our solution [tex]\( \{ KJ, KL, LJ, LK \} \)[/tex]. Therefore, based on the given possible options, the correct answer to the subset of the sample space that represents the complement of the event in which Joe plays in the first match is none of the provided options as they do not fully capture the complement set.
