Answer :
Sure, let's go through the steps needed to solve this problem.
First, we need to construct a sample space [tex]\(S\)[/tex] that contains all possible outcomes of drawing two slips simultaneously from the box of slips marked with the numbers 3, 4, 5, 6, and 7.
### Constructing the Sample Space [tex]\(S\)[/tex]:
The sample space [tex]\(S\)[/tex] is defined as the set of all possible pairs of slips that can be drawn. Since the order in which the slips are drawn does not matter (drawing slip 3 and then slip 4 is the same as drawing slip 4 and then slip 3), we list all unique pairs of slips:
1. Draw 3 and 4
2. Draw 3 and 5
3. Draw 3 and 6
4. Draw 3 and 7
5. Draw 4 and 5
6. Draw 4 and 6
7. Draw 4 and 7
8. Draw 5 and 6
9. Draw 5 and 7
10. Draw 6 and 7
Hence, the sample space [tex]\(S\)[/tex] is:
[tex]\[ S = \{ (3, 4), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7) \} \][/tex]
### The Value of [tex]\(n(S)\)[/tex]:
The value [tex]\(n(S)\)[/tex] represents the number of outcomes in the sample space [tex]\(S\)[/tex]. Since we have listed 10 pairs, we have:
[tex]\[ n(S) = 10 \][/tex]
### Are the outcomes in [tex]\(S\)[/tex] equally likely?
Yes, if the slips are mixed thoroughly and two are drawn at random, each pair should be equally likely. Therefore, the outcomes in [tex]\(S\)[/tex] are equally likely.
### Indicated Events:
Now we identify the specified events in set notation.
a. Both slips are marked with odd numbers.
The odd numbers are 3, 5, and 7. The possible pairs that satisfy this condition are:
[tex]\[ \{ (3, 5), (3, 7), (5, 7) \} \][/tex]
b. One slip is marked with an odd number and the other is marked with an even number.
The odd numbers are 3, 5, and 7. The even numbers are 4 and 6. The possible pairs that satisfy this condition are:
[tex]\[ \{ (3, 4), (3, 6), (4, 5), (4, 7), (5. 6), (6, 7) \} \][/tex]
c. Both slips are marked with the same number.
Since each slip in the box is unique, it is impossible to draw two slips with the same number. Therefore, this event is:
[tex]\[ \{ \} \][/tex] (an empty set)
### Summary:
- The sample space [tex]\(S\)[/tex] is: [tex]\( \{ (3, 4), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7) \} \)[/tex]
- The value of [tex]\(n(S)\)[/tex] is: [tex]\( 10 \)[/tex]
- The outcomes in [tex]\(S\)[/tex] are equally likely.
- Event a (both slips with odd numbers): [tex]\( \{ (3, 5), (3, 7), (5, 7) \} \)[/tex]
- Event b (one odd, one even): [tex]\( \{ (3, 4), (3, 6), (4, 5), (4, 7), (5, 6), (6, 7) \} \)[/tex]
- Event c (both slips with the same number): [tex]\( \{ \} \)[/tex]
Therefore, the correct multiple-choice answer for the sample space is:
OA. [tex]\( S = \{ (3, 4), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7) \} \)[/tex]
First, we need to construct a sample space [tex]\(S\)[/tex] that contains all possible outcomes of drawing two slips simultaneously from the box of slips marked with the numbers 3, 4, 5, 6, and 7.
### Constructing the Sample Space [tex]\(S\)[/tex]:
The sample space [tex]\(S\)[/tex] is defined as the set of all possible pairs of slips that can be drawn. Since the order in which the slips are drawn does not matter (drawing slip 3 and then slip 4 is the same as drawing slip 4 and then slip 3), we list all unique pairs of slips:
1. Draw 3 and 4
2. Draw 3 and 5
3. Draw 3 and 6
4. Draw 3 and 7
5. Draw 4 and 5
6. Draw 4 and 6
7. Draw 4 and 7
8. Draw 5 and 6
9. Draw 5 and 7
10. Draw 6 and 7
Hence, the sample space [tex]\(S\)[/tex] is:
[tex]\[ S = \{ (3, 4), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7) \} \][/tex]
### The Value of [tex]\(n(S)\)[/tex]:
The value [tex]\(n(S)\)[/tex] represents the number of outcomes in the sample space [tex]\(S\)[/tex]. Since we have listed 10 pairs, we have:
[tex]\[ n(S) = 10 \][/tex]
### Are the outcomes in [tex]\(S\)[/tex] equally likely?
Yes, if the slips are mixed thoroughly and two are drawn at random, each pair should be equally likely. Therefore, the outcomes in [tex]\(S\)[/tex] are equally likely.
### Indicated Events:
Now we identify the specified events in set notation.
a. Both slips are marked with odd numbers.
The odd numbers are 3, 5, and 7. The possible pairs that satisfy this condition are:
[tex]\[ \{ (3, 5), (3, 7), (5, 7) \} \][/tex]
b. One slip is marked with an odd number and the other is marked with an even number.
The odd numbers are 3, 5, and 7. The even numbers are 4 and 6. The possible pairs that satisfy this condition are:
[tex]\[ \{ (3, 4), (3, 6), (4, 5), (4, 7), (5. 6), (6, 7) \} \][/tex]
c. Both slips are marked with the same number.
Since each slip in the box is unique, it is impossible to draw two slips with the same number. Therefore, this event is:
[tex]\[ \{ \} \][/tex] (an empty set)
### Summary:
- The sample space [tex]\(S\)[/tex] is: [tex]\( \{ (3, 4), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7) \} \)[/tex]
- The value of [tex]\(n(S)\)[/tex] is: [tex]\( 10 \)[/tex]
- The outcomes in [tex]\(S\)[/tex] are equally likely.
- Event a (both slips with odd numbers): [tex]\( \{ (3, 5), (3, 7), (5, 7) \} \)[/tex]
- Event b (one odd, one even): [tex]\( \{ (3, 4), (3, 6), (4, 5), (4, 7), (5, 6), (6, 7) \} \)[/tex]
- Event c (both slips with the same number): [tex]\( \{ \} \)[/tex]
Therefore, the correct multiple-choice answer for the sample space is:
OA. [tex]\( S = \{ (3, 4), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7) \} \)[/tex]
