Let's analyze each statement one by one:
1. [tex]\( P(6) = P(1) \)[/tex]:
- True. Since all slips of paper have an equal chance of being drawn, the probability of drawing any specific number (including 6 or 1) is the same. Each number has a probability of [tex]\( \frac{1}{10} \)[/tex].
2. [tex]\( P(5) = \frac{1}{2} \)[/tex]:
- False. The probability of drawing any specific number from the 10 slips is [tex]\( \frac{1}{10} \)[/tex], not [tex]\( \frac{1}{2} \)[/tex].
3. [tex]\( P(>10)=0 \)[/tex]:
- True. There are only slips of papers numbered from 1 to 10, so it is impossible to draw a number greater than 10. Therefore, the probability is 0.
4. [tex]\( P(1
- False. The numbers satisfying [tex]\( 1 < x < 10 \)[/tex] are 2, 3, 4, 5, 6, 7, 8, and 9, which are 8 numbers out of a total of 10. The probability should be [tex]\( \frac{8}{10} \)[/tex] or 80%, not 100%.
5. [tex]\( S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)[/tex]:
- True. The sample space [tex]\( S \)[/tex] includes all the numbers that can possibly be drawn, which are 1 through 10.
6. If [tex]\( A \subseteq S \)[/tex]; [tex]\( A \)[/tex] could be [tex]\( \{1, 3, 5, 7, 9\} \)[/tex]:
- True. The set [tex]\( A \)[/tex] is indeed a subset of [tex]\( S \)[/tex], as all elements 1, 3, 5, 7, and 9 are contained in [tex]\( S \)[/tex].
So, the true statements are:
- [tex]\( P(6) = P(1) \)[/tex]
- [tex]\( P(>10)=0 \)[/tex]
- [tex]\( S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)[/tex]
- [tex]\( A \)[/tex] could be [tex]\( \{1, 3, 5, 7, 9\} \)[/tex]
