Let’s answer each question step by step: --- ### 1. Rule Method for Sample Space S **Sample Space:** All points in the first quadrant inside a circle of radius 4 centered at the origin. - First quadrant: \( x > 0, y > 0 \) - Circle: \( x^2 + y^2 < 16 \) (since radius = 4, so \(4^2 = 16\)) **Looking for the correct choice:** Let’s parse the options (noting that sometimes typos like “x + y <” appear, but the circle is defined by \(x^2 + y^2 < 16\)). So, the sample space is: \[ S = \{(x, y) \mid x^2 + y^2 < 16; x > 0, y > 0 \} \] So, the correct choice matches: **B. \( S = \{(x, y) \mid x^2 + y^2 < 16; x > 0, y > 0 \} \)** --- ### 2. Sets S, M, N Given: - \( S = \{x \mid 0 < x < 14\} \) - \( M = \{x \mid 2 < x < 12\} \) - \( N = \{x \mid 0 < x < 5\} \) #### (a) \( M \cup N \): Union: all \( x \) in either \( M \) or \( N \): - \( N \): \( 0 < x < 5 \) - \( M \): \( 2 < x < 12 \) So, their union is \( 0 < x < 12 \). **Answer:** A. \( M \cup N = \{x \mid 0 < x < 12\} \) #### (b) \( M \cap N \): Intersection: all \( x \) in both \( M \) and \( N \): - \( N \): \( 0 < x < 5 \) - \( M \): \( 2 < x < 12 \) Their overlap is \( 2 < x < 5 \). **Answer:** B. \( M \cap N = \{x \mid 2 < x < 5\} \) #### (c) \( M' \cap N' \): - \( M' \): complement of \( M \) in \( S \): \( 0 < x \leq 2 \) or \( 12 \leq x < 14 \) - \( N' \): complement of \( N \) in \( S \): \( 5 \leq x < 14 \) Intersection: - \( (0 < x \leq 2) \cap (5 \leq x < 14) = \emptyset \) - \( (12 \leq x < 14) \cap (5 \leq x < 14) = 12 \leq x < 14 \) **Answer:** C. \( M' \cap N' = \{x \mid 12 \leq x < 14\} \) --- ### 3. Sample Space for Die and Vowel - 6-sided die: 6 outcomes - Vowels (A, E, I, O, U): 5 outcomes Total sample points: \( 6 \times 5 = 30 \) **Answer:** There are **30** points in the sample space. --- ### 4. Arrangements of 8 People #### (a) In how many ways can 8 people be lined up? \( 8! = 40,320 \) ways. #### (b) 3 Specific Persons Must Follow Each Other Treat the 3 as a single unit (block), the rest are 5 individuals = 6 units. - Arrange the 6 units: \( 6! \) ways. - The 3 persons within the block can arrange themselves: \( 3! \) ways. Total: \( 6! \times 3! = 720 \times 6 = 4,320 \) ways. #### (c) 2 Specific Persons Refuse to Follow Each Other - Total arrangements: \( 8! = 40,320 \) - Arrangements with the 2 together: - Treat them as a block, so 7 units: \( 7! \) - The 2 can be in 2 orders: \( 2! \) - Total: \( 7! \times 2 = 5,040 \times 2 = 10,080 \) - Arrangements with the 2 **not** together: - \( 8! - 10,080 = 30,240 \) --- ## **Summary Table** | Part | Answer | |------|--------| | 1 | \( S = \{(x, y) \mid x^2 + y^2 < 16; x > 0, y > 0 \} \) (Choice B) | | 2a | \( M \cup N = \{x \mid 0 < x < 12\} \) | | 2b | \( M \cap N = \{x \mid 2 < x < 5\} \) | | 2c | \( M' \cap N' = \{x \mid 12 \leq x < 14\} \) | | 3 | 30 | | 4a | 40,320 | | 4b | 4,320 | | 4c | 30,240 |