1. **Problem Statement:** A student has two spinners: Spinner X with 2 equal sections labeled "Red" and "Blue", and Spinner Y with 3 equal sections labeled "1", "2", and "3". The student spins each spinner once. 2. **Part A: Probability Spinner X lands on "Red"** - Spinner X has 2 equal sections. - Probability of landing on "Red" is \( \frac{1}{2} \). 3. **Part B: List all possible outcomes when both spinners are spun once** - Spinner X outcomes: \{Red, Blue\} - Spinner Y outcomes: \{1, 2, 3\} - Possible combined outcomes (Spinner X, Spinner Y): \[ \{(Red,1), (Red,2), (Red,3), (Blue,1), (Blue,2), (Blue,3)\} \] 4. **Part C: Probability Spinner X lands on "Red" AND Spinner Y lands on an odd number** - Odd numbers on Spinner Y: 1 and 3 (2 out of 3 sections) - Probability Spinner Y lands on odd number: \( \frac{2}{3} \) - Probability both events happen (independent events): \[ P = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3} \approx 0.3333 \] 5. **Part D: Probability Spinner X lands on "Red" OR Spinner Y lands on an odd number** - Use formula for union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] - Here: \[ P(\text{Red}) = \frac{1}{2}, \quad P(\text{odd on Y}) = \frac{2}{3}, \quad P(\text{Red and odd}) = \frac{1}{3} \] - Calculate: \[ P = \frac{1}{2} + \frac{2}{3} - \frac{1}{3} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \approx 0.8333 \] **Final answers:** - A: \( \frac{1}{2} \) - B: \( \{(Red,1), (Red,2), (Red,3), (Blue,1), (Blue,2), (Blue,3)\} \) - C: \( \frac{1}{3} \) - D: \( \frac{5}{6} \)