1. The problem states that the probability of selecting a peach from the snack box is $\frac{1}{4}$. We need to determine which option matches this probability. 2. The formula for probability is: $$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$ 3. Here, the favorable outcomes are the number of peaches, and the total outcomes are the total fruits (peaches + apples). 4. Let's check each option: - A. Peaches = 3, Apples = 9 $$\text{Probability} = \frac{3}{3+9} = \frac{3}{12} = \frac{1}{4}$$ - B. Peaches = 5, Apples = 5 $$\text{Probability} = \frac{5}{5+5} = \frac{5}{10} = \frac{1}{2}$$ - C. Peaches = 12, Apples = 4 $$\text{Probability} = \frac{12}{12+4} = \frac{12}{16} = \frac{3}{4}$$ - D. Peaches = 2, Apples = 8 $$\text{Probability} = \frac{2}{2+8} = \frac{2}{10} = \frac{1}{5}$$ 5. Only option A has the probability $\frac{1}{4}$. **Final answer:** Option A (3 peaches and 9 apples)