To find which option meets the probability condition provided by Carlos, we need to determine the total number of bags in each scenario and calculate the probability of selecting a bag of potato chips.
The probability of selecting a bag of potato chips is given as \( \frac{1}{2} \). This means that the number of bags of potato chips must equal half of the total number of bags.
Let's evaluate each option:
A. 4 bags of potato chips and 8 bags of pretzels:
- Total bags = \( 4 + 8 = 12 \)
- Probability of potato chips = \( \frac{4}{12} = \frac{1}{3} \) (not \( \frac{1}{2} \))
B. 8 bags of potato chips and 8 bags of pretzels:
- Total bags = \( 8 + 8 = 16 \)
- Probability of potato chips = \( \frac{8}{16} = \frac{1}{2} \) (this is correct)
C. 10 bags of potato chips and 5 bags of pretzels:
- Total bags = \( 10 + 5 = 15 \)
- Probability of potato chips = \( \frac{10}{15} = \frac{2}{3} \) (not \( \frac{1}{2} \))
D. 5 bags of potato chips and 10 bags of pretzels:
- Total bags = \( 5 + 10 = 15 \)
- Probability of potato chips = \( \frac{5}{15} = \frac{1}{3} \) (not \( \frac{1}{2} \))
The only option that results in a \( \frac{1}{2} \) probability of selecting a bag of potato chips is:
B. 8 bags of potato chips and 8 bags of pretzels