Question
Carlos has a box containing bags of potato chips and pretzels. He told Sarah that the probability that she randomly selects a bag of potato chips is
1
2
. Which of the following can be the contents of the box?
A.
8 bags of potato chips and 8 bags of pretzels
B.
5 bags of potato chips and 10 bags of pretzels
C.
10 bags of potato chips and 5 bags of pretzels
D.
4 bags of potato chips and 8 bags of pretzels
1
2
. Which of the following can be the contents of the box?
A.
8 bags of potato chips and 8 bags of pretzels
B.
5 bags of potato chips and 10 bags of pretzels
C.
10 bags of potato chips and 5 bags of pretzels
D.
4 bags of potato chips and 8 bags of pretzels
Answers
GPT-4o mini
To find which option represents a probability of selecting a bag of potato chips of \( \frac{1}{2} \), we need to check if the number of bags of potato chips is equal to half the total number of bags (both potato chips and pretzels).
The formula for the probability of selecting a bag of potato chips \( P \) is:
\[
P(\text{potato chips}) = \frac{\text{Number of bags of potato chips}}{\text{Total number of bags}} = \frac{x}{x + y}
\]
where \( x \) is the number of bags of potato chips, and \( y \) is the number of bags of pretzels.
If \( P = \frac{1}{2} \), then:
\[
\frac{x}{x + y} = \frac{1}{2}
\]
Cross-multiplying gives:
\[
2x = x + y \\
2x - x = y \\
x = y
\]
This means that the number of bags of potato chips must be equal to the number of bags of pretzels.
Now, let’s assess each option:
**A. 8 bags of potato chips and 8 bags of pretzels**
P = \( \frac{8}{8 + 8} = \frac{8}{16} = \frac{1}{2} \)
This option works.
**B. 5 bags of potato chips and 10 bags of pretzels**
P = \( \frac{5}{5 + 10} = \frac{5}{15} = \frac{1}{3} \)
This option does not work.
**C. 10 bags of potato chips and 5 bags of pretzels**
P = \( \frac{10}{10 + 5} = \frac{10}{15} = \frac{2}{3} \)
This option does not work.
**D. 4 bags of potato chips and 8 bags of pretzels**
P = \( \frac{4}{4 + 8} = \frac{4}{12} = \frac{1}{3} \)
This option does not work.
The only option that gives a probability of \( \frac{1}{2} \) for selecting a bag of potato chips is:
**A. 8 bags of potato chips and 8 bags of pretzels**.
The formula for the probability of selecting a bag of potato chips \( P \) is:
\[
P(\text{potato chips}) = \frac{\text{Number of bags of potato chips}}{\text{Total number of bags}} = \frac{x}{x + y}
\]
where \( x \) is the number of bags of potato chips, and \( y \) is the number of bags of pretzels.
If \( P = \frac{1}{2} \), then:
\[
\frac{x}{x + y} = \frac{1}{2}
\]
Cross-multiplying gives:
\[
2x = x + y \\
2x - x = y \\
x = y
\]
This means that the number of bags of potato chips must be equal to the number of bags of pretzels.
Now, let’s assess each option:
**A. 8 bags of potato chips and 8 bags of pretzels**
P = \( \frac{8}{8 + 8} = \frac{8}{16} = \frac{1}{2} \)
This option works.
**B. 5 bags of potato chips and 10 bags of pretzels**
P = \( \frac{5}{5 + 10} = \frac{5}{15} = \frac{1}{3} \)
This option does not work.
**C. 10 bags of potato chips and 5 bags of pretzels**
P = \( \frac{10}{10 + 5} = \frac{10}{15} = \frac{2}{3} \)
This option does not work.
**D. 4 bags of potato chips and 8 bags of pretzels**
P = \( \frac{4}{4 + 8} = \frac{4}{12} = \frac{1}{3} \)
This option does not work.
The only option that gives a probability of \( \frac{1}{2} \) for selecting a bag of potato chips is:
**A. 8 bags of potato chips and 8 bags of pretzels**.