Question
Suppose someone spins the spinner shown in the figure once. Find the given probability.
red
red
yellow
yellow
yellow
purpl e
.
.
.
Question content area right
Part 1
The probability that the arrow will stop on a section that is labeled
enter your response here is one sixth
The probability that the arrow will stop on a section that is labeled
enter your response here is one half
The probability that the arrow will stop on a section that is labeled
enter your response here is one third
red
red
yellow
yellow
yellow
purpl e
.
.
.
Question content area right
Part 1
The probability that the arrow will stop on a section that is labeled
enter your response here is one sixth
The probability that the arrow will stop on a section that is labeled
enter your response here is one half
The probability that the arrow will stop on a section that is labeled
enter your response here is one third
Answers
GPT-4o mini
To determine the probabilities requested, we need the exact number of sections on the spinner that correspond to each color. However, based on the probabilities you've referenced (one sixth, one half, and one third), we can infer possible scenarios.
Let's denote:
- \( r \): the number of red sections
- \( y \): the number of yellow sections
- \( p \): the number of purple sections
The total number of sections is \( r + y + p \).
### Part 1: Probability for one sixth
If the probability that the arrow stops on a red section is \( \frac{1}{6} \), then:
\[
\frac{r}{r + y + p} = \frac{1}{6}
\]
From this, we can deduce:
\[
6r = r + y + p \implies 5r = y + p \implies y + p = 5r
\]
### Part 2: Probability for one half
If the probability that the arrow stops on a yellow section is \( \frac{1}{2} \), then:
\[
\frac{y}{r + y + p} = \frac{1}{2}
\]
From this, we can deduce:
\[
2y = r + y + p \implies y = r + p
\]
### Part 3: Probability for one third
If the probability that the arrow stops on a purple section is \( \frac{1}{3} \), then:
\[
\frac{p}{r + y + p} = \frac{1}{3}
\]
From this, we can deduce:
\[
3p = r + y + p \implies 2p = r + y
\]
### Solving the equations:
Now we can combine these equations:
1. From \( y + p = 5r \)
2. From \( y = r + p \)
3. From \( 2p = r + y \)
Substituting \( y \) from the second equation into the first:
\[
r + p + p = 5r \implies r + 2p = 5r \implies 2p = 4r \implies p = 2r
\]
Now substituting \( p = 2r \) back into the equation \( y = r + p \):
\[
y = r + 2r = 3r
\]
From here, substituting \( y \) and \( p \) back into the total sections:
\[
r + y + p = r + 3r + 2r = 6r
\]
Thus, we can denote the numbers of sections as:
- Red sections: \( r \)
- Yellow sections: \( 3r \)
- Purple sections: \( 2r \)
If we set \( r = 1 \) (to keep the numbers simple), then:
- Red sections: \( 1 \)
- Yellow sections: \( 3 \)
- Purple sections: \( 2 \)
Total sections: \( 1 + 3 + 2 = 6 \)
### Final Probabilities:
1. The probability of landing on red: \( \frac{1}{6} \) (1 red section)
2. The probability of landing on yellow: \( \frac{3}{6} = \frac{1}{2} \) (3 yellow sections)
3. The probability of landing on purple: \( \frac{2}{6} = \frac{1}{3} \) (2 purple sections)
To fill in the probabilities in your initial statement:
- The probability that the arrow will stop on a section that is labeled **red** is **one sixth**.
- The probability that the arrow will stop on a section that is labeled **yellow** is **one half**.
- The probability that the arrow will stop on a section that is labeled **purple** is **one third**.
Let's denote:
- \( r \): the number of red sections
- \( y \): the number of yellow sections
- \( p \): the number of purple sections
The total number of sections is \( r + y + p \).
### Part 1: Probability for one sixth
If the probability that the arrow stops on a red section is \( \frac{1}{6} \), then:
\[
\frac{r}{r + y + p} = \frac{1}{6}
\]
From this, we can deduce:
\[
6r = r + y + p \implies 5r = y + p \implies y + p = 5r
\]
### Part 2: Probability for one half
If the probability that the arrow stops on a yellow section is \( \frac{1}{2} \), then:
\[
\frac{y}{r + y + p} = \frac{1}{2}
\]
From this, we can deduce:
\[
2y = r + y + p \implies y = r + p
\]
### Part 3: Probability for one third
If the probability that the arrow stops on a purple section is \( \frac{1}{3} \), then:
\[
\frac{p}{r + y + p} = \frac{1}{3}
\]
From this, we can deduce:
\[
3p = r + y + p \implies 2p = r + y
\]
### Solving the equations:
Now we can combine these equations:
1. From \( y + p = 5r \)
2. From \( y = r + p \)
3. From \( 2p = r + y \)
Substituting \( y \) from the second equation into the first:
\[
r + p + p = 5r \implies r + 2p = 5r \implies 2p = 4r \implies p = 2r
\]
Now substituting \( p = 2r \) back into the equation \( y = r + p \):
\[
y = r + 2r = 3r
\]
From here, substituting \( y \) and \( p \) back into the total sections:
\[
r + y + p = r + 3r + 2r = 6r
\]
Thus, we can denote the numbers of sections as:
- Red sections: \( r \)
- Yellow sections: \( 3r \)
- Purple sections: \( 2r \)
If we set \( r = 1 \) (to keep the numbers simple), then:
- Red sections: \( 1 \)
- Yellow sections: \( 3 \)
- Purple sections: \( 2 \)
Total sections: \( 1 + 3 + 2 = 6 \)
### Final Probabilities:
1. The probability of landing on red: \( \frac{1}{6} \) (1 red section)
2. The probability of landing on yellow: \( \frac{3}{6} = \frac{1}{2} \) (3 yellow sections)
3. The probability of landing on purple: \( \frac{2}{6} = \frac{1}{3} \) (2 purple sections)
To fill in the probabilities in your initial statement:
- The probability that the arrow will stop on a section that is labeled **red** is **one sixth**.
- The probability that the arrow will stop on a section that is labeled **yellow** is **one half**.
- The probability that the arrow will stop on a section that is labeled **purple** is **one third**.