Question

Suppose someone spins the spinner shown in the figure once. Find the given probability.
orange
orange
yellow
yellow
yellow
gr e e n
.
.
.
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

Answer 1

To solve the probability problem, we need to understand the layout of the spinner and the number of sections labeled with each color.

From your description, it seems like the spinner has the following sections:
- 2 orange sections
- 3 yellow sections
- 1 green section

Now, we can count the total number of sections on the spinner:
- Total sections = 2 (orange) + 3 (yellow) + 1 (green) = 6 sections

Now, we can calculate the probabilities for each color:

1. **Probability of stopping on an orange section:**
\( P(\text{orange}) = \frac{\text{Number of orange sections}}{\text{Total sections}} = \frac{2}{6} = \frac{1}{3} \)

So for the first blank:
**The probability that the arrow will stop on a section that is labeled orange is** **one third.**

2. **Probability of stopping on a yellow section:**
\( P(\text{yellow}) = \frac{\text{Number of yellow sections}}{\text{Total sections}} = \frac{3}{6} = \frac{1}{2} \)

So for the second blank:
**The probability that the arrow will stop on a section that is labeled yellow is** **one half.**

3. **Probability of stopping on a green section:**
\( P(\text{green}) = \frac{\text{Number of green sections}}{\text{Total sections}} = \frac{1}{6} \)

So for the third blank:
**The probability that the arrow will stop on a section that is labeled green is** **one sixth.**

To summarize:
- The probability for orange is one third.
- The probability for yellow is one half.
- The probability for green is one sixth.