Students practice calculating probability by analyzing the sections of two different spinners.
Math worksheet asking students to calculate probability using number and color spinners.
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Show Answer Key & Explanations
Step-by-step solution for: Probability Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Probability Worksheets
Let's solve each question step by step based on the two spinners shown in the image.
---
This spinner is divided into four equal sections, labeled:
- 1
- 2
- 3
- 4
Each section has an equal chance of being landed on, so the probability of landing on any one number is:
$$
\frac{1}{4}
$$
---
#### Question 1: What is the probability of the spinner landing on a 3?
There is one section with a 3 out of four total sections.
$$
P(3) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 2: What is the probability of the spinner landing on a 1?
Same logic — one section with a 1.
$$
P(1) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 3: What is the probability of the spinner landing on a 2?
One section with a 2.
$$
P(2) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 4: Are you more likely to spin an odd number or an even number? Explain.
Odd numbers on the spinner: 1, 3 → 2 outcomes
Even numbers: 2, 4 → 2 outcomes
So:
- Probability of odd: $ \frac{2}{4} = \frac{1}{2} $
- Probability of even: $ \frac{2}{4} = \frac{1}{2} $
They are equally likely.
✔ Answer: You are equally likely to spin an odd or even number because there are two odd numbers and two even numbers.
---
This spinner has four sections, but two of them are red.
The sections are:
- Red
- Blue
- Orange
- Red
So, total sections = 4
Red appears twice, blue once, orange once.
---
#### Question 5: What is the probability of the spinner landing on red?
Number of red sections = 2
Total sections = 4
$$
P(\text{red}) = \frac{2}{4} = \frac{1}{2}
$$
✔ Answer: $ \frac{1}{2} $
---
#### Question 6: What is the probability of the spinner landing on blue?
Only one blue section.
$$
P(\text{blue}) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 7: What is the probability of the spinner landing on orange?
Only one orange section.
$$
P(\text{orange}) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 8: Mike said, "You have a fifty-fifty chance of spinning red." Explain what he means.
A "fifty-fifty chance" means a 50% probability, or $ \frac{1}{2} $. Since there are two red sections out of four, the probability is:
$$
\frac{2}{4} = \frac{1}{2}
$$
So, Mike means that you are just as likely to land on red as not to land on red — it's an equal chance.
✔ Answer: Mike means that the chance of landing on red is 50%, because half of the spinner is red (2 out of 4 sections). So, you're equally likely to land on red or not land on red.
---
1. $ \frac{1}{4} $
2. $ \frac{1}{4} $
3. $ \frac{1}{4} $
4. Equally likely; there are two odd and two even numbers.
5. $ \frac{1}{2} $
6. $ \frac{1}{4} $
7. $ \frac{1}{4} $
8. Mike means there's a 50% chance of landing on red because two out of the four sections are red.
Let me know if you'd like this formatted for printing or explained in simpler terms!
---
Spinner 1: Numbered Spinner (with numbers 1, 2, 3, 4)
This spinner is divided into four equal sections, labeled:
- 1
- 2
- 3
- 4
Each section has an equal chance of being landed on, so the probability of landing on any one number is:
$$
\frac{1}{4}
$$
---
#### Question 1: What is the probability of the spinner landing on a 3?
There is one section with a 3 out of four total sections.
$$
P(3) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 2: What is the probability of the spinner landing on a 1?
Same logic — one section with a 1.
$$
P(1) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 3: What is the probability of the spinner landing on a 2?
One section with a 2.
$$
P(2) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 4: Are you more likely to spin an odd number or an even number? Explain.
Odd numbers on the spinner: 1, 3 → 2 outcomes
Even numbers: 2, 4 → 2 outcomes
So:
- Probability of odd: $ \frac{2}{4} = \frac{1}{2} $
- Probability of even: $ \frac{2}{4} = \frac{1}{2} $
They are equally likely.
✔ Answer: You are equally likely to spin an odd or even number because there are two odd numbers and two even numbers.
---
Spinner 2: Color Spinner (with colors: red, blue, orange, red)
This spinner has four sections, but two of them are red.
The sections are:
- Red
- Blue
- Orange
- Red
So, total sections = 4
Red appears twice, blue once, orange once.
---
#### Question 5: What is the probability of the spinner landing on red?
Number of red sections = 2
Total sections = 4
$$
P(\text{red}) = \frac{2}{4} = \frac{1}{2}
$$
✔ Answer: $ \frac{1}{2} $
---
#### Question 6: What is the probability of the spinner landing on blue?
Only one blue section.
$$
P(\text{blue}) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 7: What is the probability of the spinner landing on orange?
Only one orange section.
$$
P(\text{orange}) = \frac{1}{4}
$$
✔ Answer: $ \frac{1}{4} $
---
#### Question 8: Mike said, "You have a fifty-fifty chance of spinning red." Explain what he means.
A "fifty-fifty chance" means a 50% probability, or $ \frac{1}{2} $. Since there are two red sections out of four, the probability is:
$$
\frac{2}{4} = \frac{1}{2}
$$
So, Mike means that you are just as likely to land on red as not to land on red — it's an equal chance.
✔ Answer: Mike means that the chance of landing on red is 50%, because half of the spinner is red (2 out of 4 sections). So, you're equally likely to land on red or not land on red.
---
✔ Final Answers Summary:
1. $ \frac{1}{4} $
2. $ \frac{1}{4} $
3. $ \frac{1}{4} $
4. Equally likely; there are two odd and two even numbers.
5. $ \frac{1}{2} $
6. $ \frac{1}{4} $
7. $ \frac{1}{4} $
8. Mike means there's a 50% chance of landing on red because two out of the four sections are red.
Let me know if you'd like this formatted for printing or explained in simpler terms!
Parent Tip: Review the logic above to help your child master the concept of probability worksheet pdf with solutions.