Question

select the correct answer.
sam plans to use a spinner to choose which friend to invite to the movies, and he wants each friend to have a fair chance of being selected.
spinner 1
spinner 2
which spinner should sam use?
a. neither spinner 1 nor spinner 2
b. either spinner 1 or spinner 2
c. spinner 1
d. spinner 2

Explanation:

Step1: Analyze Spinner 1

Spinner 1 has 4 sections: Dean, Alec, Bart, Carl. Each section's area: Let's assume the spinner is a circle (360°). If divided into 4 equal parts? Wait, no, looking at the colors: Dean (green), Alec (blue), Bart (yellow), Carl (red). Wait, maybe the areas: Wait, Spinner 1: how many parts? Let's count the number of each friend. Dean: 1, Alec:1, Bart:1, Carl:1? Wait no, the diagram: Spinner 1: Dean (green) is a larger section? Wait no, maybe I missee. Wait, the problem says "each friend to have a fair chance", so each friend should have equal probability. So we need to check if each friend has the same number of sections in the spinner.

Spinner 1: Let's list the friends: Dean, Alec, Bart, Carl. Wait, Spinner 1: Dean (green), Alec (blue), Bart (yellow), Carl (red). Wait, how many times does each appear? Dean:1, Alec:1, Bart:1, Carl:1? Wait no, maybe the spinner is divided into 4 unequal parts? Wait no, maybe Spinner 2: Let's check Spinner 2. Spinner 2: sections are Carl, Bart, Dean, Alec, Dean, Bart, Carl, Alec? Wait no, the labels: Spinner 2: Carl (red), Bart (yellow), Dean (green), Alec (blue), Dean (green), Bart (yellow), Carl (red), Alec (blue). So let's count each friend:

Spinner 1:
- Dean: 1 section
- Alec: 1 section
- Bart: 1 section
- Carl: 1 section? Wait no, the first spinner (Spinner 1) has 4 sections? Wait the diagram: Spinner 1 is a circle divided into 4 parts? Wait no, the colors: Dean (green) is a quarter? Alec (blue) is a quarter? Bart (yellow) is a smaller part? Carl (red) is a smaller part? Wait maybe I'm wrong. Wait the problem is about fairness: each friend should have equal probability, so the number of sections for each friend should be equal.

Spinner 2: Let's count the number of times each friend appears:

- Carl: 2 sections (red)
- Bart: 2 sections (yellow)
- Dean: 2 sections (green)
- Alec: 2 sections (blue)

So Spinner 2 has 8 sections, each friend (Carl, Bart, Dean, Alec) appears 2 times. So probability for each: 2/8 = 1/4, equal.

Spinner 1: Let's check. Spinner 1: sections: Dean (green), Alec (blue), Bart (yellow), Carl (red). Wait, maybe the sizes: Dean and Alec have larger sections, Bart and Carl have smaller? So Dean: 1 large, Alec:1 large, Bart:1 small, Carl:1 small. So probability for Dean and Alec is higher than Bart and Carl. So Spinner 1 is unfair. Wait no, maybe I misread. Wait the friends are Dean, Alec, Bart, Carl. Let's check Spinner 2: each friend has 2 sections. So Spinner 2 has equal probability for each friend. Wait, but the question is which spinner Sam should use for fairness. Wait maybe I made a mistake. Wait let's re-express:

Wait the friends are Dean, Alec, Bart, Carl. So we need each to have the same number of sections.

Spinner 1:
- Dean: 1
- Alec: 1
- Bart: 1
- Carl: 1? No, that can't be. Wait maybe Spinner 1 has 4 sections, each with different sizes, so probabilities are unequal. Spinner 2: 8 sections, with each friend (Carl, Bart, Dean, Alec) appearing 2 times. So 2/8 = 1/4 for each. So Spinner 2 is fair. Wait but the options are A. neither, B. either, C. spinner 1, D. spinner 2.

Wait wait, maybe I got Spinner 1 wrong. Let's look again. Spinner 1: the labels are Dean, Alec, Bart, Carl. So four sections, each labeled once. But are the sections equal in area? If Spinner 1 is divided into four equal areas (each 90°), then each has probability 1/4, fair. But Spinner 2: 8 sections, each friend twice, so 2/8=1/4, also fair. Wait that can't be. Wait maybe the friends are Dean, Alec, Bart, Carl, so four friends. Spinner 1: four sections, each friend once, equal area: fair. Spinner 2: eight sections, each friend twice, equal area: fair. So either spinner? But that's option B. Wait but maybe the first spinner's sections are unequal. Wait the diagram: Spinner 1: Dean (green) and Alec (blue) are larger (maybe each 120°), Bart (yellow) and Carl (red) are smaller (each 60°). So Dean: 120°, Alec:120°, Bart:60°, Carl:60°. So probability for Dean: 120/360=1/3, Alec:1/3, Bart:1/6, Carl:1/6. Unfair. Spinner 2: 8 sections, each friend 2 times, so each section is 45°, so 2*45=90°, 90/360=1/4. So Spinner 2 is fair. Wait but the options: D is spinner 2. Wait maybe I misread the friends. Wait the problem says "which friend to invite", so the friends are Dean, Alec, Bart, Carl. So Spinner 2 has each friend 2 times, so equal probability. Spinner 1: if sections are unequal, then unfair. So Sam should use Spinner 2 (option D). Wait but let's check again.

Wait the question: "he wants each friend to have a fair chance of being selected". So each friend must have the same probability.

Spinner 1:
- Dean: 1 section (size: let's say area A)
- Alec: 1 section (size A)
- Bart: 1 section (size B, B < A)
- Carl: 1 section (size B)
So probabilities: Dean: A/(2A + 2B), Alec: A/(2A + 2B), Bart: B/(2A + 2B), Carl: B/(2A + 2B). So Dean and Alec have higher probability than Bart and Carl. Unfair.

Spinner 2:
- Carl: 2 sections
- Bart: 2 sections
- Dean: 2 sections
- Alec: 2 sections
Total sections: 8. So each has 2/8 = 1/4 probability. Fair.

So Sam should use Spinner 2 (option D).

Answer:

D. spinner 2