Worksheet for calculating theoretical probability using spinners with different sections.

Let’s solve each problem step by step. We’re dealing with theoretical probability — that means we assume every section on the spinner is equally likely to be landed on.

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Problem 1: Spinner divided into 6 equal sections (labeled A, B, C, D, E, F)

Total possible outcomes = 6

- P(not B) → Not landing on B. There are 5 other sections: A, C, D, E, F → So, 5/6
- P(A) → Only one section labeled A → 1/6
- P(E and C) → Can’t land on both at once! This is impossible → 0
- P(F or D) → Landing on either F or D → 2 sections → 2/6 = 1/3

✔ Answers for Problem 1:
- P(not B) = 5/6
- P(A) = 1/6
- P(E and C) = 0
- P(F or D) = 1/3

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Problem 2: Spinner divided into 4 equal sections (A, B, C, D)

Total possible outcomes = 4

- P(not A) → Not A → B, C, D → 3 sections → 3/4
- P(B) → One section → 1/4
- P(B or C) → Two sections → 2/4 = 1/2
- P(D and C) → Can’t land on two at once → 0

✔ Answers for Problem 2:
- P(not A) = 3/4
- P(B) = 1/4
- P(B or C) = 1/2
- P(D and C) = 0

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Problem 3: Spinner divided into 5 equal sections (A, B, C, D, E)

Total possible outcomes = 5

- P(A or C) → Two sections → 2/5
- P(B) → One section → 1/5
- P(E and D) → Impossible → 0
- P(not D) → Not D → A, B, C, E → 4 sections → 4/5

✔ Answers for Problem 3:
- P(A or C) = 2/5
- P(B) = 1/5
- P(E and D) = 0
- P(not D) = 4/5

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Problem 4: Spinner has 4 equal sections: A, A, B, C

Wait — let’s look carefully: The spinner shows two A’s, one B, one C.

So total sections = 4
But labels: A appears twice, B once, C once.

We want P(spinning A or B)

Number of favorable outcomes:
- A appears in 2 sections
- B appears in 1 section
→ Total = 2 + 1 = 3

So probability = 3/4

✔ Answer for Problem 4: 3/4

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Problem 5: Spinner has 5 equal sections: A, A, A, B, B

Labels: A appears 3 times, B appears 2 times.

We want P(not spinning A) → That means spinning B.

B appears in 2 sections → So 2 out of 5 → 2/5

✔ Answer for Problem 5: 2/5

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## Final Answer:

Problem 1:
P(not B) = 5/6
P(A) = 1/6
P(E and C) = 0
P(F or D) = 1/3

Problem 2:
P(not A) = 3/4
P(B) = 1/4
P(B or C) = 1/2
P(D and C) = 0

Problem 3:
P(A or C) = 2/5
P(B) = 1/5
P(E and D) = 0
P(not D) = 4/5

Problem 4:
3/4

Problem 5:
2/5