This worksheet helps students practice finding the probability of single and compound events using spinner diagrams divided into equal sections.

Let’s solve each problem step by step. Remember: theoretical probability = (number of favorable outcomes) ÷ (total number of possible outcomes). All sections are equal, so we just count how many sections match what we’re looking for.

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Problem 1:
Spinner has 6 equal sections: A, B, C, D, E, F

- P(not B): Not B means any section except B → that’s A, C, D, E, F → 5 out of 6 → 5/6
- P(A): Only one A → 1 out of 6 → 1/6
- P(E and C): You can’t land on both E and C at once → impossible → 0
- P(F or D): F or D → two sections → 2 out of 6 → simplify to 1/3

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Problem 2:
Spinner has 4 equal sections: A, B, C, D

- P(not A): Not A → B, C, D → 3 out of 4 → 3/4
- P(B): One B → 1 out of 4 → 1/4
- P(B or C): B or C → two sections → 2 out of 4 → simplify to 1/2
- P(D and C): Can’t be both → impossible → 0

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Problem 3:
Spinner has 5 equal sections: A, B, C, D, E

- P(A or C): A or C → two sections → 2 out of 5 → 2/5
- P(B): One B → 1 out of 5 → 1/5
- P(E and D): Can’t be both → impossible → 0
- P(not D): Not D → A, B, C, E → 4 out of 5 → 4/5

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Problem 4:
Spinner has 4 equal sections: A, A, B, C
(Note: Two sections are labeled “A”)

We want P(spining A or B)

- A appears twice, B appears once → total favorable = 2 + 1 = 3
- Total sections = 4
→ Probability = 3/4

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Problem 5:
Spinner has 5 equal sections: A, A, A, B, B
(Three A’s, two B’s)

We want P(not spinning A) → that means spinning B

- B appears twice → 2 favorable
- Total = 5
→ Probability = 2/5

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Final Answer:

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

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

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

4.
P(A or B) = 3/4

5.
P(not A) = 2/5