Calculate Theoretical Probability Worksheets [PDF] (7.SP.C.7.A ... - Free Printable

I'll solve each probability problem step by step, checking my work carefully.

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

Total outcomes = 6

- P(not B) = outcomes that are NOT B / total = 5/6
- P(A) = 1/6
- P(E and C) = This means getting BOTH E and C on one spin, which is impossible = 0/6 = 0
- P(F or D) = outcomes that are F OR D / total = 2/6 = 1/3

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

Total outcomes = 4

- P(not A) = outcomes that are NOT A / total = 3/4
- P(B) = 1/4
- P(B or C) = outcomes that are B OR C / total = 2/4 = 1/2
- P(D and C) = Getting both D AND C on one spin = impossible = 0/4 = 0

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

Total outcomes = 5

- P(A or C) = outcomes that are A OR C / total = 2/5
- P(B) = 1/5
- P(E and D) = Getting both E AND D on one spin = impossible = 0/5 = 0
- P(not D) = outcomes that are NOT D / total = 4/5

Problem 4: Spinner with 4 equal sections (A appears twice, B once, C once)

Looking at the spinner: A, A, B, C
Total outcomes = 4

P(A or B) = outcomes that are A OR B / total
- A appears 2 times
- B appears 1 time
- Total favorable = 2 + 1 = 3
- P(A or B) = 3/4

Problem 5: Spinner with 5 equal sections (A appears twice, B appears twice, need to count carefully)

Looking at the spinner sections: I can see A appears 2 times, B appears 2 times, and there's 1 more section. Let me recount: The spinner shows A, B, B, A, and one more section. That's 5 total sections.

Wait, let me look more carefully. The spinner has 5 sections labeled: A, B, B, A, and what appears to be another letter. Counting: A appears 2 times, B appears 2 times, and there's 1 other section.

Actually, looking at the diagram: I see A in two positions, B in two positions, and one more position. So we have: A, A, B, B, and one more (appears to be unlabeled or another letter).

For "not spinning A":
- Total sections = 5
- A appears 2 times
- Not A = 5 - 2 = 3 sections
- P(not A) = 3/5

Let me verify all answers:

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:
- P(A or B) = 3/4

Problem 5:
- P(not A) = 3/5