Two Basic Rules of Probability Quiz
The question sheet
Reveal any answer as you study-
Which formula is the multiplication rule for P(A ∩ B)?
- P(A) - P(B)
- P(A) + P(B)
- P(A)/P(B)
- P(B)P(A|B)
Reveal answer
Answer: P(B)P(A|B)
Source evidence
PDF page 154: If A and B are two events defined on a sample space, then: P(A ∩ B) = P(B)P(A|B) . We can think of the intersection symbol as substituting for the word "and".
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Why do we subtract the intersection in the addition rule?
- To satisfy two conditions
- To make it larger
- To ensure independence
- To keep from double counting
Reveal answer
Answer: To keep from double counting
Source evidence
PDF page 155: If A and B are defined on a sample space, then: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) . We can think of the union symbol substituting for the word "or". The reason we subtract the intersection of A and B is to keep from double counting elements that are in both A and B. If A and B are mutually exclusive, then P(A ∩ B) = 0 . Then P(A ∪ B) = P(A) + P(B) - P(A ∩ B) becomes P(A ∪ B) = P(A) + P(B) .
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If A and B are mutually exclusive, then P(A ∩ B) equals:
- 0.5
- 0
- P(A)P(B)
- 1
Reveal answer
Answer: 0
Source evidence
PDF page 155: If A and B are defined on a sample space, then: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) . We can think of the union symbol substituting for the word "or". The reason we subtract the intersection of A and B is to keep from double counting elements that are in both A and B. If A and B are mutually exclusive, then P(A ∩ B) = 0 . Then P(A ∪ B) = P(A) + P(B) - P(A ∩ B) becomes P(A ∪ B) = P(A) + P(B) .
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When A and B are independent, P(A|B) equals:
- P(A ∩ B)
- P(A)
- 0
- P(B)
Reveal answer
Answer: P(A)
Source evidence
PDF page 155: If A and B are independent, then P(A|B) = P(A) . Then P(A ∩ B) = P(A|B)P(B) becomes P(A ∩ B) = P(A)(B)
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When A and B are mutually exclusive, P(A ∪ B) simplifies to:
- P(A)P(B)
- P(B)P(A|B)
- P(A) - P(B)
- P(A) + P(B)
Reveal answer
Answer: P(A) + P(B)
Source evidence
PDF page 155: If A and B are defined on a sample space, then: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) . We can think of the union symbol substituting for the word "or". The reason we subtract the intersection of A and B is to keep from double counting elements that are in both A and B. If A and B are mutually exclusive, then P(A ∩ B) = 0 . Then P(A ∪ B) = P(A) + P(B) - P(A ∩ B) becomes P(A ∪ B) = P(A) + P(B) .
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For Klaus's vacation choices, what is P(A ∪ B)?
- 0.95
- 0.35
- 0.60
- 0.05
Reveal answer
Answer: 0.95
Source evidence
PDF page 155: P(A ∪ B) = P(A) + P(B) = 0.6 + 0.35 = 0.95 . Note that the probability that he does not choose to go
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Why is P(A ∩ B) = 0 in Klaus's vacation example?
- The trips overlap
- He dislikes Alaska
- Both are equally likely
- He can only afford one vacation
Reveal answer
Answer: He can only afford one vacation
Source evidence
PDF page 155: • P(A ∩ B) = 0 because Klaus can only afford to take one vacation
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What is the probability Klaus does not vacation anywhere?
- 0.05
- 0.35
- 0.95
- 0.60
Reveal answer
Answer: 0.05
Source evidence
PDF page 155: anywhere on vacation must be 0.05.
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What is the probability that Carlos makes both goals?
- 0.585
- 0.715
- 0.65
- 0.423
Reveal answer
Answer: 0.585
Source evidence
PDF page 155: P(A) = (0.90)(0.65) = 0.585
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What is the probability Carlos makes either goal?
- 0.423
- 0.585
- 0.90
- 0.715
Reveal answer
Answer: 0.715
Source evidence
PDF page 156: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.65 + 0.65 - 0.585 = 0.715
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Are events A and B independent for Carlos's goals?
- Yes, always
- No, because 0.423 ≠ 0.585
- Cannot be determined
- Yes, since P(A)=P(B)
Reveal answer
Answer: No, because 0.423 ≠ 0.585
Source evidence
PDF page 156: c. No, they are not, because P(B ∩ A) = 0.585.
PDF page 156: 0.423 ≠ 0.585 = P(B ∩ A)
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Are A and B mutually exclusive for Carlos's goals?
- Yes, since P(A ∩ B)=0
- Cannot be determined
- No, because P(A ∩ B) = 0.585
- Yes
Reveal answer
Answer: No, because P(A ∩ B) = 0.585
Source evidence
PDF page 156: d. No, they are not because P(A ∩ B) = 0.585.
Introductory Business Statistics
Introductory Business Statistics by OpenStax, used under CC BY 4.0. Changes made by Stratacademy.
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