Two Basic Rules of Probability Quiz
The question sheet
Reveal any answer as you study-
According to the text, how many rules are considered when determining independence and mutual exclusivity?
- One
- Three
- Two
- Four
Reveal answer
Answer: Two
Source evidence
PDF page 206: In calculating probability, there are two rules to consider when you are determining if two events are independent or dependent and if they are mutually exclusive or not.
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What is the multiplication rule for events A and B on a sample space?
- P(A AND B) = P(A) − P(B)
- P(A AND B) = 0
- P(A AND B) = P(A) + P(B)
- P(A AND B) = P(B)P(A|B)
Reveal answer
Answer: P(A AND B) = P(B)P(A|B)
Source evidence
PDF page 206: If A and B are two events defined on a sample space, then P(A AND B) = P(B)P(A|B). This equation can be rewritten as P(A AND B) = P(B)P(A|B), the multiplication rule.
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If A and B are independent, then P(A AND B) equals which of the following?
- P(A) + P(B)
- 0
- P(A|B)
- P(A)P(B)
Reveal answer
Answer: P(A)P(B)
Source evidence
PDF page 206: If A and B are independent, then P(A|B) = P(A). In this special case, P(A AND B) = P(A|B)P(B) becomes P(A AND B) =
PDF page 206: P(A)P(B). A bag contains four green marbles, three red marbles, and two yellow marbles. Mark draws two marbles from the bag without replacement. The probability that he draws a yellow marble and then a green marble is
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Mark draws yellow then green without replacement. What is P(yellow and green)?
- 1/2
- 2/9
- 4/9
- 1/9
Reveal answer
Answer: 1/9
Source evidence
PDF page 206: P⎝yellow and green⎠ = P⎝yellow⎠ ⋅ P⎝green | yellow⎠ 2 4 = ⋅ 9 8 1 = 9
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After the yellow marble is drawn, how many marbles remain in the bag?
- Nine
- Ten
- Eight
- Seven
Reveal answer
Answer: Eight
Source evidence
PDF page 206: Notice that P⎝green | yellow⎠ = . After the yellow marble is drawn, there are four green marbles in the bag and eight 8 marbles in all.
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What is the addition rule for events A and B?
- P(A OR B) = P(A) − P(B)
- P(A OR B) = P(A) + P(B) − P(A AND B)
- P(A OR B) = 0
- P(A OR B) = P(A)P(B)
Reveal answer
Answer: P(A OR B) = P(A) + P(B) − P(A AND B)
Source evidence
PDF page 206: If A and B are defined on a sample space, then P(A OR B) = P(A) + P(B) − P(A AND B). Draw one card from a standard deck of playing cards. Let H = the card is a heart, and let J = the card is a jack. These events are not mutually exclusive because a card can be both a heart and a jack.
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If A and B are mutually exclusive, then P(A AND B) equals what?
- P(A)P(B)
- 0.5
- 0
- 1
Reveal answer
Answer: 0
Source evidence
PDF page 207: = + − 52 52 52 16 = 52 4 = 13 ≈ .3077 If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) − P(A AND B) becomes P(A OR B) = P(A) + P(B). Draw one card from a standard deck of playing cards. Let H = the card is a heart and S = the card is a spade. These events are mutually exclusive because a card cannot be a heart and a spade at the same time. The probability that the card is a heart or a spade is P(H or S) = P(H) + P(S) 13 13 = + 52 52 26 = 52 1 = 2 = .5
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For mutually exclusive events, P(A OR B) simplifies to which formula?
- P(A) + P(B)
- P(A)P(B)
- P(A) − P(B)
- P(A|B)
Reveal answer
Answer: P(A) + P(B)
Source evidence
PDF page 207: = + − 52 52 52 16 = 52 4 = 13 ≈ .3077 If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) − P(A AND B) becomes P(A OR B) = P(A) + P(B). Draw one card from a standard deck of playing cards. Let H = the card is a heart and S = the card is a spade. These events are mutually exclusive because a card cannot be a heart and a spade at the same time. The probability that the card is a heart or a spade is P(H or S) = P(H) + P(S) 13 13 = + 52 52 26 = 52 1 = 2 = .5
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Drawing one card, what is P(heart or spade)?
- 1/13
- 1/2
- 13/52
- 1/4
Reveal answer
Answer: 1/2
Source evidence
PDF page 207: = + − 52 52 52 16 = 52 4 = 13 ≈ .3077 If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) − P(A AND B) becomes P(A OR B) = P(A) + P(B). Draw one card from a standard deck of playing cards. Let H = the card is a heart and S = the card is a spade. These events are mutually exclusive because a card cannot be a heart and a spade at the same time. The probability that the card is a heart or a spade is P(H or S) = P(H) + P(S) 13 13 = + 52 52 26 = 52 1 = 2 = .5
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For Klaus, P(A)=.6 and P(B)=.35 with P(A AND B)=0. What is P(A OR B)?
- .05
- .21
- 1.0
- .95
Reveal answer
Answer: .95
Source evidence
PDF page 207: • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = .6 +
PDF page 207: .35 = .95. Note that the probability that he does not choose to go anywhere on vacation must be .05.
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For Klaus, what is the probability he does not go anywhere on vacation?
- .35
- .05
- .60
- .95
Reveal answer
Answer: .05
Source evidence
PDF page 207: .35 = .95. Note that the probability that he does not choose to go anywhere on vacation must be .05.
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Carlos scores 65% and P(B|A)=.90. What is P(A AND B)?
- .650
- .715
- .585
- .423
Reveal answer
Answer: .585
Source evidence
PDF page 207: a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = .90: P(B AND A) = P(B|A)
PDF page 207: P(A) = (.90)(.65) = .585.
High School Statistics
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