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The probability that Maria will eat breakfast on any given d
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05 Aug 2018, 14:45
05 Aug 2018, 14:45
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Question Stats:
59% (00:45) correct
40% (00:34) wrong based on 256 sessions
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The probability that Maria will eat breakfast on any given day is 0.5. The probability that Maria will wear a sweater on any given day is 0.3. The two probabilities are independent of each other.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability that Maria eats breakfast or wears a sweater |
0.8 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: The probability that Maria will eat breakfast on any given d
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08 Aug 2018, 14:34
08 Aug 2018, 14:34
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sandy wrote:
The probability that Maria will eat breakfast on any given day is 0.5. The probability that Maria will wear a sweater on any given day is 0.3. The two probabilities are independent of each other.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability that Maria eats breakfast or wears a sweater |
0.8 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Important concept #1: P(event A OR event B) = P(event A) + P(event B) - P(event A AND event B)
Important concept #2: If two events are independent, then P(event A AND event B) = P(event A) x P(event B)
From concept #2, P(Maria eats breakfast AND wears a sweater) = P(Maria eats breakfast) x P(Maria wears a sweater)
= 0.5 x 0.3
= 0.15
From concept #1, P(Maria eats breakfast OR wears a sweater) = P(Maria eats breakfast) + P(Maria wears a sweater) - P(Maria eats breakfast AND wears a sweater)
= 0.5 + 0.3 - P(Maria eats breakfast AND wears a sweater)
= 0.5 + 0.3 - 0.15
= 0.65
So, we get:
Quantity A: 0.65
Quantity B: 0.8
Answer: B
Cheers,
Brent
General Discussion
Re: The probability that Maria will eat breakfast on any given d
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21 Aug 2018, 18:30
21 Aug 2018, 18:30
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Expert Reply
Explanation
The problem indicates that the events occur independently of each other. Therefore, in calculating Quantity A, the first step is to calculate the “or” situation, but don’t stop there. Adding 0.5 + 0.3 = 0.8 double counts the occurrences when both events occur. To compensate, subtract out the probability of both events occurring in order to get rid of the “double counted” occurrences.
Notice that this is a Quantitative Comparison. Because the 0.8 figure includes at least one “both” occurrence, the real figure for Quantity A must be less than 0.8. Therefore, Quantity B must be greater.
To do the actual math, find the probability of both events occurring (breakfast and sweater): (0.5) (0.3) = 0.15. Subtract the “and” occurrences from the total “or” probability: 0.8 – 0.15 = 0.65.
Thus, Quantity B is greater.
The problem indicates that the events occur independently of each other. Therefore, in calculating Quantity A, the first step is to calculate the “or” situation, but don’t stop there. Adding 0.5 + 0.3 = 0.8 double counts the occurrences when both events occur. To compensate, subtract out the probability of both events occurring in order to get rid of the “double counted” occurrences.
Notice that this is a Quantitative Comparison. Because the 0.8 figure includes at least one “both” occurrence, the real figure for Quantity A must be less than 0.8. Therefore, Quantity B must be greater.
To do the actual math, find the probability of both events occurring (breakfast and sweater): (0.5) (0.3) = 0.15. Subtract the “and” occurrences from the total “or” probability: 0.8 – 0.15 = 0.65.
Thus, Quantity B is greater.
