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The probability that Maria will eat breakfast on any given d

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The probability that Maria will eat breakfast on any given d [#permalink] New post 05 Aug 2018, 14:45
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Question Stats:

60% (00:40) correct 39% (00:27) wrong based on 131 sessions
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.

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.
[Reveal] Spoiler: OA

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Re: The probability that Maria will eat breakfast on any given d [#permalink] New post 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.

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
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Re: The probability that Maria will eat breakfast on any given d [#permalink] New post 21 Aug 2018, 18:30
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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.
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Re: The probability that Maria will eat breakfast on any given d [#permalink] New post 03 Jan 2020, 17:12
i'm confused here where is the double counts happen? the both events are independent.
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Re: The probability that Maria will eat breakfast on any given d [#permalink] New post 16 Jan 2020, 08:29
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GR3A wrote:
i'm confused here where is the double counts happen? the both events are independent.


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#greprepclub The probability that Maria will eat breakfast on any given d.jpg [ 35.91 KiB | Viewed 2146 times ]


Plz see the diag above.

The intersection i.e probability of Event A & Event B is counted twice.

So in OR probability of event A or event B :- P(A) + P(B) - P(A & B)
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Re: The probability that Maria will eat breakfast on any given d [#permalink] New post 17 Jan 2020, 10:11
Expert's post
GR3A wrote:
i'm confused here where is the double counts happen? the both events are independent.


I believe you may be confusing independence with mutual exclusivity.
If two events are mutually exclusive then they cannot both occur at the same time.

Cheers,
Brent
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Re: The probability that Maria will eat breakfast on any given d [#permalink] New post 20 Jun 2020, 15:23
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We can solve this problem using M1 (probability that Maria eats breakfast) = 0.3, M2 (probability that Maria wears a sweater) = 0.5

The probability that Maria eats breakfast or wears a sweater will be: M1 + M2 - M1*M2 = 0.8 - 0.15 = 0.65
Re: The probability that Maria will eat breakfast on any given d   [#permalink] 20 Jun 2020, 15:23
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