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Portia rates all her ?rst dates as either “duds” or “dudes.
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08 Dec 2017, 21:00
08 Dec 2017, 21:00
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Question Stats:
48% (01:21) correct
51% (01:19) wrong based on 49 sessions
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Portia rates all her first dates as either “duds” or “dudes.” Her date on Wednesday night was a dud. On the next night, she went on a date with someone else who was also a dud. If the probability of her getting two duds in a row was 4/25, what is the probability that her next date will be a dude ?
A. \(\frac{1}{5}\)
B. \(\frac{9}{25}\)
C. \(\frac{2}{5}\)
D. \(\frac{3}{5}\)
E. \(\frac{21}{25}\)
A. \(\frac{1}{5}\)
B. \(\frac{9}{25}\)
C. \(\frac{2}{5}\)
D. \(\frac{3}{5}\)
E. \(\frac{21}{25}\)
Drill 2
Question: 10
Page: 533
Question: 10
Page: 533
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Re: Portia rates all her ?rst dates as either “duds” or “dudes.
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09 Dec 2017, 14:12
09 Dec 2017, 14:12
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sandy wrote:
Portia rates all her first dates as either “duds” or “dudes.” Her date on Wednesday night was a dud. On the next night, she went on a date with someone else who was also a dud. If the probability of her getting two duds in a row was 4/25, what is the probability that her next date will be a dude ?
A. \(\frac{1}{5}\)
B. \(\frac{9}{25}\)
C. \(\frac{2}{5}\)
D. \(\frac{3}{5}\)
E. \(\frac{21}{25}\)
A. \(\frac{1}{5}\)
B. \(\frac{9}{25}\)
C. \(\frac{2}{5}\)
D. \(\frac{3}{5}\)
E. \(\frac{21}{25}\)
Drill 2
Question: 10
Page: 533
Question: 10
Page: 533
Although not stated, the assumption here is that, on ANY GIVEN DATE, the probability of getting a dud is always the same.
Let y = P(getting a dud on a given date)
We're told that P(2 duds in a row) = 4/25
In other words: P(dud on 1st date AND dud on 2nd date) = 4/25
Rewrite as: P(dud on 1st date) X P(dud on 2nd date) = 4/25
So, we get: y X y = 4/25
In other words, y² = 4/25
Solve to get: y = 2/5
So, P(getting a dud on a given date) = 2/5
So, P(NEXT date is a dude) = 1 - 2/5 = 3/5
Answer:
Show: ::
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RELATED VIDEO
Re: Portia rates all her ?rst dates as either “duds” or “dudes.
[#permalink]
17 Jun 2020, 05:35
17 Jun 2020, 05:35
GreenlightTestPrep wrote:
sandy wrote:
Portia rates all her first dates as either “duds” or “dudes.” Her date on Wednesday night was a dud. On the next night, she went on a date with someone else who was also a dud. If the probability of her getting two duds in a row was 4/25, what is the probability that her next date will be a dude ?
A. \(\frac{1}{5}\)
B. \(\frac{9}{25}\)
C. \(\frac{2}{5}\)
D. \(\frac{3}{5}\)
E. \(\frac{21}{25}\)
A. \(\frac{1}{5}\)
B. \(\frac{9}{25}\)
C. \(\frac{2}{5}\)
D. \(\frac{3}{5}\)
E. \(\frac{21}{25}\)
Drill 2
Question: 10
Page: 533
Question: 10
Page: 533
Although not stated, the assumption here is that, on ANY GIVEN DATE, the probability of getting a dud is always the same.
Let y = P(getting a dud on a given date)
We're told that P(2 duds in a row) = 4/25
In other words: P(dud on 1st date AND dud on 2nd date) = 4/25
Rewrite as: P(dud on 1st date) X P(dud on 2nd date) = 4/25
So, we get: y X y = 4/25
In other words, y² = 4/25
Solve to get: y = 2/5
So, P(getting a dud on a given date) = 2/5
So, P(NEXT date is a dude) = 1 - 2/5 = 3/5
Answer:
Show: ::
D
RELATED VIDEO
Is this a real GRE question ? It is totally unclear question ! and the question isn't in the book.
