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Portia rates all her ?rst dates as either “duds” or “dudes. [#permalink]
08 Dec 2017, 21:00
Expert's post
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Question Stats:
100% (00:25) correct
0% (00:00) wrong based on 1 sessions
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 , 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. \(\frac{21}{25}\)
Drill 2 Question: 10 Page: 533
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Re: Portia rates all her ?rst dates as either “duds” or “dudes. [#permalink]
09 Dec 2017, 14:12
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Expert's post
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 , 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. \(\frac{21}{25}\)
Drill 2 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
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100% (00:25) correct
