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Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket.

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Senior Manager
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Joined: 13 Oct 2016
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket.  [#permalink]

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New post 30 Jan 2017, 01:22
Professor wrote:
Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?

(A) 1
(B) 2
(C) 5
(D) 6
(E) 7


\(P(at-least-one-quater) = 1 - P(No-quaters) = \frac{1}{2}\)

\(1 - \frac{15Cn}{21Cn} = \frac{1}{2}\)

\(\frac{15Cn}{21Cn} = \frac{1}{2}\)

\(\frac{2*15!}{n!(15-n)!} = \frac{21!}{n!(21-n)!}\)

\(2*15!*(21-n)! = 21!*(15-n)!\)

\(2(21-n)(20-n)(19-n)(18-n)(17-n)(16-n) = 21*20*19*18*17*16\)

\((21-n)(20-n)(19-n)(18-n)(17-n)(16-n) = 3*7*2*5*19*2*3^2*17*2^4\)

We need to construct 6 elements of AP starting from max 19.

\((21-n)(20-n)(19-n)(18-n)(17-n)(16-n) = 19*18*17*16*15*14\)

Our n = 2.
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Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket.  [#permalink]

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New post 12 Jul 2017, 01:29
Bunuel wrote:
WilDThiNg wrote:
Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?

1
2
5
6
7


Please follow the rules when posting a question: https://gmatclub.com/forum/rules-for-po ... 33935.html


Bunuel, shouldn't the complement event be The probability of selecting no quarter is LESS THAN 50%???

In that case, the answer will be C, not B.
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket.  [#permalink]

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New post 17 Aug 2018, 01:02
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. &nbs [#permalink] 17 Aug 2018, 01:02

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