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

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Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket.  [#permalink]

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30 Jan 2017, 02:22
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.
Senior Manager
Joined: 03 Apr 2013
Posts: 283
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket.  [#permalink]

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12 Jul 2017, 02:29
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

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

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