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

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Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her [#permalink] New post 19 Dec 2004, 00:51
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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?

Please post answer and step by step explanation on how to solve. Thanks in advance!
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 [#permalink] New post 19 Dec 2004, 06:50
9 coins.

Suppose karen drew all dimes and nickels in 9 draws, then she is will left with 12 coins. the remaining coins are 6 quarters and 6 other two (could be any combinations of dimes and nickles). now if karen draws any coin, now she has a chance of 50% getting quarters.
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 [#permalink] New post 20 Dec 2004, 09:06
Ma,

How can we make sure the first 9 coins karen draw will consist only dimes and nickles?
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 [#permalink] New post 22 Dec 2004, 22:46
jinino wrote:
Ma,

How can we make sure the first 9 coins karen draw will consist only dimes and nickles?



it is required to make the prob. of getting a quarter with 50% chance.
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Here is how I did it [#permalink] New post 23 Dec 2004, 08:31
I got the same answer as MA, but here is how I approached it. Since you want at least a 50% chance of getting a quarter, you want to be guaranteed at least a 6/12 chance. In order to be sure you would have to create the most difficult conditions to prevent you from getting a quarter, which would be if you drew non quarters as much as possible.

You start out with a 6/21 chance of drawing a quarter. You want a 6/12 chance, so you need to draw at least (21-12) or 9 coints to be sure. I hope this explanation helps.
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 [#permalink] New post 23 Dec 2004, 10:46
Hi
IMO the answer should be 3.

The prob of drawing dime+nickle=

(15/21)^ n < .5

the equation says: The prob of drawing a dime or nickle is 15/21

so how many time it should multiply it self that the prob of drawing reduces to .5

so for N=3 the LHS os the equation reduces to a number below .5

Hence at the fourth draw the prob of drawing a quater will be more than .5

So 3 draws are needed.

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Saurabh Malpani
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 [#permalink] New post 23 Dec 2004, 11:49
I also agree with 9 as toddmartin explained. Saurabh, what you said here:
Quote:
(15/21)^ n < .5
assumes that the coins are withdrawn with replacement and it would instead be the answer to the following question: Given the above coin breakdown/value and assuming that the coins are put back in her pocket after each draw, how many times will Karen have to draw a coin so that the % of not getting a quarter is less than 50%
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 [#permalink] New post 23 Dec 2004, 13:02
Hi Paul the question dosen't asy anything about the with or without replacement. So i don't think any assumption is valid. May be we have to come up with the assumption after seeing the answer choices. Moreover if you re-rad your last statement:

"a coin so that the % of not getting a quarter is less than 50%"

It also tells you that" % of getting a quater is more than 50%"

both the statement are just NEGATION of each other.

well I migh be missing something so please do let me know about my apporach.

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Saurabh Malpani
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 [#permalink] New post 23 Dec 2004, 13:26
saurabh, I agree that statements are just negation of each other but what I mean is that the way you set up the formula is assuming that it is done with replacement.
When you are saying (15/21)^n, the mere fact of having "n" as exponent means that replacement is allowed.
Consider this simpler example. Let's say you have 3 coins with different values, after 3 draws, how many outcomes are possible:
with replacement: 3^3 --> 3^n
without replacement: 3!
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 [#permalink] New post 23 Dec 2004, 13:45
I agree with todd that's how i solved , i got 9 too . Similar problem was addressed by paul in earlier probability problems .

key word is " atleast " and " 50 % " .
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 [#permalink] New post 22 May 2005, 20:19
Im sorry guys but the answer on my kaplan book says its 2 coins need to be withdrawn but i dont understand how they arrived at that answer since no explanation was given
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Re: PS tricky one [#permalink] New post 22 May 2005, 21:16
If she draws one coin, P=6/21<0.5
If she draws two coins, P=1-C(15,2)/C(21,2)=1-15*14/(21*20)=5/10=0.5

Therefore she only needs to draw two coins to make sure a 50% chance.
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Re: PS tricky one   [#permalink] 22 May 2005, 21:16
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