Author 
Message 
Intern
Joined: 10 Dec 2004
Posts: 2

Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her [#permalink]
Show Tags
19 Dec 2004, 01:51
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please repost it in the respective forum.
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!



VP
Joined: 25 Nov 2004
Posts: 1483

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.



Manager
Joined: 31 Aug 2004
Posts: 162
Location: Vancouver, BC, Canada

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



VP
Joined: 25 Nov 2004
Posts: 1483

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.



Manager
Joined: 29 Jul 2004
Posts: 61

Here is how I did it [#permalink]
Show Tags
23 Dec 2004, 09:31
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 (2112) or 9 coints to be sure. I hope this explanation helps.



Senior Manager
Joined: 19 Sep 2004
Posts: 367

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.
Thanks
Saurabh Malpani



GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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%
_________________
Best Regards,
Paul



Senior Manager
Joined: 19 Sep 2004
Posts: 367

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 rerad 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.
Thanks
Saurabh Malpani



GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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!
_________________
Best Regards,
Paul



Manager
Joined: 18 Jun 2004
Posts: 103
Location: san jose , CA

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 % " .
_________________
 Hero never chooses Destiny
Destiny chooses Him ......



Intern
Joined: 10 Dec 2004
Posts: 2

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



SVP
Joined: 03 Jan 2005
Posts: 2233

Re: PS tricky one [#permalink]
Show Tags
22 May 2005, 22:16
If she draws one coin, P=6/21<0.5
If she draws two coins, P=1C(15,2)/C(21,2)=115*14/(21*20)=5/10=0.5
Therefore she only needs to draw two coins to make sure a 50% chance.
_________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.










