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

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VP
Joined: 29 Dec 2005
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Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her [#permalink]  11 May 2006, 20:21
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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?

No OA. lets discuss and finalize the OA.
Director
Joined: 08 Jun 2004
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Location: Europe
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Prof, probability is not my strong part of the math, but anyway I will try my best shot.

She must withdraw 9 coins before she has at least 50 percent chance of withdrawing at least one quarter.
Solution: she has 6 quarters. So to have at least 50% chance she must have left with 12 coins (6 quarters and 6 any of the other).
Total 21 coins minus 9 coins left her with needed 12 coins.
Director
Joined: 13 Nov 2003
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Hallo Prof and M8,
Agree with the approach presented by M8 i think there is a worse scenario. She withdraws 5 quarters, then she will heed to withdraw 14 coins from dimes and nickels to make sure that the chance is at least 50% so in total 19 coins can be withdrawn.
Director
Joined: 08 Jun 2004
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BG wrote:
Hallo Prof and M8,
Agree with the approach presented by M8 i think there is a worse scenario. She withdraws 5 quarters, then she will heed to withdraw 14 coins from dimes and nickels to make sure that the chance is at least 50% so in total 19 coins can be withdrawn.

BG we both forgot about the possibility that she could withdraw all quarters before she left with 12 or with 2 coins (dime and nickel). What do you think?
Director
Joined: 13 Nov 2003
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Agree M8
but in this case she will have nothing to draw and the Q has no sense. Think that this type of Q asks about the worse possible scenario i. e. when the maximum numbers of attempts should be made in order to have the chance of 50/50.
Intern
Joined: 10 Jan 2006
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I think it helps to invert the problem. So whats the probability that you don't have any quarters after N coin withdrawals.

That would be F(N)=15/21*14/20.. (N terms).

The probability that you have at least one quarter is (1-F(N)).

F(1) = 15/21
F(2) = 0.5

Since 1-F(2) = 0.5, the answer is 2.
Director
Joined: 16 Aug 2005
Posts: 950
Location: France
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Kudos [?]: 8 [0], given: 0

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?

Worst chance is when she has to draw 16 coins to get a quarter. Best chance is when she has to draw 1 coin.

So Total # of outcomes = 1, 2, 3, ... 16 = 16
# of desired outcomes = 50% of total # of outcomes
which is 8.

So, she has to draw 8 coins to give her a 50% chance that she draws 1 quarter...

Now, im going to read you folks' response!!
Director
Joined: 08 Jun 2004
Posts: 502
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Hi gmatmba,

Let's wait for Prof's opinion now.
Director
Joined: 16 Aug 2005
Posts: 950
Location: France
Followers: 1

Kudos [?]: 8 [0], given: 0

BTW Prof and M8, when are you guys taking the Gmat...
mine may 25 and im doing real bad in math..

I was going through my error logs from last month for DS and an interesting thing I noticed. All the DS problems that I got wrong, about 90% of them had E as answer where I had guessed C. Result of premature guessing...without reading the questions carefully...and you guys have been very helpful so ... THANKS!
Director
Joined: 08 Jun 2004
Posts: 502
Location: Europe
Followers: 1

Kudos [?]: 10 [0], given: 0

gmatmba wrote:
BTW Prof and M8, when are you guys taking the Gmat...
mine may 25 and im doing real bad in math..

I was going through my error logs from last month for DS and an interesting thing I noticed. All the DS problems that I got wrong, about 90% of them had E as answer where I had guessed C. Result of premature guessing...without reading the questions carefully...and you guys have been very helpful so ... THANKS!

Buddy you are lucky one, you have plenty of time till May 25.
My G-Day in on May 16.
But I'm the apprentice to Professor, he is real genius in Math.
My scaled score in Math is stable - 48. It's proved by real GMAT I passed last October and by a couple of GMATPreps. Trying to improve it in the course of about 2 months but nothing can do - neither more nor less than 48 - stable.
Hope that next Thursday it will be no less than 48 or even more.
Senior Manager
Joined: 09 Aug 2005
Posts: 286
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it is 2

[6c2 + (6c1 x 15c1)]/21C2

but the more important question is should this be a trial and error
p(1), p(2) ........ approach ?

Can anyone suggest a more direct method ?
VP
Joined: 29 Dec 2005
Posts: 1351
Followers: 6

Kudos [?]: 28 [0], given: 0

M8 wrote:
gmatmba wrote:
BTW Prof and M8, when are you guys taking the Gmat...
mine may 25 and im doing real bad in math..

I was going through my error logs from last month for DS and an interesting thing I noticed. All the DS problems that I got wrong, about 90% of them had E as answer where I had guessed C. Result of premature guessing...without reading the questions carefully...and you guys have been very helpful so ... THANKS!

Buddy you are lucky one, you have plenty of time till May 25.
My G-Day in on May 16.
But I'm the apprentice to Professor, he is real genius in Math.
My scaled score in Math is stable - 48. It's proved by real GMAT I passed last October and by a couple of GMATPreps. Trying to improve it in the course of about 2 months but nothing can do - neither more nor less than 48 - stable.
Hope that next Thursday it will be no less than 48 or even more.

thanx guys for your nice words.... however i am nothing in comparision with the real genius people like honghu, laxi. if we really want to learn something in math, we should look on thier posts and approaches...

they are great.....

Last edited by Professor on 13 May 2006, 16:09, edited 1 time in total.
VP
Joined: 29 Dec 2005
Posts: 1351
Followers: 6

Kudos [?]: 28 [0], given: 0

saha wrote:
I think it helps to invert the problem. So whats the probability that you don't have any quarters after N coin withdrawals.

That would be F(N)=15/21*14/20.. (N terms).
The probability that you have at least one quarter is (1-F(N)).

F(1) = 15/21
F(2) = (15/21) (14/20) = 0.5
Since 1-F(2) = 0.5, the answer is 2.

old_dream_1976 wrote:
it is 2
[6c2 + (6c1 x 15c1)]/21C2
but the more important question is should this be a trial and error
p(1), p(2) ........ approach ?
Can anyone suggest a more direct method ?

i guess the above approaches both work. the answer should be 2 as well. remember the question says "at least 1 qt".

prob of getting "at least 1 qt" in drawing 1 qt = 6/21 = 2/7 i.e less than 50%. so not correct.

prob of getting "at least 1 qt" in 2 coins = [(prob of getting getting 1 qt and 1 dime or nickel) + (prob of getting getting 2 qts)]/(total ways of geeting 2 coins)
prob = [(6 x 15) + (6c2)]/21c2 = (90+15)/210 = 50%

the first one is also coreect way...

thanx guys...
Director
Joined: 16 Aug 2005
Posts: 950
Location: France
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Kudos [?]: 8 [0], given: 0

Prof - What is the source of this question if you have one?

I checked and my understanding of the problem was surely quite different (and incorrect) it seems. Thanks for the explanation ... all of you.
VP
Joined: 29 Dec 2005
Posts: 1351
Followers: 6

Kudos [?]: 28 [0], given: 0

gmatmba wrote:
Prof - What is the source of this question if you have one?

I checked and my understanding of the problem was surely quite different (and incorrect) it seems. Thanks for the explanation ... all of you.

i am not sure about the source. one of my friend supplied this question to me.

i thought this is a difficult question thefore i brought it into for discussion.
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