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

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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 pocket. [#permalink]

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11 May 2006, 21:21
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95% (hard)

Question Stats:

37% (02:26) correct 63% (02:00) wrong based on 254 sessions

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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?

(A) 1
(B) 2
(C) 5
(D) 6
(E) 7
[Reveal] Spoiler: OA

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

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12 May 2006, 00:22
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.

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Director
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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12 May 2006, 01:50
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to withdraw 14 coins from dimes and nickels to make sure that the chance is at least 50% so in total 19 coins can be withdrawn.

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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12 May 2006, 02:08
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?

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Director
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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12 May 2006, 02:19
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.

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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12 May 2006, 20:29
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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.

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Director
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 08:06
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!!

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Director
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 08:33
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!

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 08:50
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.

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 12:13
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 ?

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VP
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 13:16
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, 17:09, edited 1 time in total.

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VP
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 13:58
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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...

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Director
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 14:49
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.

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VP
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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13 May 2006, 16:52
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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Re: PS - 700 level - quarter [#permalink]

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07 Aug 2011, 14:20
Hey this question was posted already.

Basically there are 21 coins overall, now the odds of picking a quarter are 6/21= .286. Then it is 6/20 = .3.
When you treat each event individually you get those percentages, Adding these two probabilities together are ~.59

So the answer is B or 2.

Here is the link of the question previously posted.
probability-confusing-72435.html
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Re: PS - 700 level - quarter [#permalink]

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07 Aug 2011, 14:23
bschool83 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

i will go with choice 2nd here..
suppose that she does not with draw quarter in 1st.. then probability of dime or nickel = 15/21 = 5/7 ~ 0.72
and probability of withdrawing quarter in 1st is = 0.28

if the 1st coin is dime or nickel then the probability of withdrawing quarter 2nd time is 6/20 = 3/10 = 0.3

together - 0.28 + 0.3 = 0.58 which is greater than 0.5 or 50% chances of withdrawing quarter..
so she should withdraw quarter in 1 of the 2 chances..

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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06 Aug 2015, 08:54
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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(drawing at least one quarter) = 1 - P(drawing no quarter)

Let's assume she draws $$n$$ coins.

Total no. of ways to draw n coins without selecting any quarter $$= (10+5)Cn = 15Cn$$ (selecting n coins out of 15 coins consisting of 5 dimes and 10 nickels)

Total no. of cases to draw n coins $$= 21Cn$$

Now, we need minimum value of n for which $$15Cn/21Cn <= 1/2$$

at n = 1, value = 5/7
at n = 2, value = 1/2

as n increases value keeps decreasing

She needs to draw minimum 2 coins.

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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04 Nov 2015, 23:52
I did this a bit differently,

Finished in ~2 mins

Case 1 - Prob 1 quarter from 1 coin drawn 6C1 / 21C1 = 6/21 < 1/2 so X
Case 2 - P(1 quarter from 2 draws) = 6C1*20C1 / 21C2 = 126/196 This is > 1/2 so We need 2 coins to ensure this.

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. [#permalink]

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19 Oct 2016, 07:06
I used an easier method.

1 withdrawal -> we have 6/21 of chance to get a quarter. It falls short of the 1/2 target.
2 withdrawals ->we have (6/21*15/20*2P) + (6/21*6/21)= 3/7+4/49=21/49+4/49=25/49> 1/2

So B (2 withdrawals) is the correct answer.
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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels [#permalink]

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29 Jan 2017, 23:27
Saw this question here only, couldn't get a very convincing explanation though. My first take was; for 100℅ probability, one needs to withdraw 10+5+1. So, 50℅ should be half of that i.e. 8. Please can anyone tell me where am I going wrong. Thanks

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Re: Karen has exactly 6 quarters, 5 dimes, and 10 nickels   [#permalink] 29 Jan 2017, 23:27

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