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

Question

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

(A) 1 
(B) 2 
(C) 5 
(D) 6 
(E) 7

ravindra88 · Accepted Answer

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.

Answer

saha · Answer

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.

M8 · Answer

Prof, probability is not my strong part of the math, but anyway I will try my best shot. :wink:

My answer is 9 coins. :roll:

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.

BG · Answer

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to withdraw 14 coins from dimes and nickels to make sure that the chance is at least 50% so in total 19 coins can be withdrawn.

M8 · Answer

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?

BG · Answer

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.

gmatmba · Answer

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!

M8 · Answer

Buddy you are lucky one, you have plenty of time till May 25. My G-Day in on May 16. :wink:

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.

old_dream_1976 · Answer

it is 2 

 /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 ?

Professor · Answer

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.....

Professor · Answer

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 =  /(total ways of geeting 2 coins)
prob =  /21c2 = (90+15)/210 = 50%

the first one is also coreect way...

thanx guys...

gmatmba · Answer

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.

Professor · Answer

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.

nphelan · Answer

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

albo94 · Answer

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.

vitaliyGMAT · Answer

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.

road740 · Answer

-->   B  , 2 coins.

There are 6 quarters and 15 "non" quarters.

If she withdraws one coin, the prob. of it being a quarter is 6/15 = 2/5 = 0.4. This is not enough. 
The prob. of her withdrawing two quarters is (6/15)*(5/14) = 1/7 = 0.14...

0.4 + 0.14... > 0.5

--> She needs to withdraw at least 2 coins os that her probability of withdrawing one quarter is at least 50%

JuniqueLid · Answer

This is a "funny" question because when I first saw it, I thought the question was asking what's the minimum draws needed to get a 50% chance of getting the equivalent of a quarter's money, which could be 1 quarter, 3 dimes or 5 nickles or some combinations. That threw me off...

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

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

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