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There are n different size pairs of shoes in the box. One

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There are n different size pairs of shoes in the box. One [#permalink] New post 02 May 2008, 09:04
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There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<2n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?

Last edited by lexis on 06 May 2008, 22:20, edited 1 time in total.
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Re: Math: Probability - n Shoes [#permalink] New post 05 May 2008, 21:46
Well, no one left comments here :(
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Re: Math: Probability - n Shoes [#permalink] New post 06 May 2008, 05:47
lexis wrote:
There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?


I get nC1 * 2n-2C2k-2 / 2nC2k
= (2*k^2-k)/(2*n-1) .
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Re: Math: Probability - n Shoes [#permalink] New post 07 May 2008, 05:42
lexis; Wats the OA ?
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Re: Math: Probability - n Shoes [#permalink] New post 07 May 2008, 21:45
farend wrote:
lexis wrote:
There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?


I get nC1 * 2n-2C2k-2 / 2nC2k
= (2*k^2-k)/(2*n-1) .


u mean nC1=C(n,1)?

U should explain how did U get it. As I see, your answer is not correct.
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Re: Math: Probability - n Shoes [#permalink] New post 08 May 2008, 04:49
My contribution:

Prob=2k/(2n-1)

Reasoning:

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)

Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards
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Re: Math: Probability - n Shoes [#permalink] New post 08 May 2008, 08:58
all too often, I see a problem like this and just draw a blank.
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Re: Math: Probability - n Shoes [#permalink] New post 08 May 2008, 11:28
JohnLewis1980 wrote:
My contribution:

Prob=2k/(2n-1)

Reasoning:

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)


Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards


WELL, your answer is not correct.
-----------

Let me explain more about this statement:
For example, there are 5 pairs of shoes A,B,C,D,E
probability to take only one pair of shoes is correct and 1 incorrect pair of shoes (mean 2 different shoes)?

Mean:
Let A1, A2 is correct pair (pretend)
C1,D2 is incorrect pair or C2, E2 or... (pretend)

Hope it helps you solve the general puzzle.

@ RyanDe680: Your avatar is so interesting.
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Re: Math: Probability - n Shoes [#permalink] New post 16 May 2008, 02:19
If no one can take the correct answer, I will leave the OA in next week.
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Re: Math: Probability - n Shoes [#permalink] New post 16 May 2008, 06:14
lexis wrote:
JohnLewis1980 wrote:
My contribution:

Prob=2k/(2n-1)

Reasoning:

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)


Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards


WELL, your answer is not correct.
-----------

Let me explain more about this statement:
For example, there are 5 pairs of shoes A,B,C,D,E
probability to take only one pair of shoes is correct and 1 incorrect pair of shoes (mean 2 different shoes)?

Mean:
Let A1, A2 is correct pair (pretend)
C1,D2 is incorrect pair or C2, E2 or... (pretend)

Hope it helps you solve the general puzzle.

@ RyanDe680: Your avatar is so interesting.


I'm afraid so :oops:

but why? :?:

Don't we agree in the probability to take one complete pair of shoes? i.e. to take the right shoe once you've already take one out?

For me: 1/(2n-1)

Explanation: you take one shoe out, therefore, just 2n-1 shoes remain in the box. The probability to take the right one off is 1/(2n-1), doesn't it?

:| I need to improve my statistic skill.

Thanks for the explanation
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Re: Math: Probability - n Shoes [#permalink] New post 16 May 2008, 23:48
You should try ***. It is a really good and comprehensive website for preparing yourself for a GMAT exam. Do try it and see for yourself. Try a 24 hour free trial which you get when you sign up.
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Re: Math: Probability - n Shoes [#permalink] New post 18 May 2008, 00:52
I agree with alex1234 that *** is very helpful for preparing yourself for the math section of GMAT. I love the practice tests and the way they explain each sum with a video. I found it very useful. It is worth trying.
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Re: Math: Probability - n Shoes [#permalink] New post 18 May 2008, 09:24
JohnLewis1980 wrote:
My contribution:

Prob=2k/(2n-1)

Reasoning:

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)

Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards

===============================================

Shoudn't this be solved like this ->

Probability of taking out 2k shoes from total of 2n shoes = 2k/2n
Probability of taking out second compatible shoe = 1/(2n-1)

Hence probability of both events happening together - 2k/2n(2n-1)

Can someone confirm the OA please?

Regards,
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Re: Math: Probability - n Shoes [#permalink] New post 18 May 2008, 10:29
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Interesting problem. +1

My attempt: P=\frac{C^n_k*C^{k}_{1}*(C^2_1)^{k-1}*C^{n-k}_{k-1}}{C^{2n}_{2k}}

when 2k>n+1 P=0
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Re: Math: Probability - n Shoes [#permalink] New post 18 May 2008, 10:34
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alex1234 wrote:
You should try ***. It is a really good and comprehensive website for preparing yourself for a GMAT exam. Do try it and see for yourself. Try a 24 hour free trial which you get when you sign up.

alexsmith wrote:
I agree with alex1234 that *** is very helpful for preparing yourself for the math section of GMAT. I love the practice tests and the way they explain each sum with a video. I found it very useful. It is worth trying.


alex, please, doesn't think that all here are fools. Be frank and post advertisement in appropriate threads.
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Re: Math: Probability - n Shoes [#permalink] New post 28 May 2008, 07:46
I'm just giving my views, I'm not advertising.
Is it so that if someone gives their views about a site which I used for my preparation of GMAT is called advertisement??
I benefited from this site and I want everyone else to also.
I'll keep giving my views about the source what I used for my GMAT entrance and will ask all to just try it once.
I'm sure you will get the results that you are dreaming for.
Thats all I want to say.
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Re: Math: Probability - n Shoes [#permalink] New post 29 May 2008, 05:11
Let C(i, k) = iCk = i!/(k!(k-i)!)

Total number of ways to grab shoes:
C(2n, 2k)

Total number of ways to grab only 1 pair:
C(n, 1) * C(n-1, 2k-2) * 2^(k-2)

Answer:
C(n, 1) * C(n-1, 2k-2) * 2^(2k-2) / C(2n, 2k)
(unless 2k > n+1, then prob = 0)

This isn't an official question, right? The variables are awkwardly defined.

Either way, can I get the A, B, C, D, E answers/distractors, I would like to give this problem to a friend.
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Re: Math: Probability - n Shoes [#permalink] New post 30 May 2008, 06:42
Hey friends try *** its really a good site for all people who have problem in GMAT Math and Verbal..
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Re: Math: Probability - n Shoes [#permalink] New post 12 Jun 2008, 03:44
JingChan wrote:
Let C(i, k) = iCk = i!/(k!(k-i)!)

Total number of ways to grab shoes:
C(2n, 2k)

Total number of ways to grab only 1 pair:
C(n, 1) * C(n-1, 2k-2) * 2^(k-2)

Answer:
C(n, 1) * C(n-1, 2k-2) * 2^(2k-2) / C(2n, 2k)
(unless 2k > n+1, then prob = 0)

This isn't an official question, right? The variables are awkwardly defined.

Either way, can I get the A, B, C, D, E answers/distractors, I would like to give this problem to a friend.


Congratulation! You're correct. Your math skill is very good!
The tricky is to require ONLY ONE pair of shoes be same size.
To solve it, we separate two sequences:
1st: There is NO pair of shoes
2nd: There is ONLY ONE pair of shoes be same size
==> Combine: ONLY ONE pair of shoes be same size + NO pair of shoes in rest shoes (2n-2)
Re: Math: Probability - n Shoes   [#permalink] 12 Jun 2008, 03:44
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