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There are n different size pairs of shoes in the box. One [#permalink]

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02 May 2008, 10:04

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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, 23:20, edited 1 time in total.

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) .

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.

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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mates, please visit my profile and leave comments http://gmatclub.com/forum/johnlewis1980-s-profile-feedback-is-more-than-welcome-80538.html

I'm not linked to GMAT questions anymore, so, if you need something, please PM me

I'm already focused on my application package

My experience in my second attempt http://gmatclub.com/forum/p544312#p544312 My experience in my third attempt http://gmatclub.com/forum/630-q-47-v-28-engineer-non-native-speaker-my-experience-78215.html#p588275

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)

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

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
_________________

mates, please visit my profile and leave comments http://gmatclub.com/forum/johnlewis1980-s-profile-feedback-is-more-than-welcome-80538.html

I'm not linked to GMAT questions anymore, so, if you need something, please PM me

I'm already focused on my application package

My experience in my second attempt http://gmatclub.com/forum/p544312#p544312 My experience in my third attempt http://gmatclub.com/forum/630-q-47-v-28-engineer-non-native-speaker-my-experience-78215.html#p588275

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alexsmith wrote:

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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.

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)

gmatclubot

Re: Math: Probability - n Shoes
[#permalink]
12 Jun 2008, 04:44