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There are n different size pairs of shoes in the box. One [#permalink]
 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.
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Re: Math: Probability - n Shoes [#permalink]
05 May 2008, 22:46
Well, no one left comments here 
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Re: Math: Probability - n Shoes [#permalink]
06 May 2008, 06:47
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]
07 May 2008, 06:42
lexis; Wats the OA ?
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Re: Math: Probability - n Shoes [#permalink]
07 May 2008, 22:45
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]
08 May 2008, 05:49
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]
08 May 2008, 09:58
all too often, I see a problem like this and just draw a blank.
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Re: Math: Probability - n Shoes [#permalink]
08 May 2008, 12:28
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]
16 May 2008, 03:19
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]
16 May 2008, 07:14
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  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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630-q-47-v-28-engineer-non-native-speaker-my-experience-78215.html#p588275
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Re: Math: Probability - n Shoes [#permalink]
17 May 2008, 00:48
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]
18 May 2008, 01:52
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]
18 May 2008, 10:24
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,
Cumic
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Re: Math: Probability - n Shoes [#permalink]
18 May 2008, 11:29
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]
18 May 2008, 11:34
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]
28 May 2008, 08:46
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]
29 May 2008, 06:11
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]
30 May 2008, 07:42
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]
12 Jun 2008, 04:44
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)
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Re: Math: Probability - n Shoes
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12 Jun 2008, 04:44
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