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There are 99 workers in a mine. Each worker has a identity

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Director
Joined: 03 Sep 2006
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There are 99 workers in a mine. Each worker has a identity [#permalink]

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25 Sep 2007, 18:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are 99 workers in a mine. Each worker has a identity number from 1 to 99 and no two workers share the same identity number. During a disaster drill exercise all the workers leave the mine one-by-one in a completely random manner. What is the probability that the first four mine workers off the mine will leave in order of increasing identity numbers ( e.g., #2, then # 10, then #97, then #98, etc. ) ?

( A ) 1/64

( B) 1/48

( C ) 1/36

( D ) 1/24

( E) 1/16
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5045
Location: Singapore
Followers: 31

Kudos [?]: 376 [0], given: 0

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25 Sep 2007, 18:36
Isn't this the same question as the one mgmat posted on their challenge?
Director
Joined: 11 Jun 2007
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25 Sep 2007, 21:47
ywilfred wrote:
Isn't this the same question as the one mgmat posted on their challenge?

yes! eeerie-ly similar if you ask me..

http://www.gmatclub.com/forum/t52818
Director
Joined: 03 Sep 2006
Posts: 873
Followers: 7

Kudos [?]: 895 [0], given: 33

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26 Sep 2007, 00:34
beckee529 wrote:
ywilfred wrote:
Isn't this the same question as the one mgmat posted on their challenge?

yes! eeerie-ly similar if you ask me..

http://www.gmatclub.com/forum/t52818

Hi,

I just checked the link, and yes you are right, the questions is same. Now the point shouldn't be, it's the same or it isn't, the point for all of us being here in GClub is to be able to solve the questions. Could you guys please focus and help me reaching the solution instead of similarities between them?

Regards,
VP
Joined: 08 Jun 2005
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26 Sep 2007, 01:11
The trick is to understand that the question really asks us for the odds that in a group of four workes (w1 - w4) ---> w1< w2 < w3 < w4.

Since we have four workers, the total possibable arrangements is 4! = 24 but in only one possibable arrangement can w1 < w2 < w3< w4 so the probability is 1/24.

Hence for any four workers chosen the probability is 1/24 , no matter how many workers we started with (just ignore the 99 number).

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Joined: 07 Jul 2004
Posts: 5045
Location: Singapore
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26 Sep 2007, 01:16
LM wrote:
beckee529 wrote:
ywilfred wrote:
Isn't this the same question as the one mgmat posted on their challenge?

yes! eeerie-ly similar if you ask me..

http://www.gmatclub.com/forum/t52818

Hi,

I just checked the link, and yes you are right, the questions is same. Now the point shouldn't be, it's the same or it isn't, the point for all of us being here in GClub is to be able to solve the questions. Could you guys please focus and help me reaching the solution instead of similarities between them?

Regards,

okay, i thought there was a solution on that link... anyway, squirrel solved it ..
Manager
Joined: 13 Feb 2007
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26 Sep 2007, 09:34
this is a tough problem. i was trying all sorts of useless things and then i figured i'd just read the answer to save my time.

26 Sep 2007, 09:34
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