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PS_A man chooses an outfit

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Manager
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Joined: 01 Aug 2008
Posts: 116

Kudos [?]: 162 [0], given: 2

PS_A man chooses an outfit [#permalink]

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New post 14 Jun 2009, 09:01
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Do not answer without sharing the reasoning behind ur choice
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Working on my weakness : GMAT Verbal
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Kudos [?]: 162 [0], given: 2

Director
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Joined: 05 Jun 2009
Posts: 807

Kudos [?]: 383 [0], given: 106

WE 1: 7years (Financial Services - Consultant, BA)
Re: PS_A man chooses an outfit [#permalink]

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New post 14 Jun 2009, 10:25
3 St(Shirts), 2 Sh(Shoes), 3 P(Pants)

3 days
Probability of just shirts to be in 3 different days:
on first day can pick any of the three (3/3)
On second day can pick rest 2, i.e. not picked on the first day( so random picking probability of picking rest 2 od 3 shirts) = 2/3
Similarly 3rd day = 1/3
So probability of having 3 different St in 3 different days = \((3/3)*(2/3)*(1/3)=2*[(1/3)^2]\)

Probability of Pants will be the same as similar 3 different pants on 3 different days= \(2*[(1/3)^2]\)

Probability of same shoes on 3 different days
1st day = any 2 of 2 = 2/2
2nd day the same one of the 2 = 1/2
3rd day= 1/2
same shoes probability on 3 days=(2/2)*(1/2)*(1/2) = \((1/2)^2\)

So total probability = \(2*[(1/3)^2]*2*[(1/3)^2]*(1/2)^2\) = = \((1/3)^4\)

(C)
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Kudos [?]: 383 [0], given: 106

Senior Manager
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Re: PS_A man chooses an outfit [#permalink]

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New post 14 Jun 2009, 10:33
mbaMission wrote:
How to solve?


P= 3*2*3*2*1*2*1*1*1/3*2*3*3*2*3*3*2*3 = 1/3^4
C is my choice.

Kudos [?]: 304 [0], given: 0

Re: PS_A man chooses an outfit   [#permalink] 14 Jun 2009, 10:33
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PS_A man chooses an outfit

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