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# can some one get me the breakup of this sum ? Thanks in

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Manager
Joined: 16 Jan 2008
Posts: 102
can some one get me the breakup of this sum ? Thanks in [#permalink]

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04 Oct 2008, 20:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

can some one get me the breakup of this sum [attached]?

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SVP
Joined: 29 Aug 2007
Posts: 2467

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04 Oct 2008, 21:00
technocrat wrote:
can some one get me the breakup of this sum [attached]?

probability is more about logical understanding the sequence of events. so

prob of geting into a car for the first time = 1
prob of geting into remaining two cars = 2/3
prob of geting into remaining car = 1/3
= 1 x 2/3 x 1/3
= 2/9
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Manager
Joined: 10 Aug 2008
Posts: 74

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05 Oct 2008, 01:12
- So for 1st time, user can choose any of car, means probability = 1
- 2nd time user can sit any of 2 remaing car out of 3, so probability = 2/3
- 3rd time, user can choose only 1 remaining car out of 3, so prob = 1/3

Total probability 1*2/3*1/3 = 2/9
Intern
Joined: 04 Oct 2008
Posts: 8

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05 Oct 2008, 07:05
I got the same answer a little differently.

The rider has equal chances of getting into each car each time, so there are 3^3=27 different possible riding orders.
But there are only 3! = 6 ways of riding in each car exactly once.

6/27 = 2/9
Re: probability   [#permalink] 05 Oct 2008, 07:05
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# can some one get me the breakup of this sum ? Thanks in

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