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# GMAT Prep- Probability question

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GMAT Prep- Probability question [#permalink]  21 Jun 2008, 09:13
Can some one explain the below question? Thanks!
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GMAT PREP_Test2_21June_Maths10.JPG [ 62.66 KiB | Viewed 915 times ]

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Re: GMAT Prep- Probability question [#permalink]  21 Jun 2008, 10:20
Let’s denote the cars as 1,2,3. Total number of possible outcomes is 3*3*3 = 27.
Number of favorable outcomes is the number of permutations of a set (1,2,3) which is 3! = 6.

Hence, p=6/27 = 2/9.
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Re: GMAT Prep- Probability question [#permalink]  21 Jun 2008, 10:36
Expert's post
another way:

$$p=1*\frac23*\frac13=\frac29$$
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Re: GMAT Prep- Probability question [#permalink]  21 Jun 2008, 10:53
First ride :-

Any of the 3 cars :- 1/3

Second ride :-

Any of 2 remaining cars :- 2/3

Last ride :-

Third car :- 3/3

1/3 x 2/3 x 3/3 = 2/9
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Re: GMAT Prep- Probability question [#permalink]  21 Jun 2008, 12:49
Yes 2/9, reason same as walker.
Btw mate do you mind actually coping the question to the forum rather than pasting entire screenshot? Just think of the memory bytes that you will save by doing this. Lets go easy on the server that hosting this excellent forum and question is easier to work without screenshot.
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Re: GMAT Prep- Probability question [#permalink]  21 Jun 2008, 21:26
Thanks for the inputs!
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Mahesh Vollala

Re: GMAT Prep- Probability question   [#permalink] 21 Jun 2008, 21:26
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