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A certain roller coaster has 3 cars, and a passenger is

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A certain roller coaster has 3 cars, and a passenger is [#permalink] New post 26 Aug 2010, 06:26
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49% (02:35) correct 51% (00:34) wrong based on 61 sessions
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?
A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1
[Reveal] Spoiler: OA
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Re: Roller Coster Probability [#permalink] New post 26 Aug 2010, 06:49
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udaymathapati wrote:
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the
3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?

A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1


First time passenger can ride any car - p=1;
Second time the probability that passenger will drive another car is 2/3;
Third time the probability that the passenger will drive not the 2 cars he had already driven is 1/3.

So P=1*\frac{2}{3}*\frac{1}{3}=\frac{2}{9}.

OR:
Total # of ways the passenger can drive 3 cars (for 3 rides) is 3^3 (for each passenger he has 3 options);
# of ways the passenger can drive 3 different cars is 3! (ABC, ACB, BAC, BCA, CAB, CBA);

So P=\frac{3!}{3^3}=\frac{2}{9}.

Answer: C.

Please tag your questions correctly: added probability tag.
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Re: Roller Coster Probability [#permalink] New post 18 Sep 2010, 06:18
Bunuel wrote:
utin wrote:
didn't get this one!!!


Can you please be more specific?



Actually I get confused while solving probability questions.
In this one I am thinking that 3 cars are there and the person has to ride thrice ensuring that he sits in a different car every time.So i took it this way:

probability of getting into 1st car=1/3
probability of getting into 2nd car=1/3
probability of getting into 3rd car=1/3
so final answer=1/27

I know i am not able to understand this one...Please help me clear my concept...
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Re: Roller Coster Probability [#permalink] New post 18 Sep 2010, 11:11
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utin wrote:
Bunuel wrote:
utin wrote:
didn't get this one!!!


Can you please be more specific?



Actually I get confused while solving probability questions.
In this one I am thinking that 3 cars are there and the person has to ride thrice ensuring that he sits in a different car every time.So i took it this way:

probability of getting into 1st car=1/3
probability of getting into 2nd car=1/3
probability of getting into 3rd car=1/3
so final answer=1/27

I know i am not able to understand this one...Please help me clear my concept...


The question asks you: What is the probability that the passenger will ride in each of the 3 cars?

The order the passenger rides in the cars does not matter, he just has to ride in all three over three consecutive trips. You're trying to compute the probability as if it matters which order he chooses to ride the cars.

First ride: He has a 100% chance of riding in a car he has not yet ridden in.
Second ride: He has ridden in one of the three cars, so he has a 2/3 chance of riding in a car he has not yet ridden in.
Third ride: There's only one car he has not ridden in, so he has a 1/3 chance of choosing the remaining car.
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Re: Roller Coster Probability [#permalink] New post 18 Sep 2010, 21:27
Thanks LJ!!!Got it this time... +1
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Re: Roller Coster Probability [#permalink] New post 25 Sep 2010, 19:10
Thanks for your help Bunnel
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Re: A certain roller coaster has 3 cars, and a passenger is [#permalink] New post 27 Sep 2013, 22:22
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Re: A certain roller coaster has 3 cars, and a passenger is   [#permalink] 27 Sep 2013, 22:22
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