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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 13 Oct 2005, 12:49
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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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Last edited by Bunuel on 07 Jun 2012, 08:50, edited 2 times in total.
Edited the OA
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Re: A certain roller coaster has 3 cars, and a passenger is [#permalink] New post 21 Feb 2012, 00:30
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nakib77 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 a passenger can ride any car: p=1;
Second time the passenger should ride another car: p=2/3;
Third time the passenger should ride not the 2 cars he had already driven: p=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 ride passenger 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.
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Re: A certain roller coaster has 3 cars, and a passenger is [#permalink] New post 28 Feb 2012, 10:24
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+1 C

Let's suppose that the cars are A, B, and C.
Now, let's find the probability of picking the cars in this order: A-B-C

\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}

However, there could be more way to organize A-B-C. The numbers of ways is 3! = 6

Then, \frac{1}{27} * 6 = \frac{6}{27} = \frac{2}{9}

Answer: C
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Re: A certain roller coaster has 3 cars, and a passenger is [#permalink] New post 04 Jun 2012, 04:30
Hi,

Probability = (favorable cases)/(total number of cases)

Total number of ways in which a person can ride car = 3*3*3 = 27
(In first ride he has 3 options to sit, in second right again he has 3 seats available to sit and so on)

Number of favorable cases, i.e., when he rides on different cars;
He can choose seat car in 3 ways in his 1st ride.
He can choose seat car in 2 ways in his 2nd ride.
He can choose seat car in 1 ways in his 3rd ride.
So, 3*2*1 = 6 ways

Thus, probability of choosing different seats = 6/27 = 2/9
(C)

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Re: A certain roller coaster has 3 cars, and a passenger is   [#permalink] 04 Jun 2012, 04:30
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