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# PS - Probability

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Senior Manager
Joined: 05 Jun 2008
Posts: 304

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23 Sep 2008, 20:20
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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

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Intern
Joined: 07 Sep 2008
Posts: 11

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23 Sep 2008, 22:04
I get C as well.

There are three rounds, something specific has to happen each round for the outcome to happen:

_ * _ * _

the first time around any car will fulfill the desired outcome, the odds of getting into any car is a 1,

1 * _ * _

the second time around you've already ridden in one car, so only if you ride in one of the other 2 out of 3 cars will you satisfy the outcome and the odds of getting it are 2/3,

1 * 2/3 * _

the third time around only the last remaining car will satisfy your desired outcome, so you have a chance of 1 in 3 of getting that

1 * 2/3 * 1/3 = 2/9

Perhaps others can explain this more eloquently or formally.

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Senior Manager
Joined: 05 Jun 2008
Posts: 304

Kudos [?]: 181 [0], given: 0

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24 Sep 2008, 07:09
vivektripathi 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

I have cracked it.....
1) Total no of ways to select select 3 roller coaster is - 3*3*3 = 27
2) Favorable ways = 3 *2 * 1 = 6
Hence Probability is 6/27= 2/9....

Kudos [?]: 181 [0], given: 0

Re: PS - Probability   [#permalink] 24 Sep 2008, 07:09
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