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

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Director
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A certain roller coaster has 3 cars, and a passenger is [#permalink]  06 Feb 2006, 16:45
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?
VP
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Is it 2/9?

Total possibilities = 3*3*3 = 27 (3 choices in each ride)

There are 3! = 6 ways in which the passenger can ride each of them

123
132
213
231
312
321

So (I think) Prob = 6/27 = 2/9
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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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Director
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First time the passenger can ride on any car.. so 3/3, next time in any of the two cars out of 3, so 2/3 and the last time 1/3.. so 1*2/3*1/3 = 2/9
VP
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Re: probability [#permalink]  06 Feb 2006, 17:39
agree with 6/27=2/9
VP
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Giddi,

Unfortunately I'm completely lost here I'm not sure how else you can explain to me
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VP
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TeHCM wrote:
Giddi,

Unfortunately I'm completely lost here I'm not sure how else you can explain to me

Think of this as a die rolling problem. A die has 3 faces and it is rolled 3 times. What is the probability that each face will appear exactly once?

Ortherwise consider 2 cars and 2 rides and try writing all possibilities. It should be easier to see.
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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Intern
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i think i can explain this one.

so let's say he gets on the roller coaster..he doesnt' care what car he picks..

so that means the probability of getting any car is 1. let's say you chose car A.

the next time, you must choose either car A or car B and that probability is 2/3.

the third time, you must choose car C, therefore,

you have

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