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GMATPrep 34: Got this problem right and I did it the same

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GMATPrep 34: Got this problem right and I did it the same [#permalink] New post 24 Oct 2007, 22:40
GMATPrep 34:

Got this problem right and I did it the same way, I just thought id post it anyway.

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?

0
1/9
2/9
1/3
1


Heres what I said, 3 cars: ABC.

Prob of entering A is 1, it is certain the rider will enter A.

So we get ABC: 1*1/3*1/3 (1/3 chance this person will ride in any of the three cars)

Now we also have to consider ACB: 1*1/3*1/3

We get 1/9+1/9 --> 2/9. C.
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Re: GMATPrep 34 [#permalink] New post 24 Oct 2007, 23:15
three casrs: a, b and c.

first: any of the three = 3/3
second: any of rest two = 2/3
third: last one = 1/3
so the prob: 3/3 x 2/3 x 1/3 = 2/9
Re: GMATPrep 34   [#permalink] 24 Oct 2007, 23:15
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