A certain roller coaster has 3 cars

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ska7945: Master | Next Rank: 500 Posts; Posts: 200; Joined: Thu Aug 07, 2008 5:14 pm; Thanked: 1 times

A certain roller coaster has 3 cars

by ska7945 » Thu Sep 11, 2008 5:13 am

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?
F. 0
G. 1/9
H. 2/9
I. 1/3
J. 1

let's beat GMAT.

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Source: Beat The GMAT — Problem Solving | Thu Sep 11, 2008 5:13 am

parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 am; Thanked: 146 times; Followed by:2 members

Re: A certain roller coaster has 3 cars

by parallel_chase » Thu Sep 11, 2008 5:28 am

ska7945 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?
F. 0
G. 1/9
H. 2/9
I. 1/3
J. 1

Is the answer 2/9.

Quote

ska7945: Master | Next Rank: 500 Posts; Posts: 200; Joined: Thu Aug 07, 2008 5:14 pm; Thanked: 1 times

by ska7945 » Thu Sep 11, 2008 6:29 am

i had 1/3..what went wrong?

let's beat GMAT.

Quote

parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 am; Thanked: 146 times; Followed by:2 members

Re: A certain roller coaster has 3 cars

by parallel_chase » Thu Sep 11, 2008 6:34 am

ska7945 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?
F. 0
G. 1/9
H. 2/9
I. 1/3
J. 1

Choosing the first car is irrelevant.

Second time we have 2 choices.

Third time we only have 1 choice.

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

or the other method is

you have 3 cars ABC

ABC can be arranged in 3! ways or 6 ways.

Total permutations = 3*3*3 = 27

Probability = 6/27 = 2/9

Hope this helps.