Probability

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Sun Mar 06, 2011 7:42 am
Location: India
Thanked: 2 times

Probability

by rohanberi » Mon Sep 05, 2011 7:00 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?

a) 0 b) 1/9 c) 2/9 d) 1/3 e) 1
Source: — Problem Solving |

User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 am
Location: London
Thanked: 70 times
Followed by:3 members

by kmittal82 » Mon Sep 05, 2011 7:10 am
Lets say the cars are A,B and C

To simplify, the question asks us whats the probability that passengers goes in A, then B and then C

P(A) = 1/3
P(B) = 1/3
P(C) = 1/3

Ofcourse, there are 6 different combinations of ABC available as well, so we multiply that

Total Proability = P(A)x P(B)x P(C)x Permutaions of the cars themselves = (1/27 ) x 6 = 2/9

OA please?

Senior | Next Rank: 100 Posts
Posts: 31
Joined: Sun Mar 06, 2011 7:42 am
Location: India
Thanked: 2 times

by rohanberi » Mon Sep 05, 2011 7:12 am
Thanks a lot Mittal.
The answer is correct.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Mon Sep 05, 2011 7:18 am
rohanberi 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
P(all 3 cars) = P(any car on 1st trip) x P(different car on 2nd trip) x P(different car on 3rd trip)

P(any car on 1st trip) = 1
P(different car on 2nd trip) = 2/3 [at this point, the passenger has occupied one of the cars, which means that 2 of the 3 cars remain not ridden]
P(different car on 3rd trip) = 1/3 [at this point, the passenger has occupied two of the cars, which means that 1 of the 3 cars remains not ridden]

So, P(all 3 cars) = (1) x (2/3) x (1/3) = 2/9

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

Senior | Next Rank: 100 Posts
Posts: 31
Joined: Sun Mar 06, 2011 7:42 am
Location: India
Thanked: 2 times

by rohanberi » Mon Sep 05, 2011 8:41 am
Thanks a lot Brent.

Cheers,
Rohan

User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Fri Nov 05, 2010 5:43 pm
Thanked: 15 times
Followed by:1 members

by mehrasa » Mon Sep 05, 2011 10:04 pm
Dear Brent

Could you please explain why the probability of first event ( p(at any car at 1st time) is equal to 1?there is one person want to choose from 3 car...
thanks

Senior | Next Rank: 100 Posts
Posts: 31
Joined: Sun Mar 06, 2011 7:42 am
Location: India
Thanked: 2 times

by rohanberi » Tue Sep 06, 2011 12:29 am
Mehrasa,

That's because P(at any car on 1st day);

Favourable number of outcomes = 3 (as the passenger can choose any of the three cars), and
Total number of outcomes = 3

Hence, probability (at any car on 1st day) = Favourable number of outcomes/ Total number of outcomes = 3/3 = 1

||ly, for probability(diff car on 2nd day)= 2/3
and probability(dif car on 3rd day) = 1/3

Hence, total probability = 1 x (2/3) x (1/3) = 2/9


P.S. :- All Thanks to Brent for clarifying.