BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars...

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 2505
Joined: Sun Oct 15, 2017 1:50 pm
Followed by:6 members

A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars...

by BTGmoderatorLU » Fri Jul 24, 2020 1:01 pm

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Source: GMAT Prep

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

The OA is C
Top
Source: — Problem Solving |

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

Re: A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars...

by Brent@GMATPrepNow » Fri Jul 24, 2020 4:57 pm
BTGmoderatorLU wrote: ↑
Fri Jul 24, 2020 1:01 pm
 Source: GMAT Prep

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

The OA is C
Here's an approach that uses counting techniques.

P(3 different cars) = (# of ways to ride in 3 different cars)/(total # of ways to take 3 rides)

total # of ways to take 3 rides
For the 1st ride, there are 3 options
For the 2nd ride, there are 3 options
For the 3rd ride, there are 3 options
So, the total number of ways to take three rides = (3)(3)(3) = 27

# of ways to ride in 3 different cars
Let the cars be Car A, Car B and Car C
In how many different ways can we order cars A, B and C (e.g., ABC, CAB, BAC, etc)?
Rule: We can arrange n unique objects in n! ways.
So, we can arrange 3 unique cars in 3! ways ( = 6 ways)

P(3 different cars) = (# of ways to ride in 3 different cars)/(total # of ways to take 3 rides)
= 6/27
= 2/9
= C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Post new topic Post Reply
• Page 1 of 1