## 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

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### 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

by AAPL » Fri Jul 22, 2022 4:22 pm

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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

OA C

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### Re: 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

by [email protected] » Mon Aug 01, 2022 6:53 am
AAPL wrote:
Fri Jul 22, 2022 4:22 pm
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

OA C
APPROACH #1: Probability rules
P(rides in all 3 cars) = P(1st car is ANY car AND 2nd car is different from 1st car AND 3rd car is different from 1st and 2nd cars)
= P(1st car is ANY car) x P(2nd car is different from 1st car) x P(3rd car is different from 1st and 2nd cars)
= 3/3 x 2/3 x 1/3
= 2/9

APPROACH #2: 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

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