GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
A certain roller coaster has 3 cars, and a passenger is equally likely
[#permalink]
27 Nov 2021, 04:33
27 Nov 2021, 04:33
Expert Reply
1
Bookmarks
00:00
Question Stats:
55% (00:51) correct
44% (01:13) wrong based on 9 sessions
Hide Show timer Statistics
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
A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: A certain roller coaster has 3 cars, and a passenger is equally likely
[#permalink]
27 Nov 2021, 06:08
27 Nov 2021, 06:08
2
GeminiHeat 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
A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1
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
Answer: C
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
gmatclubot
Re: A certain roller coaster has 3 cars, and a passenger is equally likely [#permalink]
27 Nov 2021, 06:08
Moderators:
