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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
A certain roller coaster has 3 cars, and a passenger is equally likely
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27 Nov 2021, 04:33
27 Nov 2021, 04:33
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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
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
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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
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