nakib77
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
APPROACH #1: Probability rulesP(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 techniquesP(3 different cars) =
(# of ways to ride in 3 different cars)/
(total # of ways to take 3 rides)total # of ways to take 3 ridesFor 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 carsLet 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