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
This is how I solved this problem:
Probability that the person rides the first car but not the other two = (1/3)(2/3)(2/3) = 4/27
Similarly probability for riding second car and not other two = 4/27
Similarly probability for riding third car and not other two = 4/27
So total probability = 3(4)/27 = 12/27 = 4/9
But the OA is [spoiler]2/9[/spoiler]
Can anyone explain what did I do wrong? Thanks
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
This is how I solved this problem:
Probability that the person rides the first car but not the other two = (1/3)(2/3)(2/3) = 4/27
Similarly probability for riding second car and not other two = 4/27
Similarly probability for riding third car and not other two = 4/27
So total probability = 3(4)/27 = 12/27 = 4/9
But the OA is [spoiler]2/9[/spoiler]
Can anyone explain what did I do wrong? Thanks
