roller

[ketkoag]

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

please explain how to solve this question.

[kanha81]

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

please explain how to solve this question.

I think. Not Sure. [spoiler][E][/spoiler]
Person is equally likely to ride in any of 3 cars and we want to find out probab that person rides in 3 diff. cars in 3 times:
t(1) = 3/3
t(2) = 2/2
(3) = 1/1

t(1) * t(2) * t(3) = 1

[mike22629]

IMO C.

First car can be any of the three so Probability = 1

Second Car can be one of the remaining two so Probability = 2/3

Third Car there is only one left not driven so Probability = 1/3

1*(2/3)*(1/3) = 2/9

[dumb.doofus]

I am sure the above are valid solutions and correct. I am not trying to make this problem more complex than what it is, but to me the wholistic solution is as follows. Again, this is just my opinion.

Let the three cars be a, b, c
Probability of riding a car = 1/3
Probability of not riding a car = 1-1/3 = 2/3

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

but we see that there are 6 possible arrangements of the car i.e. abc, acb, bac, bca, cab, cba and there are two possibilities each for the three cars to be first, second or third. So we need to divide the final probability by 2

i.e (12/27)*(1/2) = 2/9 and that's the answer..

[sam117]

I am sure the above are valid solutions and correct. I am not trying to make this problem more complex than what it is, but to me the wholistic solution is as follows. Again, this is just my opinion.

Let the three cars be a, b, c
Probability of riding a car = 1/3
Probability of not riding a car = 1-1/3 = 2/3

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

but we see that there are 6 possible arrangements of the car i.e. abc, acb, bac, bca, cab, cba and there are two possibilities each for the three cars to be first, second or third. So we need to divide the final probability by 2

i.e (12/27)*(1/2) = 2/9 and that's the answer..

I don't understand why you divide by 2 and not by 6 for instance, can anybody help? Thanks