Problem Solving

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


[spoiler]{C}[/spoiler]

One suggestion, please don't post the OA along with the question. All the fun of doing the question goes away.

For the first ride, passenger has 3 choices, i.e., he can take any 1 of the 3. (P=1)
For the second ride, passenger has 2 choices, i.e., he can't take the roller coaster which he took in the first ride. (P=2/3)
For the third ride, passenger has only 1 choice, i.e., he can't take the roller coaster which he took in the first and second ride. (P=1/3)

Since, he has to take the first and second and third ride, so multiply all three.
Therefore, P = 1 * 2/3 * 1/3 = 2/9.

So, the answer is C

Sahil, my perception says that if you are not posting the OA then you are testing other forum members .. that is not the case here.. it was a pretty much direct question.
For tricky questions, I also avoid posting OA, initially

theCodeToGMAT 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


[spoiler]{C}[/spoiler]

Sample space :
1st ride => 3 options
2nd ride => 3 options
3rd ride => 3 options
Total number of combinations = 3*3*3 = 27

Favorable cases :
Choose the 3 cars in any order: 3! = 6

Probability = 6/27 = 2/9

Choose C

Cheers

Pls point out the error with my logic:

The passenger has to ride on ALL cars: CarA, CarB, & CarC.

Equally likely chance of picking any car= 1/3
First ride: Say, he starts with CarA(1/3)
Second ride: NotcarA(2/3)x but CarB(1/3) = 2/9.....chance of not riding=2/3
Third ride: NotcarA(2/3)x NotcarB(2/3)x but carC(1/3) = 4/27

To ride on all the 3 cars: 1/3 x 2/9x4/27 = 19/27 (not among OA)


I understand some other approaches, but I need to know the logic gap here.

Thank you

I will be happy for your corrections:

Say, had the question requested for a specific order- A, B, & C.
I think the solution would be like below?

All cars are equally likely at 1/3.
1st step: CarA = 1/3
2nd step: Not CarA(2/3), x but CarB(1/3)= 2/9
3rd step: Not CarA(2/3), x Not CarB(2/3), x but CarC(1/3) = 4/27

Passenger rides in Car A, B, & C = 1/3 x 2/9 x 4/27
= 19/27

Is this correct?

But where is the approach I used applied?

Thanks to explain and probably cite another illustration.

peter456 wrote:Pls point out the error with my logic:

The passenger has to ride on ALL cars: CarA, CarB, & CarC.

Equally likely chance of picking any car= 1/3
First ride: Say, he starts with CarA(1/3)
Second ride: NotcarA(2/3)x but CarB(1/3) = 2/9.....chance of not riding=2/3
Third ride: NotcarA(2/3)x NotcarB(2/3)x but carC(1/3) = 4/27

To ride on all the 3 cars: 1/3 x 2/9x4/27 = 19/27 (not among OA)

The letter of the first car doesn't matter, so no probability is needed.

Given the first car, the second car need only be different from the first (prob = 2/3).

Given the first two cars, the third car need only be different again (prob = 1/3).

Following your approach, we'd need something like:

ABC = (1/3)³
ACB = (1/3)³
BAC = (1/3)³
BCA = (1/3)³
CAB = (1/3)³
CBA = (1/3)³

So our solution is 6 * (1/3)³, or 6/27, or 2/9.