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Probability - Roller Coaster

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Probability - Roller Coaster

by theCodeToGMAT » Tue Oct 22, 2013 10:00 am
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]
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by sahilchaudhary » Tue Oct 22, 2013 10:36 am
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
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by Brent@GMATPrepNow » Tue Oct 22, 2013 11:06 am
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
Sahil's solution is the preferred (faster) approach, but I'll show the approach that uses 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)

So, P(3 different cars) = (# of ways to ride in 3 different cars)/(total # of ways to take 3 rides)
= 6/27
= [spoiler]2/9[/spoiler]
= C

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by Brent@GMATPrepNow » Tue Oct 22, 2013 11:27 am
sahilchaudhary wrote:One suggestion, please don't post the OA along with the question. All the fun of doing the question goes away.
Hey Sahil,

Beat The GMAT asks posters to include the OA, but use the spoiler function to hide the answers.
https://www.beatthegmat.com/mba/community-rules

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by theCodeToGMAT » Tue Oct 22, 2013 5:34 pm
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 ;)
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by ganeshrkamath » Tue Oct 22, 2013 8:22 pm
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

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by peter456 » Thu Nov 19, 2015 11:04 pm
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.

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by peter456 » Sat Nov 21, 2015 9:05 pm
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?
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by [email protected] » Sun Nov 22, 2015 8:41 am
Hi peter456,

With 3 cars, the possibility of riding in the same car more than once, and the specific order you described, the math would actually be more straight-forward.

Probability of sitting in Car A FIRST = 1/3
Probability of sitting in Car B SECOND = 1/3
Probability of sitting in Car C THIRD = 1/3

Probability of that order (A, B, then C) = (1/3)(1/3)(1/3) = 1/27

You could also physically prove this by writing out all 27 options.

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by peter456 » Mon Nov 23, 2015 2:30 am
But where is the approach I used applied?

Thanks to explain and probably cite another illustration.
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by [email protected] » Mon Nov 23, 2015 10:06 am
Hi peter 456,

The math you attempted to use would be more appropriate on questions that involve probabilities and multiple independent events. (e.g. What's the probability that events A and B happen, but that event C does not happen?).

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by Matt@VeritasPrep » Fri Nov 27, 2015 2:33 am
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.
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