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Rollar coaster probability

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by student22 » Thu May 20, 2010 5:57 pm
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

OA:C

I've searched for this problem, and all of the posts about it are simply formulas of (3/3)(2/3)(1/3) = 2/9.

However, none of the posts explain the logic behind why it's done this way.

Why isn't it like standard probability methods of (1/3)(1/3)(1/3)?

Can somebody please walk me through the logic of this problem?
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by Stuart@KaplanGMAT » Thu May 20, 2010 6:29 pm
student22 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

OA:C

I've searched for this problem, and all of the posts about it are simply formulas of (3/3)(2/3)(1/3) = 2/9.

However, none of the posts explain the logic behind why it's done this way.

Why isn't it like standard probability methods of (1/3)(1/3)(1/3)?

Can somebody please walk me through the logic of this problem?
Hi,

if you do (1/3)(1/3)(1/3), then you're ignoring the different orders in which you can ride the cars.

Let's call the 3 cars A, B and C. To ride in all 3 exactly once, the order could be:

ABC
ACB
BAC
BCA
CAB
CBA

In other words, there are 3! different ways to arrange the 3 cars.

So, another way you can solve is:

(1/3)(1/3)(1/3) * 3! = (1/3)(1/3)(1/3)*6 = 6/27 = 2/9

Here's the logic behind the (3/3)(2/3)(1/3):

The first car can be any of the 3 to satisfy what we want, so the probability of picking a unique car on the 1st ride is 3/3.

Once we take our first ride, there are 2 cars left in which we haven't ridden. So, there's a 2/3 chance of riding in a new car.

After the second ride, there's only 1 car left that we haven't ridden. So, the chance of getting that car on the 3rd ride is 1/3.

Accordingly, the probability of getting 3 different cars is:

3/3 * 2/3 * 1/3 = 6/27 = 2/9.
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by frank1 » Thu May 20, 2010 6:30 pm
I think the logic behind that is
1st time =3/3 (she can ride in any....3 fav cases/3 total cases)
2nd time=2/3 (one is already gone)
3rd time =1/3 (he has no other 2 choices left as she has already ridden over ....it)
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by frank1 » Thu May 20, 2010 6:35 pm
Sorry the comprehensive reply from Stuart Kovinsky was not there when i posted my logic(answer)....
(May be we posted simulataneosuly with difference of fraction of seconds...)

Just explaining as some might argue,why my reply was needed after such an comprehensive explanation...
once again sorry....
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by student22 » Thu May 20, 2010 8:11 pm
Thanks Stuart, I get the explanation.

This seems like one of those problems where you break it up in to slots and you have 4 people for the first slot, 3 for the 2nd, etc. Except with probabilities.

When I was doing 1/3 * 1/3 * 1/3 , I completely forgot that order mattered.


Also, Frank1, don't worry about it, thanks for replying to my topic with an explanation, as well!
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by student22 » Fri May 21, 2010 4:20 pm
Thinking about this problem some more, I think that the fractions that Stuart used, could also be derived with combinations:

Desired outcomes / possible outcomes

Slot 1: 3C1/3C1 = 3/3 choosing 1 car out of 3
Slot 2: 2C1/3C1 = 2/3 choosing 1 car out of 2
Slot 3: 1C1/3C1 = 1/3 choosing 1 car out of 1


Possible outcomes will always be 3C1.

Please correct me if I'm wrong.
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by kstv » Fri May 21, 2010 10:27 pm
First Ride - Probability of riding any one car be it A,B and C = 1/3 if it is A
Second Ride - Prob of choosing either B or C = 2/3, if it is B
Third Ride - Has to be C , so prob is 1
1/3*2/3 = 2/9
Same result, but the approach is diff. Here I not assuming that the passenger deliberately chooses a particular car. Rather it can be any car A, B or C, but having choosen a car on his second ride he has to avoid that car.
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by evsistr » Sat May 22, 2010 2:15 am
I have one maybe more complicated solution, but nevertheless:

We have three cars A, B, and C. We need to obtain the number of combinations with unique cars (it would mean that the passenger has chosen every time the different car), i.e. ABC, ACB, BAC, BCA, CAB, CBA. The number of such combinations is given by 3! = 6.

The number of all possible non-unique combinations (all possible ways of riding three cars), i.e. AAB, ACB, CBB, etc. is given by 3^3 = 27.

Hence, the probability of riding every of 3 cars is: 6 / 27 = 2 / 9. Answer C.
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