Probability: A certain roller coaster...

[a123]

A certain roller coaster has 3 cars, and a pasenger is equally likely to ride in any 1 of 3 cars each time he/she rides. If a certain passenger is to ride the roller coaster 3x, what's the probability that the passenger will ride in each of the cars?

Answer = 2/9

The probability of the 1st car = 1/3
But then I get confused on how to calculate the prob of the 2nd and 3rd car.

It's wrong but I tried = (1/3)*(1/3)*(1/3) since that's the probability of 1 in 3 cars for each ride??

How should I be thinking about this?

[george.kourdin]

let's say 3 cards are car A, B and C. we want to ride once in each of the cards. it doesn't matter which car we ride in first, as long as the subsequent cars are all different. in other words, whether we ride in A, B or C is irrelevant as long as our 2nd and 3rd rides will be in cars we have not ridden yet.


P (any car) = 3/3 -> we are riding in either A, B or C. lets say A.
P ( B or C) = 2/3
P (C) = 1/3

3/3*2/3*1/3 = 2/9

another way using combinations:

3c1 * 2c1 * 1c1 = 6 ways we can ride in 3 diff cars
3c1*3c1*3c1 = 27 total ways we can ride in 3 cars

7/27 = 2/9

[RonPurewal]

let's say 3 cards are car A, B and C. we want to ride once in each of the cards. it doesn't matter which car we ride in first, as long as the subsequent cars are all different. in other words, whether we ride in A, B or C is irrelevant as long as our 2nd and 3rd rides will be in cars we have not ridden yet.


P (any car) = 3/3 -> we are riding in either A, B or C. lets say A.
P ( B or C) = 2/3
P (C) = 1/3

what you're doing here is perfectly correct (you must have some sort of higher math background -- this seems like a "without loss of generality" argument). however, the notation may be a little confusing for forum readers who lack such a background.

this approach can be simplified as follows:
P(any car) = 3/3
P(either of the two remaining cars) = 2/3
P(last car in which we haven't ridden yet) = 1/3
so, yes.

another way using combinations:

3c1 * 2c1 * 1c1 = 6 ways we can ride in 3 diff cars
3c1*3c1*3c1 = 27 total ways we can ride in 3 cars

7/27 = 2/9

perfect, although the combination formulas are a bit on the bulky side -- the former quantity is just 3! (the total number of ways of arranging any 3 things), and the latter is just 3 x 3 x 3.

but, yeah. well done.

[jigar24]

I tried using the following approach:

prob of sitting in any one car

p(sitting in one car) * p(not sitting in 2nd car) * p(not sitting in 3rd car) = (1/3*2/3*2/3) = 4/27

no 3 cars can be arranged in 3! ways = 6

6*4/27 =8/9

what's wrong with this approach (of course there is something wrong)

could you also please elaborate of 'without loss of generality' principle?

Thanks

[RonPurewal]

I tried using the following approach:

prob of sitting in any one car

p(sitting in one car) * p(not sitting in 2nd car) * p(not sitting in 3rd car) = (1/3*2/3*2/3) = 4/27

no 3 cars can be arranged in 3! ways = 6

6*4/27 =8/9

what's wrong with this approach (of course there is something wrong)

i can't actually figure out *what* you are trying to calculate here.

could you please explain (in words) what this calculation is supposed to represent?

could you also please elaborate of 'without loss of generality' principle?

i'd rather not discuss it on the forum, because you don't need to know it for the gmat. but, you can read about it on the internet:
http://en.wikipedia.org/wiki/Without_loss_of_generality

[ffearth]

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 2/9

Can anyone explain what did I do wrong? Thanks

[RonPurewal]

samymakhlouf, i can't tell what you are doing. more specifically, i can't understand the phrases you're using (such as "rides in first car but not other two"). you'll have to elaborate on those, please.

if someone can prove (via a screenshot) that this problem is actually from the official software, then the discussion will continue. if not, the post below this one will be the end of the discussion.
thanks.

[RonPurewal]

ok, guys, we are going to kill further discussion on this thread unless and until someone can prove that the problem is actually from the OFFICIAL GMAT PREP SOFTWARE (= the only source allowed in this folder).

if anyone has a screenshot of this problem, please post it -- either directly here or by posting at an image hosting site such as postimage.org.
thanks.

--

in the meantime, note that one can solve this problem rather easily by just listing all the possibilities for the three rides. if the cars are #1, #2, and #3, then here are all of those possibilities:
1, 1, 1
1, 1, 2
1, 1, 3
1, 2, 1
1, 2, 2
1, 2, 3 *
1, 3, 1
1, 3, 2 *
1, 3, 3
2, 1, 1
2, 1, 2
2, 1, 3 *
2, 2, 1
2, 2, 2
2, 2, 3
2, 3, 1 *
2, 3, 2
2, 3, 3
3, 1, 1
3, 1, 2 *
3, 1, 3
3, 2, 1 *
3, 2, 2
3, 2, 3
3, 3, 1
3, 3, 2
3, 3, 3
there are only twenty-seven possibilities in this list, so just about everyone here should be able to make the whole list in substantially less than the allotted time.

of the possibilities in the list, exactly six (marked by asterisks) satisfy the desired condition (= each of cars #1, #2, #3 appears exactly one time), so the requisite probability is 6/27, which reduces to 2/9.