Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 14 Feb 2016, 00:11

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A certain roller coaster has 3 cars, and a passenger is

Author Message
TAGS:
SVP
Joined: 28 May 2005
Posts: 1723
Location: Dhaka
Followers: 6

Kudos [?]: 142 [0], given: 0

A certain roller coaster has 3 cars, and a passenger is [#permalink]  13 Oct 2005, 12:49
5
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

58% (02:05) correct 42% (00:57) wrong based on 319 sessions
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
[Reveal] Spoiler: OA

_________________

hey ya......

Last edited by Bunuel on 07 Jun 2012, 08:50, edited 2 times in total.
Edited the OA
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1713
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 82

Kudos [?]: 628 [6] , given: 109

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]  28 Feb 2012, 10:24
6
KUDOS
+1 C

Let's suppose that the cars are A, B, and C.
Now, let's find the probability of picking the cars in this order: A-B-C

$$\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}$$

However, there could be more way to organize A-B-C. The numbers of ways is 3! = 6

Then, $$\frac{1}{27} * 6 = \frac{6}{27} = \frac{2}{9}$$

_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Math Expert
Joined: 02 Sep 2009
Posts: 31304
Followers: 5365

Kudos [?]: 62574 [0], given: 9457

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]  21 Feb 2012, 00:30
Expert's post
6
This post was
BOOKMARKED
nakib77 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

First time a passenger can ride any car: p=1;
Second time the passenger should ride another car: p=2/3;
Third time the passenger should ride not the 2 cars he had already driven: p=1/3.

So $$P=1*\frac{2}{3}*\frac{1}{3}=\frac{2}{9}$$.

OR:
Total # of ways the passenger can drive 3 cars (for 3 rides) is 3^3 (for each ride passenger has 3 options);
# of ways the passenger can drive 3 different cars is 3! (ABC, ACB, BAC, BCA, CAB, CBA);

So $$P=\frac{3!}{3^3}=\frac{2}{9}$$.

_________________
Senior Manager
Joined: 28 Mar 2012
Posts: 287
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 23

Kudos [?]: 291 [0], given: 23

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]  04 Jun 2012, 04:30
Hi,

Probability = (favorable cases)/(total number of cases)

Total number of ways in which a person can ride car = 3*3*3 = 27
(In first ride he has 3 options to sit, in second right again he has 3 seats available to sit and so on)

Number of favorable cases, i.e., when he rides on different cars;
He can choose seat car in 3 ways in his 1st ride.
He can choose seat car in 2 ways in his 2nd ride.
He can choose seat car in 1 ways in his 3rd ride.
So, 3*2*1 = 6 ways

Thus, probability of choosing different seats = 6/27 = 2/9
(C)

Regards,
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 8242
Followers: 419

Kudos [?]: 111 [0], given: 0

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]  01 Dec 2014, 05:12
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 8242
Followers: 419

Kudos [?]: 111 [0], given: 0

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]  29 Dec 2015, 11:15
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 07 Jul 2014
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 25

A certain roller coaster has 3 cars, and a passenger is [#permalink]  08 Feb 2016, 01:48
Bunuel wrote:
nakib77 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

First time a passenger can ride any car: p=1;
Second time the passenger should ride another car: p=2/3;
Third time the passenger should ride not the 2 cars he had already driven: p=1/3.

So $$P=1*\frac{2}{3}*\frac{1}{3}=\frac{2}{9}$$.

OR:
Total # of ways the passenger can drive 3 cars (for 3 rides) is 3^3 (for each ride passenger has 3 options);
# of ways the passenger can drive 3 different cars is 3! (ABC, ACB, BAC, BCA, CAB, CBA);

So $$P=\frac{3!}{3^3}=\frac{2}{9}$$.

I did -

For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride he has to choose one out of the two left, so the probability would be - 1/2
And last because he is left with only one ride to choose from to have ridden each once, probability would be - 1/1

Total probability = 1/3*1/2*1 = 1/6

I understood the solution but need to understand why this is incorrect. Thanks.
Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 2071
Followers: 83

Kudos [?]: 972 [0], given: 37

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]  08 Feb 2016, 01:59
Expert's post
ParmarKP wrote:

I did -

For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride he has to choose one out of the two left, so the probability would be - 1/2
And last because he is left with only one ride to choose from to have ridden each once, probability would be - 1/1

Total probability = 1/3*1/2*1 = 1/6

I understood the solution but need to understand why this is incorrect. Thanks.

HI,
why you are going wrong is because you are not applying the info of Q correctly..
Quote:
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

so each time he plans to take the ride he can take any of the three again...
BUT in your solution, you assume that a ride once taken cannot be taken again..

so the solution on your lines would be:-
For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride too, he has to choose one out of the three, so the probability would be - 1/3
And last because he is left with only one ride to choose, but he has to choose again from all three, probability would be - 1/3

Total probability = 1/3*1/3*1/3 = 1/27..
but within these three the order can differ and there will be 3! ways..

hope it helps
_________________

BACK IN CLUB AFTER A GAP OF 6 YEARS

http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

Re: A certain roller coaster has 3 cars, and a passenger is   [#permalink] 08 Feb 2016, 01:59
Similar topics Replies Last post
Similar
Topics:
5 40% of all high school students hate roller coasters; the rest love th 6 01 Jul 2015, 01:18
In a certain parking lot, 3% of the cars are towed for parki 4 02 Jan 2014, 18:41
10 A parking garage has places for a certain number of cars. If 10 17 Sep 2010, 00:05
5 A certain roller coaster has 3 cars, and a passenger is 6 26 Aug 2010, 06:26
20 A certain roller coaster has 3 cars, and a passenger is 8 19 Nov 2007, 17:55
Display posts from previous: Sort by