It is currently 19 Nov 2017, 07:46

### 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
Your Progress

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 20 Sep 2006
Posts: 58

Kudos [?]: 20 [2], given: 0

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

### Show Tags

19 Nov 2007, 18:55
2
This post received
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

57% (00:42) correct 43% (00:36) wrong based on 410 sessions

### HideShow timer Statistics

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-roller-coaster-has-3-cars-and-a-passenger-is-21226.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jul 2013, 10:37, edited 1 time in total.
Edited the question and added the OA.

Kudos [?]: 20 [2], given: 0

Current Student
Joined: 02 Apr 2012
Posts: 77

Kudos [?]: 60 [6], given: 155

Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]

### Show Tags

17 Jul 2013, 10:33
6
This post received
KUDOS
4
This post was
BOOKMARKED
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?

I couldn´t solve it using the reverse combination approach. Anyone knows how?. But here are the other 3 ways:

Probability approach:

First ride: he can choose between 3 cars: 1/3
Second ride: he can choose only (3 less 1) cars: 2/3
Third ride: he can choose the only one in which he hadn´t ride.

Then:
$$\frac{1}{3}*\frac{2}{3} *1=\frac{2}{9}$$

Reverse probability approach:

P = 1-q.
q = probability that he rides only in one or only in two cars, but not in the three.

$$q = 1 * \frac{1}{3} * \frac{1}{3}* 3 + 1 * \frac{1}{3} * \frac{1}{3} * 4 = \frac{1}{9} *3 + \frac{1}{9} * 4 = \frac{7}{9}$$

First term I multiply by 3 because he could use A, B or C. Second term multuply by four because he could use A&B, A&C, C&B or B&C.

$$P=1-\frac{7}{9}= \frac{2}{9}$$

Combinatory approach:

$$C^3_1$$ * $$C^2_1$$ * $$C^1_1$$ = $$3*2*1= 6$$

Total combinations: $$(C^3_1)^3 = 27$$

$$\frac{6}{27}=\frac{2}{9}$$
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Kudos [?]: 60 [6], given: 155

Math Expert
Joined: 02 Sep 2009
Posts: 42247

Kudos [?]: 132680 [6], given: 12331

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]

### Show Tags

17 Jul 2013, 10:38
6
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
Maxirosario2012 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?

I couldn´t solve it using the reverse combination approach. Anyone knows how?. But here are the other 3 ways:

Probability approach:

First ride: he can choose between 3 cars: 1/3
Second ride: he can choose only (3 less 1) cars: 2/3
Third ride: he can choose the only one in which he hadn´t ride.

Then:
$$\frac{1}{3}*\frac{2}{3} *1=\frac{2}{9}$$

Reverse probability approach:

P = 1-q.
q = probability that he rides only in one or only in two cars, but not in the three.

$$q = 1 * \frac{1}{3} * \frac{1}{3}* 3 + 1 * \frac{1}{3} * \frac{1}{3} * 4 = \frac{1}{9} *3 + \frac{1}{9} * 4 = \frac{7}{9}$$

First term I multiply by 3 because he could use A, B or C. Second term multuply by four because he could use A&B, A&C, C&B or B&C.

$$P=1-\frac{7}{9}= \frac{2}{9}$$

Combinatory approach:

$$C^3_1$$ * $$C^2_1$$ * $$C^1_1$$ = $$3*2*1= 6$$

Total combinations: $$(C^3_1)^3 = 27$$

$$\frac{6}{27}=\frac{2}{9}$$

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

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-roller-coaster-has-3-cars-and-a-passenger-is-21226.html
_________________

Kudos [?]: 132680 [6], given: 12331

SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1088 [2], given: 5

Location: New York
Re: Probability [#permalink]

### Show Tags

26 Aug 2008, 22:00
2
This post received
KUDOS
saviop 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

GOOD QUESTION.

SAY 123 are car numbers..
possible combinations 123,132,231,213,321,312

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

C
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Kudos [?]: 1088 [2], given: 5

Manager
Joined: 27 Oct 2008
Posts: 185

Kudos [?]: 166 [2], given: 3

Re: Probability [#permalink]

### Show Tags

27 Sep 2009, 23:27
2
This post received
KUDOS
3
This post was
BOOKMARKED
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

Soln:
If he is to ride 3 times and since he can choose any of the 3 cars each time, total number of ways is
= 3 * 3 * 3
= 27

Now the number of ways if he is to choose a different car each time is
= 3 * 2 * 1
= 6

So the probability is
= 6/27
= 2/9

Ans is C

Kudos [?]: 166 [2], given: 3

Manager
Joined: 03 Sep 2006
Posts: 233

Kudos [?]: 26 [1], given: 0

### Show Tags

19 Nov 2007, 19:37
1
This post received
KUDOS
1
This post was
BOOKMARKED
My ans is C:

1st ride - prob is 1/3
2nd ride - prob is 2/3
3rd ride - prob is 1 (as 1 car left ...)

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

Kudos [?]: 26 [1], given: 0

VP
Joined: 09 Jul 2007
Posts: 1098

Kudos [?]: 144 [0], given: 0

Location: London
Re: Probability [#permalink]

### Show Tags

19 Nov 2007, 19:18
1
This post was
BOOKMARKED
saviop 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

prob that any car 3/3
prob that any of left 2 cars 2/3
prob of last one 1/3

(3/3)*(2/3)*(1/3)=2/9

Kudos [?]: 144 [0], given: 0

Director
Joined: 03 Sep 2006
Posts: 864

Kudos [?]: 1105 [0], given: 33

Re: Probability [#permalink]

### Show Tags

26 Aug 2008, 22:15
$$(3*2*1)/3^3 = 2/9$$

Kudos [?]: 1105 [0], given: 33

Manager
Joined: 09 Aug 2010
Posts: 105

Kudos [?]: 58 [0], given: 7

Re: Probability [#permalink]

### Show Tags

02 May 2011, 21:35
1
This post was
BOOKMARKED
Everytime the passenger rides a car, he has 3 choices each time. 3 x 3 x 3

To ride 3 times without repeating a car = 3 x 2 x 1

So (3 x 2 x 1)/(3 x 3 x 3) = 2/9

Kudos [?]: 58 [0], given: 7

Re: Probability   [#permalink] 02 May 2011, 21:35
Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.