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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

There are 3 boogies in a fun-train and a passenger is equall [#permalink]

Show Tags

16 Jun 2004, 20:57

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

88% (01:02) correct 13% (02:13) wrong based on 13 sessions

HideShow timer Statistics

There are 3 boogies in a fun-train and a passenger is equally to ride in any one of the 3 boogies each time he rides the fun-train. If a certain passenger is to ride the fun-train 3 times, what is the probability that the client would ride in each of the three boogies.

Mark his attempts as 1,2,3
For each attempts he can choose among cars a,b,c
So there are 3 vriations for each attempt possible
Total attempts = 9
out of this only one combination makes him ride different cars
p = 1/9

Sorry, thats not the answer I have here.....could you give another try.
I have changed the question a little bit, instead of cars, i have made it boogies in a train.

What is a better way to do ... use the way you have done or use the short-cut (formulae) as Halle has done. Well, i do not know the formulae... whats the approcah for such problems., If someone sees a problem like this, where do we start with. A tip in this, would clear my fear on this pal.

I told you about brute force. What would you do if there were 10 cars.
You cannot use brute force. Just go through these problems at the club and make notes. You should be able to do good. More problems you solve more you will learn. You can try to remember the formulae but problems will generally be twisted in which case u need to apply more than one formula.

- Anand.

Let me know if you still need explanation of this problem.

Yeah it should not matter how many cars are there. The only way I can explain is you try to pick a problem and solve it. I will try to find where u are going wrong and advise you.

For the above problem assume u have 10 cars (A,B,C....J ) or you could number them from 0 to 9
Each time you try to rent a car I will give you whatever is available.
It is possible that in three times you get car number 5 or 4 or anything from 0 to 9
So no of combinations is simply 10 * 10 * 10
If you had rented 5 times then no of combinations is simply
10*10*10*10*10 or 10^5

So the formula is (no of variations) ^ ( no of repeatations )
Something like how many different numbers can you make using 2 digits.

Now you have to choose 3 different cars from 10. You do that in 10C3 ways.
Say you chose 3,4,5
But in first attempt you can get 3 second 4 and third 5 or
first 5, second 3 and third 4
Therefore there are actually 3! variations.
So Desired combinations = (Choosing 3 cars ) * (arranging 3 cars )
= 10C3 * 3!
Total combinations = 10^3

Please explain your answer step-by-step... thanks guys

There are 3 boogies in a fun-train and a passenger is equally to ride in any one of the 3 boogies each time he rides the fun-train. If a certain passenger is to ride the fun-train 3 times, what is the probability that the client would ride in each of the three boogies.

Again, please explain the steps to obtain the solution.

Regards

P(first time he rides in 1st boogie, second time he rides in the second, third time he rides in the third boogie) = 1/3*1/3*1/3 = 1/27.

P(each time he rides in different boogies) = 3!*P(first = 1, second = 2, third = 3) = 6/27 = 2/9.