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 350,000 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:

Atleast 2 are dice need to have same number, which means 2,3,4,5 and 6 can have same.

So we can compute the prob of all have different and subtract that from 1 which will give us the answer.

For the first dice, the prob is 1 since any value is fine.
For the second, the prob has to be other than first. So 5/6
For third, the prob has to be other than first and second. So 4/6.
For fourth, 3/6
For fifth. 2/6
For sixth, only one remains so 1/6.

So the probability is (1).(5/6).(4/6).(3/6).(2/6).(1/6) = 5/324.
So the answer is 1-5/324

why do not we start with 6/6 and end with 2/6, instead we start with 5/6 and end with 1/6 ?

for example:

to find number of ways 6 objects can be organized in group of 6 (order matters) 6*5*4*3*2*1

to find number of ways 6 objects can be organized in group of 4 (order matters) 6*5*4*3

from here i thought we should start with 6/6 and end with 2/6...?

Please explain y not, thank you.

I think what defenestrate is trying to get at is the fact that 6/6 is just 1, i.e. doesn't matter what number we get in the first roll. What we are concerned with is the fact that this number not show up again, hence starting with 5/6 (the prob of 2nd roll) and so on and so forth. In the case of 6 rolls, the last roll has a prob of 1/6 [5/6*4/6*3/6*2/6*1/6] = 5!/6^5 which is the same as 6!/6^6 if you put 6/6 as the first roll instead of 1.

In the case of 5 rolls, the last roll has a prob of 2/6 [5/6*4/6*3/6*2/6]= 5!/6^4 which is the same as 6!/6^5 if you put 6/6 as the first roll instead of 1.

Additionally, for the sake of completeness, the prob of 5 dice is [1- (120/1296)] = 49/54 not 8/9.

For my Cambridge essay I have to write down by short and long term career objectives as a part of the personal statement. Easy enough I said, done it...