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:

Simplify this further for the lack of A/C's.. = (5^4 + 4*5^3)/6^4
= (5*5^3 + 4*5^3)/6^4
= (5+4)*5^3/6^4
= 9*5^3//(3*2)^4
= 3^2*5^3/3^4*2^4
= 5^3/3^2*2^4
= 125/9*16
= 125/144
very high probability indeed

Yes I am reviewing probability from a college text book. I am trying random problems in the back - some may not be direct gmat material but its practice nonetheless.

What would be the probability of getting atleast one 3. I am just trying to clear my concepts on alteast, atmost.. for probability.. its confusing..
_________________

Success is my only option, failure is not -- Eminem

bewakoof,
The probability of getting at least one 3 in three rolls is the same as:

1 - the probability of getting no 3's in 3 rolls...

Since there is a 1/6 probability that you will get at least one 3, the probability that you will get no 3s is 5/6. Thus, the prob that on all 3 rolls the die will not yield a 3 is:

5/6 * 5/6 * 5/6 = 125/216

Now we need to subtract this from one. The prob that at least one of the rolls will be a 3 includes every outcome, except when three consecutive non-3s are rolled. Therefore:

1- 125/216 = 91/216 = the prob that at least one 3 will be rolled

(I have illustrated the idea using three rolls) The question above involves four. I thought it was three and i don't want to go back and change it The same concept can be used though.
Hope this helps!

Simplify this further for the lack of A/C's.. = (5^4 + 4*5^3)/6^4 = (5*5^3 + 4*5^3)/6^4 = (5+4)*5^3/6^4 = 9*5^3//(3*2)^4 = 3^2*5^3/3^4*2^4 = 5^3/3^2*2^4 = 125/9*16 = 125/144 very high probability indeed

I am getting 125/216. Add p(xxxx) and p(one 3)
125/216