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:

it is: consider the most expensive 2 books....than we have 6 slots free to combine the books. the total number of combs is 8C3 because we have a total of 8 books to combine in groups of three...prob=6/8*7=3/28

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]
09 Jan 2008, 05:44

i thought about it like this.

probability = desired outcomes / total out comes

where desired outcomes is the number of outcomes of picking 3 out of 8 books, but the two most expensive have to be in there, so ive got (1C1)*(1C1)*(6C1) / 8C3 = 6/56 = 3/28

I think 2! is permutation: P^2_2. But you use C^8_3 for all variants that mean ABC and BAC are the same variant. If we distinguish between ABC and BAC, we use P^8_3 and P^3_3

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]
17 Feb 2010, 02:44

pretttyune wrote:

Please help me to understand this... I am always terrified when i face PROBABILITY questions...

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Please helpe me kindly and thank you in advance...

Out of 8 .. 2 books (which are most exp) should be considered in your selection.

No of ways the 2 Exp books can be selected = 2c2 = 1

remaining we need to choose one more book from the 6 left over books = 6c1 = 6

Total ways of selecting 3 books out of 8 = 8c3

Prob = 1X6 / 8c3 = 3/28 _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]
05 Aug 2013, 20:14

Probability of selecting most expensive out of 8 =1/8 Probability of selecting next most expensive out of 7 =1/7 And out of 6 left out can select any one out of 6 = 6c1/6c1 now probability of selecting 2 most expensive = 1/8*1/7*1*3! =3/8(3! is because of arrangements)

gmatclubot

Re: (probability) 3 out of 8, the most expensive two included...
[#permalink]
05 Aug 2013, 20:14

It’s here. Internship season. The key is on searching and applying for the jobs that you feel confident working on, not doing something out of pressure. Rotman has...