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:

There are six balls in a black sack. Among those six are [#permalink]
27 Jun 2003, 01:56

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

There are six balls in a black sack. Among those six are black, white, and yellow balls. Three balls are taken at random without repetition. What is the probability that the white and yellow balls ARE taken, but the black one IS NOT?

1C1 + 1C1 + 1C1/ 6C3 = 1/20 The 3 grey balls are identical. Therefore, 1 of the 3 grey balls can be chosen in only 1 way. And since the balls have been chosen all at once WITHOUT REPETITION, the ordering does not matter. Therefore, IMO, it should be 1/20.

1C1 + 1C1 + 1C1/ 6C3 = 1/20 The 3 grey balls are identical. Therefore, 1 of the 3 grey balls can be chosen in only 1 way. And since the balls have been chosen all at once WITHOUT REPETITION, the ordering does not matter. Therefore, IMO, it should be 1/20.

I don't agree with you.

It should be 1C1 *1C1*3C1 / 6C3 = 3/20. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

There are six balls in a black sack. Among those six are 1 black, 1 white, 1 yellow and rest grey balls. Three balls are taken at random without repetition. What is the probability that the white and yellow balls ARE taken, but the black one IS NOT?

its pretty straight:

total possibilitites = 6c3 = 20 possibilities for white and yellow but a black = 3

Actually I meant to edit my response, but somehow it did not go through. Absolutely, as we have always done in the past, the answer should be 3/20. The one question I did have was, if Repetition vs. No Repetition should make a difference in our approach. In the same question, if balls are drawn one by one, the approach would be like this: 1/6 * 1/5 * 3/4 * 3! = 3/20. However, if all balls are drawn together, how do we assign 1st, 2nd or 3rd position to any of the balls? The approach should then be: 1C1 * 1C1 * 1C1/6C3 = 1/20. Even if we numbered the 3 grey balls for the sake of our clarification, W-Y-G1 is the same as W-Y-G2 or W-Y-G3. And there is only one chance that the person takes to draw the 3 balls. But we do know that in a total of 6C3 = 20 combinations, there are 3 combinations of W-Y-G. Hence, the prob of getting that combination is 1/20 * 3 = 3/20. All I am saying is that 3C1 sounds like" choosing 1 thing out of 3, which can be done in 3 ways", whereas there is only 1 way to choose the color.

there are 6*5*4 = 120 possible arrangements of the selected balls. Of those, 6 include white and yellow but not black..so 6/120

Black balls ARE part of that 6.

Here is what I think: From 1black, 1 white, 1 yellow, and 3 gray balls ---> With 3 draws (no repetition)

(1/6)(1/5)(3/4)= 3/120 = 1/40

where 1/6= prob of 1 white where 1/5 = prob of 1 yellow (after white is drawn) where 3/4= prob of selecting not black on final draw

am i right? or where did I go wrong?

here you have to multiply by3c2 , as there are 6 ways of picking . i.e the white can be first or second or third. similarly for the other colours. so the requirement is wyg, this can be picked wyg wgy gyw gwy ywg ygw. here order is important because the balls are not being taken out together, rather one by one. if they were to be taken out together , then the order would not matter.

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...