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:

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
16 Nov 2007, 06:46

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

81% (02:10) correct
19% (02:08) wrong based on 228 sessions

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

I'm assuming you did the following: The 1st chosen can be any color and it doesn't really matter so let's say the first chosen is Blue.

The probability that the 2nd chosen is not blue = 1-2/8 = 6/8

Now what?? How do I proceed? Thanks!

You are right about the first marble , you can choose any marble out of nine = 9/9 = 1 (nine out of nine).

For the second marble you are also correct - since you removed one marble from the basket you now have eight left but you can't choose the color already been choosen so 6/8 (six out of eight - three for one color and three for the other color).

For the third marble the logic is the same - you can choose one of the three marbles from the color not already choosen and you have seven mables to choose from meaning 3/7 (three out seven).

All of this is true only when the marbles taken out are not returned to the basket - if they were the probability then was ?? - can you solve this ?

Re: Probability - Marbles of each color [#permalink]
17 Nov 2007, 00:15

bmwhype2 wrote:

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

Re: Probability - Marbles of each color [#permalink]
28 Sep 2009, 03:16

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
23 Dec 2011, 01:36

1

This post received KUDOS

We can choose 3 balls from 9 balls. Total possibilities: \(C^9_3\) We should choose 1 ball from each color. Favorite possibilities: \(C^3_1*C^3_1*C^3_1\)

\(p=\frac{C^3_1*C^3_1*C^3_1}{C^9_3}=\frac{9}{28}\) D

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
16 Jul 2012, 22:56

Is ' picking of 3 marbles at random' same as 'picking 3 marbles one by on without replacement' ? Although looking at the solution these two scenarios appear to be similar. However, when 3 marbles are picked up together (in one shot) then should the problem not be treated differently. Please clarify .

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
17 Jul 2012, 02:08

3

This post received KUDOS

Expert's post

malikshilpa wrote:

Is ' picking of 3 marbles at random' same as 'picking 3 marbles one by on without replacement' ? Although looking at the solution these two scenarios appear to be similar. However, when 3 marbles are picked up together (in one shot) then should the problem not be treated differently. Please clarify .

Mathematically the probability of picking 3 marbles simultaneously, or picking them one at a time (without replacement) is the same.

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted? A. 2/21 B. 3/25 C. 1/6 D. 9/28 E. 11/24

We need to find the probability of BRY (a marble of each color).

Probability approach:

\(P(BRY)=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}\), we are multiplying by 3! since BRY scenario can occur in several different ways: BRY, RYB, RBY (# of permutations of 3 distinct letters BRY is 3!).

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
29 Mar 2014, 16:19

Just wanted to clarify here, as I feel like I'm looking at this wrong n should be 3 (because this is the total set) r should be 1 (this is the number of objects being extracted)

so shouldn't the combination formula be C 3 over 1? Instead of C 1 over 3?

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
30 Mar 2014, 10:42

1

This post received KUDOS

Expert's post

TheBookie wrote:

Just wanted to clarify here, as I feel like I'm looking at this wrong n should be 3 (because this is the total set) r should be 1 (this is the number of objects being extracted)

so shouldn't the combination formula be C 3 over 1? Instead of C 1 over 3?

\(C^1_3\), \(C^3_1\), 3C1, mean the same thing choosing 1 out of 3, just different ways to write. Can it be choosing 3 out of 1??? _________________

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
30 Mar 2014, 12:15

Bunuel wrote:

TheBookie wrote:

Just wanted to clarify here, as I feel like I'm looking at this wrong n should be 3 (because this is the total set) r should be 1 (this is the number of objects being extracted)

so shouldn't the combination formula be C 3 over 1? Instead of C 1 over 3?

\(C^1_3\), \(C^3_1\), 3C1, mean the same thing choosing 1 out of 3, just different ways to write. Can it be choosing 3 out of 1???

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
20 Apr 2014, 11:47

Bunuel wrote:

malikshilpa wrote:

Is ' picking of 3 marbles at random' same as 'picking 3 marbles one by on without replacement' ? Although looking at the solution these two scenarios appear to be similar. However, when 3 marbles are picked up together (in one shot) then should the problem not be treated differently. Please clarify .

Mathematically the probability of picking 3 marbles simultaneously, or picking them one at a time (without replacement) is the same.

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted? A. 2/21 B. 3/25 C. 1/6 D. 9/28 E. 11/24

We need to find the probability of BRY (a marble of each color).

Probability approach:

\(P(BRY)=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}\), we are multiplying by 3! since BRY scenario can occur in several different ways: BRY, RYB, RBY (# of permutations of 3 distinct letters BRY is 3!).

I have a question with the methodology here -- I can follow the work you're doing but i'm having a hard time understanding why. Can you please clarify the reasoning?

a) In your statement below, can you please elaborate WHY we multiply by 3!? -- [\(P=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}\)

If I reference the GMAT club book, I don't see it referencing the multiplication by a factorial(Page 109 of the book).

b) What is the difference between the method you used vs. doing the following? (9/9)(6/8)(3/7)

c) lastly, regarding your combinatorial approach, isn't that technically a reverse approach(pg 109 of the book)? Sorry if i'm being trivial here but i'm having a hard time connecting the dots hence the trivial questions. How can you use the 1C3 approach because don't you have 9 to begin with? Sorry - i'm going nuts here.

gmatclubot

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3
[#permalink]
20 Apr 2014, 11:47

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

As part of our focus on MBA applications next week, which includes a live QA for readers on Thursday with admissions expert Chioma Isiadinso, we asked our bloggers to...