It is currently 10 Dec 2017, 23:07

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

1 KUDOS received
CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1085 [1], given: 4

Location: New York City
A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 16 Nov 2007, 07:46
1
This post received
KUDOS
9
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

83% (01:13) correct 17% (01:57) wrong based on 375 sessions

HideShow timer Statistics

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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jul 2012, 02:43, edited 1 time in total.
Edited the question and added the OA.

Kudos [?]: 1085 [1], given: 4

2 KUDOS received
VP
VP
User avatar
Joined: 08 Jun 2005
Posts: 1141

Kudos [?]: 257 [2], given: 0

 [#permalink]

Show Tags

New post 16 Nov 2007, 15:58
2
This post received
KUDOS
1
This post was
BOOKMARKED
9/9*6/8*3/7 = 9/28

the answer is (D)

:)

Kudos [?]: 257 [2], given: 0

1 KUDOS received
Manager
Manager
avatar
Joined: 26 Sep 2007
Posts: 65

Kudos [?]: 29 [1], given: 5

 [#permalink]

Show Tags

New post 16 Nov 2007, 16:50
1
This post received
KUDOS
1
This post was
BOOKMARKED
Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28
So answer is D

Kudos [?]: 29 [1], given: 5

1 KUDOS received
Director
Director
avatar
Joined: 09 Aug 2006
Posts: 754

Kudos [?]: 265 [1], given: 0

 [#permalink]

Show Tags

New post 17 Nov 2007, 00:19
1
This post received
KUDOS
KillerSquirrel wrote:
9/9*6/8*3/7 = 9/28

the answer is (D)

:)


KS, can you please explain how you get 3/7?

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!

Kudos [?]: 265 [1], given: 0

VP
VP
User avatar
Joined: 08 Jun 2005
Posts: 1141

Kudos [?]: 257 [0], given: 0

 [#permalink]

Show Tags

New post 17 Nov 2007, 01:05
1
This post was
BOOKMARKED
GK_Gmat wrote:
KillerSquirrel wrote:
9/9*6/8*3/7 = 9/28

the answer is (D)

:)


KS, can you please explain how you get 3/7?

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 ?

:)

Kudos [?]: 257 [0], given: 0

SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2470

Kudos [?]: 867 [0], given: 19

Re: Probability - Marbles of each color [#permalink]

Show Tags

New post 17 Nov 2007, 01:15
1
This post was
BOOKMARKED
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?

2/21
3/25
1/6
9/28
11/24


alternatively:

= (3c1 x 3c1 x 3c1)/9c3
= 9/28

Kudos [?]: 867 [0], given: 19

CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1085 [0], given: 4

Location: New York City
 [#permalink]

Show Tags

New post 17 Nov 2007, 13:23
sevenplus wrote:
Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28
So answer is D


yep. most intuitive approach.

desired/total

OA is D

Kudos [?]: 1085 [0], given: 4

1 KUDOS received
Director
Director
avatar
Joined: 01 May 2007
Posts: 793

Kudos [?]: 390 [1], given: 0

 [#permalink]

Show Tags

New post 17 Nov 2007, 15:50
1
This post received
KUDOS
For me it was easier to just write it out.

I need 3 different colors so, we'll say I wanted a red, blue, yellow.

So chance of picking a red = 3/9

On 2nd pick chance of picking a blue = 3/8

On 3rd pick chance of picking a yellow = 3/7.

Then I listed out the different 3 ball combos I could have.

BRY
BYR
RBY
RYB
YRB
YBR

So 6 different combos. So the prob of pick r,b,y is 3/9 * 3/8 * 3/7 = 3/56

Multiply that by 6 and you have 9/28 prob of picking three different colored balls.

Kudos [?]: 390 [1], given: 0

Manager
Manager
avatar
Joined: 27 Oct 2008
Posts: 184

Kudos [?]: 166 [0], given: 3

Re: Probability - Marbles of each color [#permalink]

Show Tags

New post 28 Sep 2009, 04:16
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?

2/21
3/25
1/6
9/28
11/24

Soln:
= 3C1 * 3C1 * 3C1/9C3
= 27/(9 * 8 * 7/6)
= 27/84
= 9/28

Kudos [?]: 166 [0], given: 3

Senior Manager
Senior Manager
User avatar
Joined: 08 Nov 2010
Posts: 388

Kudos [?]: 133 [0], given: 161

WE 1: Business Development
GMAT ToolKit User
Re: Probability - Marbles of each color [#permalink]

Show Tags

New post 24 Aug 2011, 14:56
Hey guys,

I understand the solution, but i need someone to explain me why the following is not working:

total groups of 3 - 9C3

total options of 3 different color:
abc
acb
bac
bca
cab
cba

so - 6 options to get 3 different colors.

6/84 is not the answer.

please help me find the mistake. thanks.
_________________

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 133 [0], given: 161

1 KUDOS received
Manager
Manager
avatar
Joined: 25 May 2011
Posts: 144

Kudos [?]: 75 [1], given: 71

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 23 Dec 2011, 02:36
1
This post received
KUDOS
1
This post was
BOOKMARKED
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

Kudos [?]: 75 [1], given: 71

Intern
Intern
avatar
Joined: 04 Jun 2012
Posts: 2

Kudos [?]: [0], given: 0

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 16 Jul 2012, 23:56
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 .

Kudos [?]: [0], given: 0

Expert Post
7 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135179 [7], given: 12671

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 17 Jul 2012, 03:08
7
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
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!).


Combination approach:

\(P(BRY)=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}\).

Answer: D.

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135179 [7], given: 12671

Intern
Intern
avatar
Joined: 29 Mar 2014
Posts: 9

Kudos [?]: 7 [0], given: 0

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 29 Mar 2014, 17:19
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?

Kudos [?]: 7 [0], given: 0

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135179 [1], given: 12671

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 30 Mar 2014, 11:42
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???
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135179 [1], given: 12671

Intern
Intern
avatar
Joined: 29 Mar 2014
Posts: 9

Kudos [?]: 7 [0], given: 0

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 30 Mar 2014, 13:15
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???


Agreed, but for the sake of consistency, I see the combination formula written as C n over r
http://www.mathwords.com/c/combination_formula.htm

Thanks

Kudos [?]: 7 [0], given: 0

Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 298

Kudos [?]: 83 [0], given: 23

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 20 Apr 2014, 12:47
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!).


Combination approach:

\(P(BRY)=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}\).

Answer: D.

Hope it's clear.


Hi Bunuel,

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.

Kudos [?]: 83 [0], given: 23

Intern
Intern
avatar
B
Joined: 20 Feb 2016
Posts: 6

Kudos [?]: [0], given: 27

Premium Member CAT Tests
Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 23 Jun 2016, 21:08
shahideh wrote:
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



I'm trying to understand the basics, please excuse me if you find the question too silly:
Why don't we multiple by 3! in combinatorics method?
So first it's choosing i.e. 3C1 * 3C1* 3C1
Then why aren't the chosen balls arranged amongst themselves in 3! ways?

Kudos [?]: [0], given: 27

Expert Post
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5333

Kudos [?]: 6083 [0], given: 121

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 23 Jun 2016, 22:33
salonipatil wrote:
shahideh wrote:
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



I'm trying to understand the basics, please excuse me if you find the question too silly:
Why don't we multiple by 3! in combinatorics method?
So first it's choosing i.e. 3C1 * 3C1* 3C1
Then why aren't the chosen balls arranged amongst themselves in 3! ways?


hi,
We are not doing so, because it will not effect the answer..
if you consider ORDER important, even the TOTAL ways will be multiplied by 3!..
ans = \(p=\frac{C^3_1*C^3_1*C^3_1*3!}{C^9_3*3!}=\frac{9}{28}\)
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6083 [0], given: 121

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 14954

Kudos [?]: 287 [0], given: 0

Premium Member
Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

Show Tags

New post 29 Jun 2017, 15:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 287 [0], given: 0

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3   [#permalink] 29 Jun 2017, 15:18
Display posts from previous: Sort by

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.