Author 
Message 
TAGS:

Hide Tags

CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
Show Tags
16 Nov 2007, 07:46
1
This post received KUDOS
5
This post was BOOKMARKED
Question Stats:
81% (02:14) correct
19% (02:01) wrong based on 384 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
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 17 Jul 2012, 02:43, edited 1 time in total.
Edited the question and added the OA.



VP
Joined: 08 Jun 2005
Posts: 1145

2
This post received KUDOS
9/9*6/8*3/7 = 9/28
the answer is (D)



Manager
Joined: 26 Sep 2007
Posts: 65

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



Director
Joined: 09 Aug 2006
Posts: 754

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 = 12/8 = 6/8
Now what?? How do I proceed? Thanks!



VP
Joined: 08 Jun 2005
Posts: 1145

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 = 12/8 = 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 ?



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: Probability  Marbles of each color [#permalink]
Show Tags
17 Nov 2007, 01:15
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?
2/21 3/25 1/6 9/28 11/24
alternatively:
= (3c1 x 3c1 x 3c1)/9c3
= 9/28



CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

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



Director
Joined: 01 May 2007
Posts: 793

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.



Manager
Joined: 27 Oct 2008
Posts: 185

Re: Probability  Marbles of each color [#permalink]
Show Tags
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



Senior Manager
Joined: 08 Nov 2010
Posts: 408
WE 1: Business Development

Re: Probability  Marbles of each color [#permalink]
Show Tags
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



Manager
Joined: 25 May 2011
Posts: 152

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
Show Tags
23 Dec 2011, 02:36
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



Intern
Joined: 04 Jun 2012
Posts: 2

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



Math Expert
Joined: 02 Sep 2009
Posts: 39743

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
Show Tags
17 Jul 2012, 03:08
7
This post received KUDOS
Expert's post
1
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: 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? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 39743

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
Show Tags
14 Aug 2013, 02:11



Intern
Joined: 29 Mar 2014
Posts: 9

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



Math Expert
Joined: 02 Sep 2009
Posts: 39743

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
Show Tags
30 Mar 2014, 11:42



Intern
Joined: 29 Mar 2014
Posts: 9

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



Senior Manager
Joined: 15 Aug 2013
Posts: 311

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



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16017

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
Show Tags
14 Oct 2015, 17:09
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



Intern
Joined: 20 Feb 2016
Posts: 6

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




Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3
[#permalink]
23 Jun 2016, 21:08



Go to page
1 2
Next
[ 21 posts ]





Similar topics 
Author 
Replies 
Last post 
Similar Topics:


1


There are 3 red and 3 blue marbles

rsgooga 
0 
27 Jun 2017, 01:15 



A bag contains 3 green marbles, three red marbles and 2 blue marbles.

Bunuel 
4 
15 Jun 2017, 16:26 

9


A jar contains 16 marbles, of which 4 are red, 3 are blue, a

banksy 
6 
28 May 2016, 03:35 

15


A jar contains only red, yellow, and orange marbles. If there are 3 re

vanidhar 
6 
04 Jun 2016, 09:06 

98


A bag of 10 marbles contains 3 red marbles and 7 blue

bmwhype2 
37 
21 Jan 2017, 20:42 



