M15#18 : Retired Discussions [Locked]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 23:00

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M15#18

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Intern
Joined: 19 Jun 2008
Posts: 20
Followers: 0

Kudos [?]: 21 [0], given: 0

M15#18 [#permalink]

### Show Tags

19 Nov 2008, 18:24
7
This post was
BOOKMARKED
A basket contains 3 blue marbles, 3 red marbles, 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) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
Intern
Joined: 11 Jan 2010
Posts: 6
Followers: 0

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

Re: M15#18 [#permalink]

### Show Tags

08 Mar 2011, 22:44
4
This post received
KUDOS
The faster way to calculate the probability is to follow events manually:
1. The first marble will be of any one color. Probability (P1) is 100% or 1;
2. The second marble should be of other two colors (any of 6 out of 8 remaining marbles). Probability (P2) is 6/8 or 3/4;
3. The third marble should be of the last color (any of 3 out of 7 remaining marbles). Probability (P3) is 3/7;

P=P1*P2*P3=1*3/4*3/7=9/28

Answer is D
Intern
Joined: 11 Jan 2010
Posts: 39
Followers: 1

Kudos [?]: 53 [2] , given: 9

Re: M15#18 [#permalink]

### Show Tags

02 Mar 2010, 07:25
2
This post received
KUDOS
Out of 3 blue marbles, 3 red marbles, and 3 yellow marbles, probability of selecting one marble of each color is as follows:

B R Y
(3/9) * (3/8) * (3/7)

B, R, and Y can be arranged in factorial (3) = 3 * 2 * 1 ways.

Thus, resulting probability is:

(3/9) * (3/8) * (3/7) * (3 * 2 * 1) = 9/28

D is the answer.
SVP
Joined: 17 Jun 2008
Posts: 1569
Followers: 11

Kudos [?]: 250 [1] , given: 0

Re: M15#18 [#permalink]

### Show Tags

19 Nov 2008, 23:26
1
This post received
KUDOS
(3C1 * 3C1 * 3C1) / 9C3
Intern
Joined: 10 Aug 2012
Posts: 19
Location: India
Concentration: General Management, Technology
GPA: 3.96
Followers: 0

Kudos [?]: 6 [1] , given: 15

Re: M15#18 [#permalink]

### Show Tags

11 Mar 2013, 01:41
1
This post received
KUDOS
For first ball...we can pick any one out of nine..(lets say first is blue)

For second ball of different color...we have choose one ball out of 8 balls
and color of ball should not blue..it may be of red or yellow color...
so probability is 6(3 red and 3 yellow)/8...(lets say we pick a red one here)

For third ball of different color...we have choose one ball out of 7 balls
and color of ball should yellow..
so probability is 3 (yellow)/7

So probability that a marble of each color is among the extracted
$$= \frac{6}{8}*\frac{3}{7}$$
$$= \frac{9}{28}$$
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7103

Kudos [?]: 93596 [1] , given: 10579

Re: M15#18 [#permalink]

### Show Tags

11 Mar 2013, 01:49
1
This post received
KUDOS
Expert's post
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, 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) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

A basket contains 3 blue marbles, 3 red marbles, 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. $$\frac{2}{21}$$
B. $$\frac{3}{25}$$
C. $$\frac{1}{6}$$
D. $$\frac{9}{28}$$
E. $$\frac{11}{24}$$

$$P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}$$. Numerator represents ways to select 1 blue marble out of 3, 1 red marble out of 3 and 1 yellow marble out of 3. Denominator represents total ways to select 3 marbles out of 9.

Answer: D.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7103

Kudos [?]: 93596 [1] , given: 10579

Re: M15#18 [#permalink]

### Show Tags

13 Mar 2013, 01:35
1
This post received
KUDOS
Expert's post
ranjitaarao wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.

Instead if this was a case of 'OR', we would have added the events.

Hope that helped.

Can someone tell me what is the difficulty level of this question?

The difficulty level is ~650.
_________________
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 68

Kudos [?]: 736 [0], given: 19

Re: M15#18 [#permalink]

### Show Tags

19 Nov 2008, 22:44
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, 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?

(C) 2008 GMAT Club - m15#18

* $$\frac{2}{21}$$
* $$\frac{3}{25}$$
* $$\frac{1}{6}$$
* $$\frac{9}{28}$$
* $$\frac{11}{24}$$

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

= (9c1/9c1) (6c1/8c1) (3c1/7c1)
= 1 (6/8) (3/7)
= 9/28
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Intern
Joined: 19 Jun 2008
Posts: 20
Followers: 0

Kudos [?]: 21 [0], given: 0

Re: M15#18 [#permalink]

### Show Tags

20 Nov 2008, 10:08
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 911 [0], given: 334

Re: M15#18 [#permalink]

### Show Tags

17 Dec 2008, 07:27
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.

Hope this helps.
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Please read this before posting in GMAT Club Tests forum
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 01 Feb 2010
Posts: 267
Followers: 1

Kudos [?]: 55 [0], given: 2

Re: M15#18 [#permalink]

### Show Tags

02 Mar 2010, 06:17
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, 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) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

p = (3c1*3c1*3c1)/9c3 = 9/28 hence D.
Intern
Joined: 06 Jan 2010
Posts: 19
Schools: Wake Forest Evening
WE 1: ~12 years total
Followers: 0

Kudos [?]: 3 [0], given: 33

Re: M15#18 [#permalink]

### Show Tags

02 Mar 2010, 06:58
Sorry, how is (3c1*3c1*3c1)/9c3 = 9/28. I keep getting 9/84? 3c1 cubed is 9. Why is 9c3 28? I am sure it has to do with order or something or am I just missing something?

Thanks in advance.

Edit: Crap because 3 cubed is not 9. 27/84 == 9/28. I really hope I don't make stupid mistakes like this on the 17th when I take the test. (It is taking some will power not to edit my question above.)
Intern
Joined: 30 Oct 2009
Posts: 4
Followers: 0

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

Re: M15#18 [#permalink]

### Show Tags

02 Mar 2010, 08:18
3/9 * 3/8 * 3/7 * 3! = 9/28
Intern
Joined: 23 Dec 2009
Posts: 49
Schools: HBS 2+2
WE 1: Consulting
WE 2: Investment Management
Followers: 1

Kudos [?]: 40 [0], given: 7

Re: M15#18 [#permalink]

### Show Tags

02 Mar 2010, 21:41
I'm still a bit confused. I got the denominator correct (9C3), which enumerates all the possible outcomes. But I only see 6 (3!) desired outcomes...not 27:

BRY
BYR
YRB
YBR
RBY
RYB

What am I missing?

EDIT: I think this is only taking into account only one of three of each color...i.e. B1 Y1 R1, etc.

Let me know if I am right!
_________________

My GMAT quest...

...over!

Senior Manager
Joined: 13 Dec 2009
Posts: 263
Followers: 10

Kudos [?]: 184 [0], given: 13

Re: M15#18 [#permalink]

### Show Tags

13 Mar 2010, 08:15
(3C1 * 3C1 * 3C1)/9C3

D is the right answer.
_________________

My debrief: done-and-dusted-730-q49-v40

Manager
Status: A continuous journey of self-improvement is essential for every person -Socrates
Joined: 02 Jan 2011
Posts: 72
Followers: 1

Kudos [?]: 3 [0], given: 14

Re: M15#18 [#permalink]

### Show Tags

07 Mar 2011, 20:36
dzyubam wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.

Hope this helps.

Please explain why we have written 9C3 in the denominator.
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 911 [0], given: 334

Re: M15#18 [#permalink]

### Show Tags

08 Mar 2011, 00:30
Denominator contains the number of total possible outcomes. In this case, the number of possible outcomes equals the number of ways 3 marbles can be drawn from the total of 9 marbles available (9C3).
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Please read this before posting in GMAT Club Tests forum
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 15 Apr 2011
Posts: 49
Followers: 0

Kudos [?]: 7 [0], given: 8

Re: M15#18 [#permalink]

### Show Tags

10 Mar 2012, 00:35
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, 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) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

Got D!!!

First Draw - (9 marbles/ 9 marbles) = 1
Second Draw - Since one color is selected & we dont want that same one again... (6 marbles/ 8 remaining marbles) = 3/4
Third Draw - Since two colors are selected & we dont want the same ones again... (3 marbles/ 7 remaining marbles) = 3/7

Multiply all three to get the probability...
1 * (3/4) * (3/7) = 9/28
Intern
Joined: 04 Sep 2012
Posts: 15
Location: India
WE: Marketing (Consumer Products)
Followers: 0

Kudos [?]: 18 [0], given: 14

Re: M15#18 [#permalink]

### Show Tags

12 Mar 2013, 22:36
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.

Instead if this was a case of 'OR', we would have added the events.

Hope that helped.

Can someone tell me what is the difficulty level of this question?
Current Student
Joined: 28 Feb 2013
Posts: 15
Location: United States
Concentration: Statistics, Strategy
GMAT Date: 03-12-2014
GRE 1: 324 Q154 V170
WE: Information Technology (Other)
Followers: 1

Kudos [?]: 11 [0], given: 18

Re: M15#18 [#permalink]

### Show Tags

24 Feb 2014, 10:03
nevermind... i realize i was forgetting the parenthetical part of the selection. in the denominator... its follish lapses like these that affect my scores and keep me in the 600s
Re: M15#18   [#permalink] 24 Feb 2014, 10:03

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by

# M15#18

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

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO 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®.