NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 10:11 It is currently 26 Apr 2024, 10:11
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

There are 5 red marbles, 3 blue marbles and 2 green marbles. If a use

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
mariyea
Manager
Manager
Joined: 30 Nov 2010
Posts: 191
Own Kudos [?]: 266 [3]
Given Kudos: 66
Schools:UC Berkley, UCLA
GMAT 2: 540 GMAT 3: 530
Send PM
There are 5 red marbles, 3 blue marbles and 2 green marbles. If a use [#permalink] New post   05 Feb 2011, 17:41
3
Bookmarks
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (00:36) correct 0% (00:00) wrong based on 12 sessions
Hide Show timer Statistics
There are 5 red marbles, 3 blue marbles and 2 green marbles. If a user chooses to use two marbles,what is the probability that the two marbles will be a different color?

The approach that I used was:

5/10*4/9=2/9
3/10*2/9=1/15
2/10*1/9=1/45

I added the results and came up with 9/45 then when that is simplified I get 1/5.
Then
1-1/5=4/5 this is the result I got but the answer says 31/45 after using the combination approach.

Shouldn't the two approaches bring about the same result? Or did I do something wrong?

Mari
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92946
Own Kudos [?]: 619200 [1]
Given Kudos: 81609
Send PM
CAT Tests Forum Quiz Mode
There are 5 red marbles, 3 blue marbles and 2 green marbles. If a use [#permalink] New post   05 Feb 2011, 17:53
1
Bookmarks
Expert Reply
mariyea wrote:
There are 5 red marbles, 3 blue marbles and 2 greeen marbles. If a user chooses to use two marbles,what is the probability that the two marbles will be a different color?

The approach that I used was:

5/10*4/9=2/9
3/10*2/9=1/15
2/10*1/9=1/45

I added the results and came up with 9/45 then when that is simplified I get 1/5.
Then
1-1/5=4/5 this is the result I got but the answer says 31/45 after using the combination approach.

Shouldn't the two approaches bring about the same result? Or did I do something wrong?

Mari


You did everything right except the addition:
\(\frac{2}{9}+\frac{1}{15}+\frac{1}{45}=\frac{10}{45}+\frac{3}{45}+\frac{1}{45}=\frac{14}{45}\) (not 9/45);

--> \(1-\frac{14}{45}=\frac{31}{45}\).
User avatar
fluke
Retired Moderator
Joined: 20 Dec 2010
Posts: 1114
Own Kudos [?]: 4702 [1]
Given Kudos: 376
Send PM
Re: There are 5 red marbles, 3 blue marbles and 2 green marbles. If a use [#permalink] New post   05 Feb 2011, 23:44
1
Bookmarks
Mariyea did it by finding out probability of selecting two marbles of same colors and then taking the complement.

The other way using combination;

The total number of ways of choosing any 2 marbles out of 10 is
\(C^{10}_2 = \frac{10*9}{2} = 45\)

5R(red marbles), 3B(blue marbles) and 2G(greeen marbles)

Favorable ways of selecting 2 distinct colored marbles:
1R out of 5R AND 1B out of 3B
OR
1R out of 5R AND 1G out of 2G
OR
1B out of 3B AND 1G out of 2G

\(C^5_1*C^3_1+C^5_1*C^2_1+C^3_1*C^2_1\)
\(=5*3+5*2+3*2\)
\(=15+10+6=31\)

Ans: \(P=\frac{31}{45}\)
User avatar
mariyea
Manager
Manager
Joined: 30 Nov 2010
Posts: 191
Own Kudos [?]: 266 [0]
Given Kudos: 66
Schools:UC Berkley, UCLA
GMAT 2: 540 GMAT 3: 530
Send PM
Re: There are 5 red marbles, 3 blue marbles and 2 green marbles. If a use [#permalink] New post   07 Feb 2011, 06:40
Bunuel wrote:
mariyea wrote:
There are 5 red marbles, 3 blue marbles and 2 greeen marbles. If a user chooses to use two marbles,what is the probability that the two marbles will be a different color?

The approach that I used was:

5/10*4/9=2/9
3/10*2/9=1/15
2/10*1/9=1/45

I added the results and came up with 9/45 then when that is simplified I get 1/5.
Then
1-1/5=4/5 this is the result I got but the answer says 31/45 after using the combination approach.

Shouldn't the two approaches bring about the same result? Or did I do something wrong?

Mari


You did everything right except the addition: 2/9+1/15+1/45=10/45+3/45+1/45=14/45 (not 9/45) --> 1-14/45=31/45.

Unbelievable! I can't believe I did that... Thanks Bunuel! I don't know what's wrong with me..
User avatar
mariyea
Manager
Manager
Joined: 30 Nov 2010
Posts: 191
Own Kudos [?]: 266 [0]
Given Kudos: 66
Schools:UC Berkley, UCLA
GMAT 2: 540 GMAT 3: 530
Send PM
Re: There are 5 red marbles, 3 blue marbles and 2 green marbles. If a use [#permalink] New post   07 Feb 2011, 06:41
fluke wrote:
Mariyea did it by finding out probability of selecting two marbles of same colors and then taking the complement.

The other way using combination;

The total number of ways of choosing any 2 marbles out of 10 is
\(C^{10}_2 = \frac{10*9}{2} = 45\)

5R(red marbles), 3B(blue marbles) and 2G(greeen marbles)

Favorable ways of selecting 2 distinct colored marbles:
1R out of 5R AND 1B out of 3B
OR
1R out of 5R AND 1G out of 2G
OR
1B out of 3B AND 1G out of 2G

\(C^5_1*C^3_1+C^5_1*C^2_1+C^3_1*C^2_1\)
\(=5*3+5*2+3*2\)
\(=15+10+6=31\)

Ans: \(P=\frac{31}{45}\)

Yes this is the approach that I saw was being used...
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32688
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: There are 5 red marbles, 3 blue marbles, and 2 green [#permalink] New post   07 Feb 2021, 14:39
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 Club Bot
gmatclubot
Re: There are 5 red marbles, 3 blue marbles, and 2 green [#permalink]
07 Feb 2021, 14:39
Moderators:
Bunuel
Math Expert
92945 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions