Last visit was: 15 Dec 2024, 12:45 It is currently 15 Dec 2024, 12:45
Home
GMAT
Messages
Tests
Schools
Reviews
Social
Chat
My Profile
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
avatar
stolyar
avatar
Joined: 03 Feb 2003
Last visit: 06 May 2014
Posts: 1,012
Own Kudos:
1,754
 []
Posts: 1,012
Kudos: 1,754
 []
A certain box holds 3 green balls, 2 white balls, and 1 blue 15 Jul 2003, 09:44
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

History

Date
Time
Result
Not Attempted Yet
A certain box holds 3 green balls, 2 white balls, and 1 blue ball. Three balls are taken at random without repetition. What is the probability of having the blue ball among the taken ones?



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
avatar
stolyar
avatar
Joined: 03 Feb 2003
Last visit: 06 May 2014
Posts: 1,012
Own Kudos:
1,754
 []
Posts: 1,012
Kudos: 1,754
 []
A certain box holds 3 green balls, 2 white balls, and 1 blue 15 Jul 2003, 10:01
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The only thing that I can add is that it is 1/2.

Let us reformulate the question: what is the probability of NOT having the blue ball among the taken ones?
User avatar
evensflow
User avatar
Joined: 08 Apr 2003
Last visit: 04 Feb 2004
Posts: 66
Own Kudos:
155
Posts: 66
Kudos: 155
A certain box holds 3 green balls, 2 white balls, and 1 blue Updated on: 16 Jul 2003, 02:24
Originally posted by evensflow on 15 Jul 2003, 10:09.
Last edited by evensflow on 16 Jul 2003, 02:24, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Select all three which are not blue..

So, 5C3/6C3.

Opps!! it was a typo. I corrected the mistake.
User avatar
RK73
User avatar
Joined: 07 Jul 2003
Last visit: 07 Aug 2003
Posts: 33
Own Kudos:
1
Posts: 33
Kudos: 1
A certain box holds 3 green balls, 2 white balls, and 1 blue 15 Jul 2003, 19:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Stop! i stuck!

let's do it slowly -
when we choose three balls, we have 6c3 possibilities.
from these choices we have 3c1 outcomes with blue one.
where I am wrong?
User avatar
AkamaiBrah
User avatar
GMAT Instructor
Joined: 07 Jul 2003
Last visit: 24 Jun 2009
Posts: 392
Own Kudos:
497
 []
Location: New York NY 10024
Concentration: Finance
Schools:Haas, MFE; Anderson, MBA; USC, MSEE
Posts: 392
Kudos: 497
 []
A certain box holds 3 green balls, 2 white balls, and 1 blue 16 Jul 2003, 01:18
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
RK73
Stop! i stuck!

let's do it slowly -
when we choose three balls, we have 6c3 possibilities.
from these choices we have 3c1 outcomes with blue one.
where I am wrong?


There are 6c3 or 20 possible combinations of 3 balls.

To calculate the number of possible arrangement of 3 balls which include a blue ball assume that the arrangement has a blue ball. Now calculate the number of ways that 2 balls can be combined of the 5 left to accompany the blue one. This is 5c2 = 10. Hence, the probability is 1/2.

Another way to do this is say:
Suppose I pull the balls out 1 by 1 (w/o replacement). For the 3 balls to have the blue one, it must be the 1st, 2nd, or 3rd ball.

Pr(1st) = 1/6
pr(2nd) = 5/6*1/5 = 1/6
pr(3rd) = 5/6*4/5*1/4 = 1/6

These are mutually exclusive events, so we can add them together to get a total probability of 1/2.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 15 Dec 2024
Posts: 97,886
Own Kudos:
686,153
Given Kudos: 88,273
Products:
GMAT Club Tests Forum Quiz
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 97,886
Kudos: 686,153
A certain box holds 3 green balls, 2 white balls, and 1 blue 08 Nov 2024, 06:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Testing...............................................
Moderators:
Bunuel
Math Expert
97886 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3116 posts