Last visit was: 23 Apr 2026, 08:57 It is currently 23 Apr 2026, 08:57
Home
Messages
Tests
Schools
Events
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
Back to Forum
Create Topic
User avatar
slingfox
User avatar
Joined: 18 Oct 2009
Last visit: 29 Dec 2010
Posts: 41
Own Kudos:
1,562
 [1]
Given Kudos: 3
Concentration: Finance
Schools:Kellogg
Posts: 41
Kudos: 1,562
 [1]
A bag contains a total of four billiard balls. The billiard Updated on: 09 Jul 2013, 10:43
Originally posted by slingfox on 26 Oct 2009, 00:39.
Last edited by Bunuel on 09 Jul 2013, 10:43, edited 1 time in total.
Renamed the topic and edited the question.
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:

33% (02:10) correct 67% (01:52) wrong based on 18 sessions

History

Date
Time
Result
Not Attempted Yet
A bag contains a total of four billiard balls. The billiard balls are numbered as follows: 2, 3, 5 and 7.

Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd?

Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,778
Own Kudos:
810,800
 [1]
Given Kudos: 105,853
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,778
Kudos: 810,800
 [1]
A bag contains a total of four billiard balls. The billiard 26 Oct 2009, 00:54
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
slingfox
Can you guys help me figure out the most efficient way to do this type of problem?:

A bag contains a total of four billiard balls. The billiard balls are numbered as follows: 2, 3, 5 and 7.

Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd?

Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd?
Show more

Scenario 1:

P(sum odd) = (Favorable outcomes)/(Total # of outcomes)
Total # of outcomes = 4C2 = 6, # of selection of 2 balls out of 4 balls.
Favorable outcome: two numbers can total odd if one number is even 2 and another odd 3, 5, 7. So two ball must be (2,3), (2,5), (2,7). So we have 3 favorable outcomes.

P(sum odd) = (Favorable outcomes)/(Total # of outcomes) = 3/6 = 1/2.

Scenario 2:

Different situation. But again the odd sum can occur if either of drawing gives even 2 and another odd 3, 5, 7.

This can happen in TWO ways:
even (2), odd (3,5,7); OR
odd(3,5,7), even(2).

P(sum odd) = 2(as we have 2 ways favorable sum can occur)*1/4(probability of picking even 2, 1 ball out of 4 balls)*3/4(probability of picking odd 3,5 or 7, 3 out of 4 balls) = 2*1/4*3/4 = 3/8
User avatar
slingfox
User avatar
Joined: 18 Oct 2009
Last visit: 29 Dec 2010
Posts: 41
Own Kudos:
1,562
Given Kudos: 3
Concentration: Finance
Schools:Kellogg
Posts: 41
Kudos: 1,562
A bag contains a total of four billiard balls. The billiard 26 Oct 2009, 05:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks Bunuel. You are my hero!
User avatar
rvthryet
User avatar
Joined: 02 Oct 2009
Last visit: 14 Mar 2016
Posts: 15
Own Kudos:
273
Given Kudos: 5
Posts: 15
Kudos: 273
A bag contains a total of four billiard balls. The billiard 26 Oct 2009, 07:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
slingfox
Can you guys help me figure out the most efficient way to do this type of problem?:

A bag contains a total of four billiard balls. The billiard balls are numbered as follows: 2, 3, 5 and 7.

Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd?

Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd?

Scenario 1: P(sum odd)=Favorable outcomes/Total # of outcomes
Total # of outcomes=4C2=6, # of selection of 2 balls out of 4 balls.
Favorable outcome: two numbers can total odd if one number is even 2 and another odd 3,5,7. So two ball must be (2,3),(2,5),(2,7). So we have 3 favorable outcomes.

P(sum odd)=Favorable outcomes/Total # of outcomes=3/6=1/2

Scenario 2: Different situation. But again the odd sum can occur if either of drawing gives even 2 and another odd 3,5,7.

This can happen in TWO ways:
even (2), odd (3,5,7); OR
odd(3,5,7), even(2).

P(sum odd)=2(as we have 2 ways favorable sum can occur)*1/4(probability of picking even 2, 1 ball out of 4 balls)*3/4(probability of picking odd 3,5 or 7, 3 out of 4 balls)=2*1/4*3/4=3/8
Show more

In the first case we consider (2,3),(2,5),(2,7) but why do we not consider (3,2), (5,2), (7,2). But then total no. of cases will be 6 and Probability cannot be 6/6 = 1. I m too confused :shock: Help!!!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,778
Own Kudos:
810,800
 [1]
Given Kudos: 105,853
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,778
Kudos: 810,800
 [1]
A bag contains a total of four billiard balls. The billiard 26 Oct 2009, 09:03
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rvthryet

In the first case we consider (2,3),(2,5),(2,7) but why do we not consider (3,2), (5,2), (7,2). But then total no. of cases will be 6 and Probability cannot be 6/6 = 1. I m too confused :shock: Help!!!
Show more

Yes, that is confusing. But consider this:

Scenario 1: P(sum odd)=Favorable outcomes/Total # of outcomes
Total # of outcomes=4C2=6, # of selection of 2 balls out of 4 balls.
Favorable outcome: two numbers can total odd if one number is even 2 and another odd 3,5,7. So two ball must be (2,3),(2,5),(2,7). So we have 3 favorable outcomes.

P(sum odd)=Favorable outcomes/Total # of outcomes=3/6=1/2

Favorable outcome is the PAIR of balls, therefore it doesn't matter it'll be (2,3) or (3,2), it's still ONE pair. On the other hand for the total # of outcomes we are also counting number of drawing of TWO balls out of 4.

BUT we can count this in another way (the way we did for scenario 2):
First ball 2, seond any odd =1/4*3/3(as after drawing the even ball there are 3 balls left and all are odd)=1/4
First ball any odd, second ball 2 =3/4*1/3(as after drawing the odd ball there are 3 balls left and only one is even)=1/4.
Final P=1/4+1/4=1/2

OR: 2*1/4*3/3=1/2



Scenario 2: Different situation. But again the odd sum can occur if either of drawing gives even 2 and another odd 3,5,7.

This can happen in TWO ways:
even (2), odd (3,5,7); OR
odd(3,5,7), even(2).

P(sum odd)=2(as we have 2 ways favorable sum can occur)*1/4(probability of picking even 2, 1 ball out of 4 balls)*3/4(probability of picking odd 3,5 or 7, 3 out of 4 balls)=2*1/4*3/4=3/8


We also can count the probability here with combination:
2*1C1*3C1/4C1*4C1=2*(1*3)/(4*4)=3/8

1C1=# of selections of even out of one
3C1=# of selections of odd out of three
4C1=TOTAL # of selections of first ball out of 4
4C1=TOTAL # of selections of second ball out of 4 (as after drawing the first one we put it back, so there are still 4 balls in the bag)
Multiplying this by 2 as there are two ways: ood/even or even/odd.

Hope it's clear.
User avatar
rvthryet
User avatar
Joined: 02 Oct 2009
Last visit: 14 Mar 2016
Posts: 15
Own Kudos:
273
Given Kudos: 5
Posts: 15
Kudos: 273
A bag contains a total of four billiard balls. The billiard 26 Oct 2009, 16:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel


BUT we can count this in another way (the way we did for scenario 2):
First ball 2, seond any odd =1/4*3/3(as after drawing the even ball there are 3 balls left and all are odd)=1/4
First ball any odd, second ball 2 =3/4*1/3(as after drawing the odd ball there are 3 balls left and only one is even)=1/4.
Final P=1/4+1/4=1/2

Show more

I like this method more. :) And i know u get this a lot these days but this is awesome stuff.. Thnks a ton..
User avatar
amit2k9
User avatar
Joined: 08 May 2009
Last visit: 18 Jun 2017
Posts: 535
Own Kudos:
646
Given Kudos: 10
Status:There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Posts: 535
Kudos: 646
A bag contains a total of four billiard balls. The billiard 09 May 2011, 00:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1 3/6 = 1/2

2 let 1st ball be 2 then 2nd ball has to be 3,4,5 thus probability = 1/4 * 3/4

let 1st ball be either 3,4 or 5 and 2nd ball be 2 then = 3/4 * 1/4

thus 2* 1/4 * 3/4
User avatar
subhashghosh
User avatar
Retired Moderator
Joined: 16 Nov 2010
Last visit: 25 Jun 2024
Posts: 894
Own Kudos:
1,302
Given Kudos: 43
Location: United States (IN)
Concentration: Strategy, Technology
Products:
My Reviews
Posts: 894
Kudos: 1,302
A bag contains a total of four billiard balls. The billiard 10 May 2011, 07:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Scenario 1:

1 ball has to be 2

(1 * 3)/(4C2) = 3/6 = 1/2

Scenario 2:

P(2)*P(any other number) + P(Any other number) * P(2)

= 2 * 1/4 * 3/4

= 6/16

= 3/8
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,965
Own Kudos:
1,117
Posts: 38,965
Kudos: 1,117
A bag contains a total of four billiard balls. The billiard 14 Dec 2017, 05:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109778 posts
Krunaal
Tuck School Moderator
853 posts