Find all School-related info fast with the new School-Specific MBA Forum

It is currently 02 Sep 2015, 19:25
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A recent married couple decided to have four kids. What is

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
VP
VP
avatar
Joined: 10 Jun 2007
Posts: 1463
Followers: 6

Kudos [?]: 153 [0], given: 0

A recent married couple decided to have four kids. What is [#permalink] New post 25 Aug 2007, 07:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A recent married couple decided to have four kids. What is the probability that they will have exactly 2 boys and 2 girls?
VP
VP
User avatar
Joined: 08 Jun 2005
Posts: 1146
Followers: 6

Kudos [?]: 139 [0], given: 0

 [#permalink] New post 25 Aug 2007, 08:04
1
This post was
BOOKMARKED
the probability for G or B is 1/2

1/2*1/2*1/2*1/2 = 1/16

BBGG = 1/16
BGBG = 1/16
BGGB = 1/16
GBGB = 1/16
GBBG = 1/16
GGBB = 1/16

1/16+1/16+1/16+1/16+1/16+1/16 = 6/16 = 3/8

:-D
VP
VP
avatar
Joined: 10 Jun 2007
Posts: 1463
Followers: 6

Kudos [?]: 153 [0], given: 0

 [#permalink] New post 25 Aug 2007, 08:29
KillerSquirrel wrote:
the probability for G or B is 1/2

1/2*1/2*1/2*1/2 = 1/16

BBGG = 1/16
BGBG = 1/16
BGGB = 1/16
GBGB = 1/16
GBBG = 1/16
GGBB = 1/16

1/16+1/16+1/16+1/16+1/16+1/16 = 6/16 = 3/8

:-D


I'm trying to see what's wrong with my method...if you can help.
I got the same answer using the method above. However, when I use this method, I get different answer:
Say if you have the first 2 child, you have have either BB,BG,GB,GG. This means that the first 2 child doesn't affect the probability. So, wouldn't the last two child pick would be 1/2 * 1/2 = 1/4?
VP
VP
User avatar
Joined: 08 Jun 2005
Posts: 1146
Followers: 6

Kudos [?]: 139 [0], given: 0

 [#permalink] New post 25 Aug 2007, 09:18
bkk145 wrote:
KillerSquirrel wrote:
the probability for G or B is 1/2

1/2*1/2*1/2*1/2 = 1/16

BBGG = 1/16
BGBG = 1/16
BGGB = 1/16
GBGB = 1/16
GBBG = 1/16
GGBB = 1/16

1/16+1/16+1/16+1/16+1/16+1/16 = 6/16 = 3/8

:-D


I'm trying to see what's wrong with my method...if you can help.
I got the same answer using the method above. However, when I use this method, I get different answer:
Say if you have the first 2 child, you have have either BB,BG,GB,GG. This means that the first 2 child doesn't affect the probability. So, wouldn't the last two child pick would be 1/2 * 1/2 = 1/4?


I don't think you can solve this problem two kids at a time.

:-D
Director
Director
User avatar
Joined: 03 May 2007
Posts: 888
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 86 [0], given: 7

Re: Basic Probability [#permalink] New post 25 Aug 2007, 15:42
bkk145 wrote:
A recent married couple decided to have four kids. What is the probability that they will have exactly 2 boys and 2 girls?


2 b and 2 g = 4!/2!2! = 6
total = 2^4 = 16
prob = 6/16 = 3/8
VP
VP
User avatar
Joined: 08 Jun 2005
Posts: 1146
Followers: 6

Kudos [?]: 139 [0], given: 0

 [#permalink] New post 25 Aug 2007, 20:51
There is a faster way to solve those kind of questions, but I think this is out of scope.

Applying the binomial distribution formula:

http://www.answers.com/topic/binomial-distribution

C(n,k)*p^k*(1-p)^(n-k)

p = the probability for a certain event = 1/2 (getting a boy)

1-P = the completing event = 1/2 (getting a girl)

n = total items in the group = 4

k = desired outcome = two boys = 2

solving

C(4,2)*(1/2)^(2)*(1/2)*(4-2) = 6*1/4*1/4 = 6/16 = 3/8

:-D
Director
Director
User avatar
Joined: 03 May 2007
Posts: 888
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 86 [0], given: 7

 [#permalink] New post 25 Aug 2007, 22:33
KillerSquirrel wrote:
There is a faster way to solve those kind of questions, but I think this is out of scope.

Applying the binomial distribution formula:

http://www.answers.com/topic/binomial-distribution

C(n,k)*p^k*(1-p)^(n-k)

p = the probability for a certain event = 1/2 (getting a boy)

1-P = the completing event = 1/2 (getting a girl)

n = total items in the group = 4

k = desired outcome = two boys = 2

solving

C(4,2)*(1/2)^(2)*(1/2)*(4-2) = 6*1/4*1/4 = 6/16 = 3/8

:-D


yup i was looking for the binomial formula.
  [#permalink] 25 Aug 2007, 22:33
Display posts from previous: Sort by

A recent married couple decided to have four kids. What is

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.