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

It is currently 30 Aug 2015, 22:07
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

From a group of 3 boys and 3 girls, 4 children are to be

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 12 Mar 2009
Posts: 318
Followers: 1

Kudos [?]: 139 [0], given: 1

GMAT ToolKit User
From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 22 Jan 2010, 23:42
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

74% (02:13) correct 26% (01:22) wrong based on 116 sessions
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3
[Reveal] Spoiler: OA
Manager
Manager
avatar
Joined: 09 May 2009
Posts: 202
Followers: 1

Kudos [?]: 143 [0], given: 13

Re: probability [#permalink] New post 23 Jan 2010, 00:58
prob reqd= 1- probab of not having eq numbers

prob of not having eq no's is = prob of having 3 Boys and 1 girl or prob of having 3 girls and 1 boy
=(1* 3/15) +(3/15 * 1)=2/5

reqd prob= 1-2/5=3/5
_________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 29151
Followers: 4729

Kudos [?]: 49835 [1] , given: 7498

Re: probability [#permalink] New post 23 Jan 2010, 01:24
1
This post received
KUDOS
Expert's post
3
This post was
BOOKMARKED
xcusemeplz2009 wrote:
prob reqd= 1- probab of not having eq numbers

prob of not having eq no's is = prob of having 3 Boys and 1 girl or prob of having 3 girls and 1 boy
=(1* 3/15) +(3/15 * 1)=2/5

reqd prob= 1-2/5=3/5


In this case it's easier to calculate directly what is asked. Equal # of boys and girls out of 4 means 2 boys and 2 girls.

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3

\(\frac{3C2*3C2}{6C4}=\frac{3}{5}\)

OR: \(\frac{4!}{2!2!}*\frac{3}{6}*\frac{2}{5}*\frac{3}{4}*\frac{2}{3}=\frac{3}{5}\).

Here we are counting the probability of BBGG, this combination can occur in different # of ways: BGBG, GGBB, ... Total # of ways would be the # permutations of the letters BBGG, which is \(\frac{4!}{2!2!}\).

Answer: D.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

Intern
Intern
avatar
Joined: 02 Jun 2010
Posts: 29
Followers: 0

Kudos [?]: 2 [0], given: 4

GMAT ToolKit User
Re: Probability; Equal number of boys & girls [#permalink] New post 06 Jul 2010, 11:06
Hussain15 wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A.1/10
B.4/9
C.1/2
D.3/5
E.2/3



So we are asked - what is probability of selecting 2 boys and 2 girls right:

We were not told about the actual selection, hence Before finding the probability, we should find what the different ways in which we can select 4 children.
So select 1 boy, then 1 boy, then 1 girl and then 1 girl or you can also select 1 B, 1 G, 1 B and 1 G.
Basically selecting 4 from BBGG. ie. 4 * 3 * 2 *1 / (2! * 2!) ways of making the selection. = 6.

Probability of selecting 1st boy = 3/6
Probability of selecting 2nd boy = 2/5 (as we have already selected one boy above)
Probability of selecting 1st Girl = 3/4
Probability of selecting 2nd Girl = 2/3

Multiply all of above with different ways of selecting the children =
P (selecting 2 boys and 2 girls) = 3/6 * 2/5 * 3/4 * 2/3 * 6 = 3/5

Answer : D
Intern
Intern
avatar
Joined: 02 Jun 2010
Posts: 29
Followers: 0

Kudos [?]: 2 [0], given: 4

GMAT ToolKit User
Re: Probability; Equal number of boys & girls [#permalink] New post 06 Jul 2010, 12:34
surjoy wrote:
Hussain15 wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A.1/10
B.4/9
C.1/2
D.3/5
E.2/3



So we are asked - what is probability of selecting 2 boys and 2 girls right:

We were not told about the actual selection, hence Before finding the probability, we should find what the different ways in which we can select 4 children.
So select 1 boy, then 1 boy, then 1 girl and then 1 girl or you can also select 1 B, 1 G, 1 B and 1 G.
Basically selecting 4 from BBGG. ie. 4 * 3 * 2 *1 / (2! * 2!) ways of making the selection. = 6.

Probability of selecting 1st boy = 3/6
Probability of selecting 2nd boy = 2/5 (as we have already selected one boy above)
Probability of selecting 1st Girl = 3/4
Probability of selecting 2nd Girl = 2/3

Multiply all of above with different ways of selecting the children =
P (selecting 2 boys and 2 girls) = 3/6 * 2/5 * 3/4 * 2/3 * 6 = 3/5

Answer : D


There is actually much simpler approach for this problem.

P (selecting 2 boys and 2 girls) = (No. of ways of selecting 2 boys out of 3 * no. of ways of selecting 2 girls out of 3) / Total ways of selecting 4 out of 6 children
= 3C2 * 3C2 / 6C4 = 3/5 (D)
Expert Post
Ms. Big Fat Panda
Ms. Big Fat Panda
User avatar
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1923
Concentration: General Management, Nonprofit
Followers: 394

Kudos [?]: 1621 [0], given: 210

GMAT ToolKit User
Re: Probability; Equal number of boys & girls [#permalink] New post 06 Jul 2010, 12:41
Expert's post
Fairly straightforward question I think.

Ways to select 2 boys out of 3 = 3C2 = Ways to select 2 girls out of 3

Total ways to select 4 children = 6C4

So probability = \(\frac{3C2*3C2}{6C4} = \frac{3}{5}\)

Hope this helps!
Manager
Manager
avatar
Joined: 21 Feb 2010
Posts: 214
Followers: 1

Kudos [?]: 19 [0], given: 1

Re: Probability; Equal number of boys & girls [#permalink] New post 06 Jul 2010, 13:11
any bible for the GMAT probability and combination? i read the MGMAT, it helps a little...any better books out there?
Retired Moderator
User avatar
Status: The last round
Joined: 18 Jun 2009
Posts: 1316
Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 66

Kudos [?]: 667 [0], given: 157

GMAT ToolKit User
Re: Probability; Equal number of boys & girls [#permalink] New post 08 Jul 2010, 23:56
tt11234 wrote:
any bible for the GMAT probability and combination? i read the MGMAT, it helps a little...any better books out there?


Probability & Combination questions are not so common in GMAT. Hardly one can see 2 or max 3. So try to give your valuable time to the remaining 98% area of GMAT Quantative section. The concepts covered in MGMAT probability section are sufficient to answer a normal GMAT question. If you will go for a bible of GMAT probability, you will merely waste your time. So use this time to cover the topics which are most common in GMAT like number properties, Word problems & inequalities.

Best of luck.
_________________

[ From 470 to 680-My Story ] [ My Last Month Before Test ]
[ GMAT Prep Analysis Tool ] [ US. Business School Dashboard ] [ Int. Business School Dashboard ]

I Can, I Will

GMAT Club Premium Membership - big benefits and savings

2 KUDOS received
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 87

Kudos [?]: 639 [2] , given: 25

GMAT ToolKit User Reviews Badge
Re: Probability problem [#permalink] New post 24 Oct 2010, 23:14
2
This post received
KUDOS
Merging similar topics ....

monirjewel wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal number of boys and girls will be selected?


Total ways to pick children = C(6,4) = 15

Ways to pick 2 boys & 2 girls = C(3,2) x C(3,2) = 9

Probability = 9/15 = 3/5
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

GMAT Club Premium Membership - big benefits and savings

1 KUDOS received
Manager
Manager
avatar
Status: Meh, I can't take the GMAT before 2017.
Joined: 20 Aug 2011
Posts: 145
Followers: 3

Kudos [?]: 74 [1] , given: 0

Re: probability.... [#permalink] New post 30 Nov 2011, 00:34
1
This post received
KUDOS
There is only 1 way of selecting equal number of boys and girls i.e. 2 boys and 2 girls.

Boys- 3C2 Girls- 3C2
Total possible selections: 6C4

Probability= (3C2*3C2)/6C4 = 3/5
_________________

Hit kudos if my post helps you.
You may send me a PM if you have any doubts about my solution or GMAT problems in general.

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 6121
Followers: 342

Kudos [?]: 70 [0], given: 0

Premium Member
Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 14 Jan 2015, 00:44
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Senior Manager
Senior Manager
User avatar
Joined: 23 Jan 2013
Posts: 374
Schools: Cambridge'16
Followers: 2

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

CAT Tests
Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 14 Feb 2015, 01:11
The only way not presentsed is reverse combination, which is:

denominator: 6C4=15

numerator: (3C3*1C3)+(3C3*1C3)=6

so 1 - 6/15=9/15=3/5

D
Manager
Manager
User avatar
Joined: 18 Aug 2014
Posts: 120
Location: Hong Kong
Followers: 1

Kudos [?]: 54 [0], given: 33

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 14 Feb 2015, 05:31
Can somebody explain why we need to use combinatorics here and not solely probability (1/6 * 4) ?
Intern
Intern
avatar
Joined: 15 Feb 2015
Posts: 15
Followers: 0

Kudos [?]: 3 [0], given: 1

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 27 Feb 2015, 08:53
1st step: randomly select 4 from 6 = C(6 4) = 6! / 4!2! = 15
2nd step: select 2 girls from 3 girls = C(3 2) = 3! / 2!1! = 3
3rd step: select 2 boys from 3 boys = C(3 2) = 3! / 2!1! = 3
(3+3) / 15 = 3/5
Answer is D
Expert Post
1 KUDOS received
EMPOWERgmat Instructor
User avatar
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 3458
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Followers: 148

Kudos [?]: 931 [1] , given: 57

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 27 Feb 2015, 13:15
1
This post received
KUDOS
Expert's post
Hi LaxAvenger,

You COULD calculate the answer to this question using 'probability math', but the calculation would be far MORE complicated than what you wrote down. Here's how it would work....

First, you have to account for all of the different 'ways' to get 2 boys and 2 girls. Assuming that the children are chosen one at a time, here are the options that "fit" what we're looking for:

BBGG
BGBG
BGGB
GBBG
GBGB
GGBB

Using the first example, here is the probability of THAT EXACT sequence occurring:
BBGG = (3/6)(2/5)(3/4)(2/3) = 36/360 = 1/10

Each of the other 5 options will yield the exact SAME probability....
eg
BGBG = (3/6)(3/5)(2/4)(2/3) = 36/360 = 1/10

So we have 6 different options that each produce a 1/10 chance of occurring.

6(1/10) = 6/10 = 3/5

Final Answer:
[Reveal] Spoiler:
D


GMAT assassins aren't born, they're made,
Rich
_________________

Official Guide 2016 Question Breakdown:
http://gmatclub.com/forum/empowergmat-blog-198415.html#p1527977

Rich Cohen
Rich.C@empowergmat.com
http://www.empowergmat.com

GMAT Club Verified Reviews for EMPOWERgmat & Special Discount

EMPOWERgmat Podcast - A Wild Secret About The GMAT Algorithm


Expert Post
Optimus Prep Instructor
User avatar
Joined: 06 Nov 2014
Posts: 487
Followers: 6

Kudos [?]: 91 [0], given: 2

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 28 Feb 2015, 01:32
Expert's post
vaivish1723 wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3


Total children = 6
We need 2 boys and 2 girls.
Total ways of selecting boys = 3C2 = 3
Total ways of selecting girls = 3C2 = 3
Total ways of selecting 4 children = 6C4 = 15

Required probability = (3 * 3)/15
= 3/5
Hence option D.

--
Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimus-prep.com/gmat-on-demand-course
_________________

Cassandra Wolff
Customer Support | Optimus Prep
 
Facebook Linkedin Youtube Twitter slideshare Google+
 

Intern
Intern
avatar
Joined: 14 Apr 2015
Posts: 6
Concentration: Human Resources, Technology
GPA: 3.5
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 0 [0], given: 75

Re: Probability; Equal number of boys & girls [#permalink] New post 17 Jun 2015, 02:18
selecting equal no of boys and girl
selecting girl--> 3c2
selecting boy--> 3c2
total outcomes-->6c4
probab=3c2*3c2/6C4=3/5

Ans-->D
_________________

Regards,
YS
(I can,I WIL!!!)

Expert Post
Director
Director
User avatar
Joined: 08 Jul 2010
Posts: 762
GMAT: INSIGHT
Followers: 22

Kudos [?]: 416 [0], given: 30

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] New post 17 Jun 2015, 03:58
Expert's post
vaivish1723 wrote:
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3


ALTERNATIVELY

Probability = 1 -(Unfavorable Outcomes / Total Outcomes)

Total Outcomes = ways of selecting 4 out of 6 children = 6C4 = 15

Favorable Outcomes = 2 boys and 2 girls selected out of 3 boys and 3 girls
i.e. Unfavorable Outcomes = 3 boys and 1 girls selected out of 3 boys and 3 girls OR 1 boys and 3 girls selected out of 3 boys and 3 girls

i.e. Unfavorable Outcome_1 = 3 boys and 1 girls selected out of 3 boys and 3 girls = 3C3 * 3C1 = 1*3 = 3
and Unfavorable Outcome_2 = 1 boys and 3 girls selected out of 3 boys and 3 girls = 3C1 * 3C3 = 3*1 = 3

Probability = 1 -[(3+3) / 15] = 1 - [6/15] = 9/15 = 3/5
_________________

Prosper!!!

GMATinsight

Bhoopendra Singh and Dr.Sushma Jha

e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772

http://www.GMATinsight.com/testimonials.html


Contact for One-on-One LIVE Online (SKYPE Based) Quant/Verbal FREE Demo Class



______________________________________________________
Please press the Image if you appreciate this post !!

Re: From a group of 3 boys and 3 girls, 4 children are to be   [#permalink] 17 Jun 2015, 03:58
    Similar topics Author Replies Last post
Similar
Topics:
3 Experts publish their posts in the topic Group A has 2 boys and 3 girls,group B has 3 boys and 4 girls and grou kanigmat011 1 25 Jun 2015, 21:24
18 Experts publish their posts in the topic 5 girls and 3 boys are arranged randomly in a row Chembeti 8 19 Feb 2012, 05:08
1 3 boys and 3 girls shrive555 9 19 Aug 2011, 22:59
35 Experts publish their posts in the topic The ratio of boys to girls in Class A is 3 to 4. The ratio changhiskhan 29 03 Apr 2010, 11:57
7 Experts publish their posts in the topic From a group of 3 boys and 3 girls, 4 children are to be ran hrish88 10 19 Aug 2009, 08:24
Display posts from previous: Sort by

From a group of 3 boys and 3 girls, 4 children are to be

  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®.