From a group of 3 boys and 3 girls, 4 children are to be : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 01:00

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 12 Mar 2009
Posts: 311
Followers: 3

Kudos [?]: 322 [0], given: 1

From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

22 Jan 2010, 23:42
3
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

72% (02:08) correct 28% (01:19) wrong based on 208 sessions

### HideShow timer Statistics

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
Joined: 09 May 2009
Posts: 203
Followers: 1

Kudos [?]: 225 [0], given: 13

### Show Tags

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

Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93390 [2] , given: 10557

### Show Tags

23 Jan 2010, 01:24
2
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 $$\frac{4!}{2!2!}$$.

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

Kudos [?]: 6 [4] , given: 4

Re: Probability; Equal number of boys & girls [#permalink]

### Show Tags

06 Jul 2010, 11:06
4
KUDOS
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

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

Kudos [?]: 6 [0], given: 4

Re: Probability; Equal number of boys & girls [#permalink]

### Show Tags

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

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)
Ms. Big Fat Panda
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1922
Concentration: General Management, Nonprofit
Followers: 447

Kudos [?]: 1979 [0], given: 210

Re: Probability; Equal number of boys & girls [#permalink]

### Show Tags

06 Jul 2010, 12:41
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
Joined: 21 Feb 2010
Posts: 212
Followers: 1

Kudos [?]: 28 [0], given: 1

Re: Probability; Equal number of boys & girls [#permalink]

### Show Tags

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
Status: The last round
Joined: 18 Jun 2009
Posts: 1310
Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 79

Kudos [?]: 1005 [0], given: 157

Re: Probability; Equal number of boys & girls [#permalink]

### Show Tags

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.
_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 958 [2] , given: 25

### Show Tags

24 Oct 2010, 23:14
2
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
_________________
Manager
Joined: 20 Aug 2011
Posts: 146
Followers: 3

Kudos [?]: 95 [1] , given: 0

### Show Tags

30 Nov 2011, 00:34
1
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
Joined: 09 Sep 2013
Posts: 13494
Followers: 576

Kudos [?]: 163 [0], given: 0

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

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.
_________________
Director
Joined: 23 Jan 2013
Posts: 579
Schools: Cambridge'16
Followers: 1

Kudos [?]: 42 [0], given: 40

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

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
Joined: 18 Aug 2014
Posts: 132
Location: Hong Kong
Schools: Mannheim
Followers: 1

Kudos [?]: 66 [0], given: 36

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

14 Feb 2015, 05:31
Can somebody explain why we need to use combinatorics here and not solely probability (1/6 * 4) ?
Intern
Joined: 15 Feb 2015
Posts: 14
Followers: 0

Kudos [?]: 5 [0], given: 1

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

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
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 8314
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Followers: 381

Kudos [?]: 2467 [1] , given: 163

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

27 Feb 2015, 13:15
1
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

[Reveal] Spoiler:
D

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

# Rich Cohen

Co-Founder & GMAT Assassin

# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Optimus Prep Instructor Joined: 06 Nov 2014 Posts: 1782 Followers: 51 Kudos [?]: 393 [0], given: 21 Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] ### Show Tags 28 Feb 2015, 01:32 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
_________________

# Janielle Williams

Customer Support

Special Offer: $80-100/hr. Online Private Tutoring GMAT On Demand Course$299
Free Online Trial Hour

Intern
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]

### Show Tags

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!!!)

VP
Joined: 08 Jul 2010
Posts: 1443
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 68

Kudos [?]: 1412 [0], given: 42

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

17 Jun 2015, 03:58
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

Feel free to give a Kudos if it is a useful post .

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13494
Followers: 576

Kudos [?]: 163 [0], given: 0

Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink]

### Show Tags

06 Jul 2016, 15:41
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.
_________________
Re: From a group of 3 boys and 3 girls, 4 children are to be   [#permalink] 06 Jul 2016, 15:41
Similar topics Replies Last post
Similar
Topics:
2 In a group of 9 children, there are twice as many girls as boys, and 3 28 Mar 2016, 07:34
3 Group A has 2 boys and 3 girls,group B has 3 boys and 4 girls and grou 1 25 Jun 2015, 21:24
25 5 girls and 3 boys are arranged randomly in a row 9 19 Feb 2012, 05:08
72 The ratio of boys to girls in Class A is 3 to 4. The ratio 42 03 Apr 2010, 11:57
10 From a group of 3 boys and 3 girls, 4 children are to be ran 10 19 Aug 2009, 08:24
Display posts from previous: Sort by