Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 12 Mar 2009
Posts: 301

From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
23 Jan 2010, 00:42
Question Stats:
73% (01:05) correct 27% (01:18) wrong based on 243 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
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 49271

Re: probability
[#permalink]
Show Tags
23 Jan 2010, 02:24
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= 12/5=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 the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  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? Extrahard Quant Tests with Brilliant Analytics




Manager
Joined: 09 May 2009
Posts: 187

Re: probability
[#permalink]
Show Tags
23 Jan 2010, 01:58
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= 12/5=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



Intern
Joined: 03 Jun 2010
Posts: 22

Re: Probability; Equal number of boys & girls
[#permalink]
Show Tags
06 Jul 2010, 12:06
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
Joined: 03 Jun 2010
Posts: 22

Re: Probability; Equal number of boys & girls
[#permalink]
Show Tags
06 Jul 2010, 13:34
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)



SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1868
Concentration: General Management, Nonprofit

Re: Probability; Equal number of boys & girls
[#permalink]
Show Tags
06 Jul 2010, 13:41
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: 187

Re: Probability; Equal number of boys & girls
[#permalink]
Show Tags
06 Jul 2010, 14:11
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: 1234
Concentration: Strategy, General Management

Re: Probability; Equal number of boys & girls
[#permalink]
Show Tags
09 Jul 2010, 00:56
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 680My 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



Retired Moderator
Joined: 02 Sep 2010
Posts: 772
Location: London

Re: Probability problem
[#permalink]
Show Tags
25 Oct 2010, 00:14
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 writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 20 Aug 2011
Posts: 129

Re: probability....
[#permalink]
Show Tags
30 Nov 2011, 01:34
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.



Director
Joined: 23 Jan 2013
Posts: 591

Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
14 Feb 2015, 02:11
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: 120
Location: Hong Kong

Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
14 Feb 2015, 06:31
Can somebody explain why we need to use combinatorics here and not solely probability (1/6 * 4) ?



Intern
Joined: 15 Feb 2015
Posts: 13

Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
27 Feb 2015, 09:53
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



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12415
Location: United States (CA)

Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
27 Feb 2015, 14:15
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: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************



SVP
Joined: 06 Nov 2014
Posts: 1887

Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
28 Feb 2015, 02:32
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.optimusprep.com/gmatondemandcourse



Intern
Joined: 14 Apr 2015
Posts: 5
Concentration: Human Resources, Technology
GPA: 3.5
WE: Information Technology (Computer Software)

Re: Probability; Equal number of boys & girls
[#permalink]
Show Tags
17 Jun 2015, 03:18
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!!!)



SVP
Joined: 08 Jul 2010
Posts: 2313
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
17 Jun 2015, 04:58
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 ALTERNATIVELYProbability = 1 (Unfavorable Outcomes / Total Outcomes)Total Outcomes = ways of selecting 4 out of 6 children = 6 C4 = 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 = 3 C3 * 3 C1 = 1*3 = 3 and Unfavorable Outcome_2 = 1 boys and 3 girls selected out of 3 boys and 3 girls = 3 C1 * 3 C3 = 3*1 = 3 Probability = 1 [(3+3) / 15] = 1  [6/15] = 9/15 = 3/5
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835

Re: From a group of 3 boys and 3 girls, 4 children are to be
[#permalink]
Show Tags
07 Dec 2017, 10:32
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 We are given that from a group of 3 boys and 3 girls, 4 children are to be randomly selected. We need to determine the probability that an equal numbers of boys and girls will be selected, that is, the probability that two boys and two girls are selected. We can use combinations to determine the number of favorable outcomes (that 2 boys and 2 girls are selected) and the total number of outcomes (that 4 children are selected from 6 children). Let’s first determine the number of ways we can select 2 boys from 3 boys and 2 girls from 3 girls. # of ways to select 2 boys from a total of 3 boys: 3C2 = 3 # of ways to select 2 girls from a total of 3 girls: 3C2 = 3 Thus, the number of ways to select 2 girls and 2 boys = 3 x 3 = 9. Now we can determine the total number of ways to select 4 children from a total of 6 children. 6C4 = (6 x 5 x 4 x 3)/(4 x 3 x 2 x 1) = 3 x 5 = 15 Thus, the probability of selecting an equal number of girls and boys is 9/15 = 3/5. Answer: D
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: From a group of 3 boys and 3 girls, 4 children are to be &nbs
[#permalink]
07 Dec 2017, 10:32






