GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Nov 2018, 18:10

### 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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # From a group of 3 boys and 3 girls, 4 children are to be  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Senior Manager Joined: 12 Mar 2009 Posts: 282 From a group of 3 boys and 3 girls, 4 children are to be [#permalink] ### Show Tags 22 Jan 2010, 23:42 1 2 00:00 Difficulty: 25% (medium) Question Stats: 75% (01:24) correct 25% (01:46) wrong based on 249 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 ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 50585 Re: probability [#permalink] ### Show Tags 23 Jan 2010, 01:24 2 3 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. _________________ ##### General Discussion Manager Joined: 09 May 2009 Posts: 181 Re: probability [#permalink] ### 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 Intern Joined: 02 Jun 2010 Posts: 21 Re: Probability; Equal number of boys & girls [#permalink] ### Show Tags 06 Jul 2010, 11:06 4 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: 02 Jun 2010 Posts: 21 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 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: 1856 Concentration: General Management, Nonprofit 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: 184 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: 1223 Concentration: Strategy, General Management GMAT 1: 680 Q48 V34 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: 769 Location: London Re: Probability problem [#permalink] ### Show Tags 24 Oct 2010, 23:14 2 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: 128 Re: probability.... [#permalink] ### Show Tags 30 Nov 2011, 00:34 1 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: 570 Schools: Cambridge'16 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: 119 Location: Hong Kong Schools: Mannheim 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: 13 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 Answer is D EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12856 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: From a group of 3 boys and 3 girls, 4 children are to be [#permalink] ### Show Tags 27 Feb 2015, 13:15 1 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 Co-Founder & 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: 1882
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
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, 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!!!)

CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2698
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, 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 I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: From a group of 3 boys and 3 girls, 4 children are to be  [#permalink]

### Show Tags

07 Dec 2017, 09:32
1
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.

_________________

Jeffery Miller

GMAT Quant Self-Study 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, 09:32
Display posts from previous: Sort by