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

 It is currently 15 Dec 2018, 01:43

### 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

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free GMAT Strategy Webinar

December 15, 2018

December 15, 2018

07:00 AM PST

09:00 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.
• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # A couple decides to have 4 children. If they succeed in havi  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 51215 A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 13 Mar 2014, 01:29 8 77 00:00 Difficulty: 85% (hard) Question Stats: 51% (01:43) correct 49% (01:38) wrong based on 1472 sessions ### HideShow timer Statistics The Official Guide For GMAT® Quantitative Review, 2ND Edition A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys? (A) 3/8 (B) 1/4 (C) 3/16 (D) 1/8 (E) 1/16 Problem Solving Question: 160 Category: Arithmetic Probability Page: 83 Difficulty: 700 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! _________________ ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51215 Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 13 Mar 2014, 01:29 7 20 SOLUTION A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys? (A) 3/8 (B) 1/4 (C) 3/16 (D) 1/8 (E) 1/16 $$P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}$$, we should mulitply by $$\frac{4!}{2!2!}$$ as the scenario GGBB can occur in several ways: GGBB, BBGG, GBGB, ... # of scenarios possible is # of permutations of 4 letters GGBB out of which 2 G's and 2 B's are identical and equals to $$\frac{4!}{2!2!}$$. Answer: A. _________________ ##### Most Helpful Community Reply Manager Status: suffer now and live forever as a champion!!! Joined: 01 Sep 2013 Posts: 111 Location: India Dheeraj: Madaraboina GPA: 3.5 WE: Information Technology (Computer Software) Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 25 Apr 2014, 00:06 36 12 no of ways of getting P(GGBB) is 4!/2!*2!; Total no of ways is 2^n =2^4 =16; 6/16 = 3/8; We can consider this question to a coin that is flipped for 4 times . what is the probability of getting exactly two heads . P(all out comes) = 1/2 *1/2 *1/2 *1/2 =1/16; P(favorable outcomes) = 4!/(2! * 2!) = 6/16 =3/8; (OR) Second Approach GBGB GGBB BBGG BGBG GBBG BGGB 6 possible ways . total no of ways is Baby can be a boy or a girl. For each baby the probability is 1/2 ;for 4 babies it's 1/16; 6/16 = 3/8; Press Kudos if this helps ##### General Discussion Math Expert Joined: 02 Sep 2009 Posts: 51215 Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 15 Mar 2014, 09:59 3 18 Retired Moderator Joined: 29 Oct 2013 Posts: 260 Concentration: Finance GPA: 3.7 WE: Corporate Finance (Retail Banking) Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 10 Aug 2014, 01:40 9 9 By fundamental counting principle, Total No of out comes: 2^4 = 16 Total Desired outcomes: No of ways to arrange BBGG by MISSISSIPPI rule= 4!/(2!*2!) = 6 Probability is 6/16 = 3/8 _________________ Please contact me for super inexpensive quality private tutoring My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876 Intern Joined: 07 Feb 2015 Posts: 46 Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags Updated on: 14 Mar 2015, 23:07 5 4 So here we have four positions and there are two options to fill each position so total number of cases=2x2x2x2=16 now we need 2boys and 2 girls OR we can say that we simply need 2 boys because if its not a boy it has to be a girl favourable cases=4C2=6 probability=6/16=3/8 (A) _________________ Kudos if it helped you in any way Originally posted by masoomdon on 14 Mar 2015, 00:30. Last edited by masoomdon on 14 Mar 2015, 23:07, edited 1 time in total. Manager Joined: 18 Aug 2014 Posts: 119 Location: Hong Kong Schools: Mannheim Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 14 Mar 2015, 07:14 2 1/2 is the chance of boy or girl 1/2 * 1/2 * 1/2 * 1/2 = 1/16 (4 children, boy or girl) Possible ways of 2 boys, 2 girls: GGBB BBGG GBGB BGBG BGGB GBBG = 6 ways we need "OR" --> 1/16 OR 1/16 .... 1/16 + 1/16 + 1/16 ....... +1/16 = 6/16 = 3/8 Answer A Intern Joined: 23 Sep 2011 Posts: 47 Location: Singapore Concentration: Finance, Entrepreneurship GPA: 3.44 WE: Information Technology (Investment Banking) Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 16 Jul 2015, 19:06 Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery. We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls. So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong? Intern Joined: 01 Jun 2013 Posts: 8 Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 03 Nov 2015, 20:17 1 2 Binomial Probability The probability of achieving exactly k successes in n trials is shown below. Formula: P(Probability of K successes in n trials) = nCk p^k q^n-k n = number of trials k = number of successes n – k = number of failures p = probability of success in one trial q = 1 – p = probability of failure in one trial According to our question n(4 children) = 4 k( we want exactly 2 girls) = 2 n – k = 2 p (probability of getting a girl in one trial) = 1/2 q = 1 – p = 1/2 4C2 * 1/2^2 * 1/2^2 =3/8 Binomial can also be used for problem like coins. Binomial is used when following conditions are satisfied. Fixed number of trials Independent trials Two different classifications Probability of success stays the same for all trials Director Joined: 17 Dec 2012 Posts: 632 Location: India Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 04 Nov 2015, 00:08 4 2 Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND Edition A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys? (A) 3/8 (B) 1/4 (C) 3/16 (D) 1/8 (E) 1/16 Let us handle a single case. First boy, second boy, third girl, fourth girl. The probability of this case is 1/2*1/2*1/2*1/2 = 1/16 Two boys or two girls can be born among 4 children in 4C2 ways= 6 ways So the required probability is: total number of ways * probability of one such way= 6* (1/16)=3/8 _________________ Srinivasan Vaidyaraman Sravna Holistic Solutions http://www.sravnatestprep.com Holistic and Systematic Approach Current Student Joined: 23 Feb 2015 Posts: 166 Schools: Duke '19 (M) Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 04 Feb 2016, 15:50 donkadsw wrote: Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery. We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls. So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong? Bump seeking an answer to the question above: I got 1/5 as well. BBBB BBBG BBGG GGGB GGGG 1/5 outcomes The question doesn't say anything about the order. How do we know when order matters? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8678 Location: Pune, India Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 04 Feb 2016, 21:35 2 Dondarrion wrote: donkadsw wrote: Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery. We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls. So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong? Bump seeking an answer to the question above: I got 1/5 as well. BBBB BBBG BBGG GGGB GGGG 1/5 outcomes The question doesn't say anything about the order. How do we know when order matters? In this case we are saying that the probability of BGGG is the same as the probability of BBGG. But that is not so. You can have 1 boy and 3 girls in 4 ways: BGGG, GBGG, GGBG, GGGB But you can have 2 boys and 2 girls in 6 ways: BBGG, BGGB, GGBB, BGBG, GBGB, GBBG So the probability depends on the number of ways in which you can get 2 boys and 2 girls. Think of it this way: if you throw two dice, is the probability of getting a sum of 2 same as the probability of getting a sum of 8? No. For sum fo 2, you must get 1 + 1 only. For sum of 8, you could get 4 + 4 or 3 + 5 or 2 + 6 etc. So probability of getting sum of 8 would be higher. In the same way, the order matters in this question. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Intern Joined: 01 Nov 2014 Posts: 16 Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 06 Aug 2016, 14:17 Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8678 Location: Pune, India Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 08 Aug 2016, 02:22 1 1 ManonZ wrote: Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help? Do you mean in the total number of cases? If yes, then it is actually 2^4. First child can happen in 2 ways (boy or girl). Second, third and fourth kids can also happen in 2 ways. Total number of ways = 2*2*2*2 = 16 _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13087 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A couple decides to have 4 children. If they succeed in havi [#permalink] ### Show Tags 23 Feb 2018, 12:22 Hi All, There are a couple of ways to answer this question: 1) You could make a table of all the options (there are only 16 possible outcomes) and count up all the ways to get 2 boys and 2 girls or 2) You can do the math Here's the math approach: Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question... 4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls Final Answer = 6/16 = 3/8 = A 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!*****

Intern
Status: preparing for GMAT
Joined: 29 Nov 2017
Posts: 22
GPA: 3.55
WE: Military Officer (Military & Defense)
Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

### Show Tags

18 Mar 2018, 10:27
How do you arrive at this being a "600 level difficulty" problem? It's the 8th from the last problem in the OG 18 Supplemental Quant Guide.

I believe it'd be considered at least a 700 level difficulty by GMAC, but I could be wrong.

Could you explain how you determined it to be 600 level difficulty?

Thanks, Bunuel! You're the best
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

### Show Tags

18 Mar 2018, 11:07
destinyawaits wrote:
How do you arrive at this being a "600 level difficulty" problem? It's the 8th from the last problem in the OG 18 Supplemental Quant Guide.

I believe it'd be considered at least a 700 level difficulty by GMAC, but I could be wrong.

Could you explain how you determined it to be 600 level difficulty?

Thanks, Bunuel! You're the best

You can check difficulty level of a question along with the stats on it in the first post. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. For this particular question it's 700-level.
_________________
Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: A couple decides to have 4 children. If they succeed in havi  [#permalink]

### Show Tags

07 May 2018, 09:28
2
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16

We need to determine the probability of P(B-B-G-G).

P(B-B-G-G) = (1/2)^4 = 1/16

The number of ways to arrange B-B-G-G is 4C2 = 4!/(2! x 2!) = 3 x 2 = 6.

Thus, the total probability is 6/16 = 3/8.

Answer: A
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: A couple decides to have 4 children. If they succeed in havi &nbs [#permalink] 07 May 2018, 09:28
Display posts from previous: Sort by

# A couple decides to have 4 children. If they succeed in havi

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.