It is currently 17 Oct 2017, 12:27

### 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

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

Author Message
TAGS:

### Hide Tags

Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 615

Kudos [?]: 1133 [0], given: 39

A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

18 Feb 2011, 12:56
10
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

54% (01:08) correct 46% (01:07) wrong based on 276 sessions

### HideShow timer Statistics

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/a-couple-dec ... 68730.html
[Reveal] Spoiler: OA

_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Kudos [?]: 1133 [0], given: 39

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1964

Kudos [?]: 2050 [0], given: 376

Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

18 Feb 2011, 13:03
Baten80 wrote:
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

Probability of having BOY or GIRL is equally likely= 1/2

Having 2 girls out of 4 child can happen in C(4,2) ways

$$C^4_2*(\frac{1}{2})^4$$

$$\frac{12}{2}*\frac{1}{16}$$

$$\frac{3}{8}$$

Ans: "a"
_________________

Kudos [?]: 2050 [0], given: 376

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128582 [0], given: 12180

Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

18 Feb 2011, 13:08
Expert's post
8
This post was
BOOKMARKED
Baten80 wrote:
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 $$\frac{4!}{2!2!}$$.

_________________

Kudos [?]: 128582 [0], given: 12180

Intern
Joined: 24 Sep 2011
Posts: 18

Kudos [?]: 161 [2], given: 52

Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

04 Aug 2013, 00:54
2
KUDOS
21
This post was
BOOKMARKED
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

I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?

Kudos [?]: 161 [2], given: 52

Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 627

Kudos [?]: 1355 [0], given: 136

Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

04 Aug 2013, 00:59
ksung84 wrote:
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

I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?

If that was the case, the question would be similar to arranging B,B,G -->$$\frac{3!}{2!}$$.
_________________

Kudos [?]: 1355 [0], given: 136

Current Student
Joined: 24 Nov 2012
Posts: 176

Kudos [?]: 357 [7], given: 73

Concentration: Sustainability, Entrepreneurship
GMAT 1: 770 Q50 V44
WE: Business Development (Internet and New Media)
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

04 Aug 2013, 01:07
7
KUDOS
6
This post was
BOOKMARKED
Since there is an equal probability of having a boy or a girl, Any possible combination of 4 children has a probability of 1/16. Now there are 4!/2!*2! ways of having exactly 2 boys and 2 girls... which is 6. So total probability of that event happening is 6*1/16 = 3/8 = A

The answer would not change for 2 boys and 1 girl

Which will be 3*1/8 = 3/8
_________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013
Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013
Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013
GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013
Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013
Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013
GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013
My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

Kudos [?]: 357 [7], given: 73

Manager
Joined: 09 Apr 2013
Posts: 148

Kudos [?]: 119 [5], given: 24

Location: India
WE: Supply Chain Management (Consulting)
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

25 Dec 2013, 07:28
5
KUDOS
7
This post was
BOOKMARKED
Since probability that the couple will have a boy or a girl is 1/2, likelihood of having 2 boys and 2 girls is 1/2* 1/2 * 1/2 * 1/2 = 1/16

Now, 2 boys out of 4 can be chosen in 4c2 i.e 6 ways. Choosing 2 boys out of 4 leaves 2 girls, hence no additional selection of girls is required.

Finally, probability of choosing exactly 2 boys and 2 girls is 1/16 * 6 = 3/8.

Hope that helps.
_________________

+1 KUDOS is the best way to say thanks

"Pay attention to every detail"

Kudos [?]: 119 [5], given: 24

Senior Manager
Joined: 29 Oct 2013
Posts: 294

Kudos [?]: 474 [5], given: 197

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, 02:40
5
KUDOS
8
This post was
BOOKMARKED
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
_________________

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Kudos [?]: 474 [5], given: 197

Intern
Joined: 25 Sep 2014
Posts: 7

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

Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

24 Nov 2014, 14:26
Referring to the question: What if the question was 1 girls and 2 boys? What will be the nominator?

You can easily verify this by using the same methodology. There are _ _ _ --> 3 possible spots, 1 for every child. Each of these spots has only 2 possible outcomes occurring with equal likelihood (1/2).

This leaves us with a probability of (1/2)^3 =1/8 for every single possible outcome.

The next thing to ask yourself is how many of all possible cases include 2 boys (B) and 1 girl (G). The answer is 3 and turns out to be very intuitive if you start to write down a few possible outcomes.

GBB
BGB
BBG

All other scenarios will either include 2 girls and 1 boy, 3 boys or 3 girls.

Final step is to multiply the probability of each individual outcome with the number of outcomes that satisfy the condition we are looking for. --> 3x(1/8)=3/8

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

Director
Joined: 23 Jan 2013
Posts: 600

Kudos [?]: 21 [0], given: 41

Schools: Cambridge'16
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

24 Nov 2014, 22:21
One case is 1/2*1/2*1/2*1/2=1/16

counting symmetry to know how many cases we can have, so 4!/2!*2!=6

1/16*6=6/16=3/8

A

Kudos [?]: 21 [0], given: 41

Intern
Joined: 23 Sep 2011
Posts: 48

Kudos [?]: 16 [0], given: 21

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, 20:06
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?

Kudos [?]: 16 [0], given: 21

Senior Manager
Joined: 28 Jun 2015
Posts: 300

Kudos [?]: 107 [0], given: 47

Concentration: Finance
GPA: 3.5
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

16 Jul 2015, 21:14
ksung84 wrote:
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

I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?

Sample space = 2^4 = 16.

Favourable events = {bbgg}, {bgbg}, {bggb}, {ggbb}, {gbgb}, {gbbg}.

Probability = 6/16 = 3/8. Ans (A).
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

Kudos [?]: 107 [0], given: 47

Retired Moderator
Joined: 23 Feb 2015
Posts: 168

Kudos [?]: 56 [0], given: 5

GMAT 1: 690 Q46 V38
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

04 Feb 2016, 16:50
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?

Kudos [?]: 56 [0], given: 5

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7670

Kudos [?]: 17334 [2], given: 232

Location: Pune, India
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

04 Feb 2016, 22:35
2
KUDOS
Expert's post
Dondarrion 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
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17334 [2], given: 232 Intern Joined: 01 Nov 2014 Posts: 16 Kudos [?]: [0], given: 0 Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 06 Aug 2016, 15:17 Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help? Kudos [?]: [0], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7670 Kudos [?]: 17334 [0], given: 232 Location: Pune, India Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink] ### Show Tags 08 Aug 2016, 03:22 Expert's post 1 This post was BOOKMARKED 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17334 [0], given: 232

Manager
Joined: 03 Jan 2017
Posts: 197

Kudos [?]: 9 [0], given: 4

Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

23 Mar 2017, 08:17
probability question: total number of outcomes is 2*2*2*2=16
the # of desired outcomes can be found 2C4, or 4!/2!2!=6
6/16=3/8

Kudos [?]: 9 [0], given: 4

Director
Joined: 17 Dec 2012
Posts: 608

Kudos [?]: 516 [0], given: 16

Location: India
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

31 May 2017, 19:33
ksung84 wrote:
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

I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?

1.This requires ordering and repetition. So the formula is n^r for the total number of cases. By ordering I mean bbgg is different from ggbb and repetition happens because b or g can happen more than once.
2. Favorable cases are : bbgg, bgbg, bggb, ggbb, gbgb, gbbg
Probability is therefore 6/16=3/8
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/regularcourse.php

Pay After Use
Standardized Approaches

Kudos [?]: 516 [0], given: 16

Manager
Joined: 23 Dec 2013
Posts: 235

Kudos [?]: 12 [0], given: 21

Location: United States (CA)
GMAT 1: 760 Q49 V44
GPA: 3.76
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

02 Jun 2017, 10:44
ksung84 wrote:
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

I see that we can use combination rule here [4! / 2! (2!)]/2^4. What if the question was 1 girls and 2 boys? What will be the nominator?

We can solve this problem in a couple of ways.

First, realize that there are 2^4 = 16 total possibilities (boy/girl = 2 options * 4 instances).

Next, you can realize that there is 1 way to have all boys and 1 way to have all girls. So the total number of invalid options is 2.

Then, if you have 1 boy and 3 girls or if you have 1 girl and 3 boys, then you the chances of these outcomes occurring should be equal. And there are 4 ways to have 1 boy, so there must be 4 ways to have one girl.

The total number of invalid cases is 2+4+4 = 10 out of 16 total options. So the number we care about is the remaining number of options with 2 boys and 2 girls = 16-10 = 6.

6/16 = 3/8.

Kudos [?]: 12 [0], given: 21

Manager
Joined: 01 Dec 2016
Posts: 115

Kudos [?]: 13 [0], given: 32

Location: Cote d'Ivoire
Concentration: Finance, Entrepreneurship
WE: Investment Banking (Investment Banking)
Re: A couple decides to have 4 children. If they succeed in having 4 child [#permalink]

### Show Tags

19 Jun 2017, 02:56
I did it the binomial way.
Proba of having a girl is same as proba of having a boy, which is 1/2.
Proba of having exactly 2 girls and 2 boys is given by 4C2 * (1/2)^2 * (1/2)^2 = 6/16 or 3/8.

The question would be much more interesting if proba of having a girl was 1/3 for instance, and we were asked to find the proba of having exactly 3 girls and 1 boy. In this case the answer would have been: 4C3 * (1/3)^3 * (2/3)^1 = 8/81.

Binomial Proba is fast and easy...
_________________

What was previously thought to be impossible is now obvious reality.
In the past, people used to open doors with their hands. Today, doors open "by magic" when people approach them

Kudos [?]: 13 [0], given: 32

Re: A couple decides to have 4 children. If they succeed in having 4 child   [#permalink] 19 Jun 2017, 02:56

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by