Last visit was: 18 Jul 2025, 15:43 It is currently 18 Jul 2025, 15:43
Close
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Kushchokhani
Joined: 05 Jan 2020
Last visit: 03 Apr 2024
Posts: 513
Own Kudos:
631
 [32]
Given Kudos: 692
Status:Admitted to IIM Shillong (PGPEx 2023-24)
Affiliations: CFA Institute; ICAI; BCAS
Location: India
WE 2: EA to CFO (Consumer Products)
GPA: 3.78
WE:Corporate Finance (Commercial Banking)
Products:
Posts: 513
Kudos: 631
 [32]
Kudos
Add Kudos
31
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
sujoykrdatta
Joined: 26 Jun 2014
Last visit: 18 Jul 2025
Posts: 539
Own Kudos:
1,046
 [9]
Given Kudos: 13
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
Expert
Expert reply
Posts: 539
Kudos: 1,046
 [9]
4
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 18 Jul 2025
Posts: 11,294
Own Kudos:
41,823
 [1]
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,294
Kudos: 41,823
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
avatar
ZulfiquarA
Joined: 06 Apr 2021
Last visit: 11 Jun 2025
Posts: 65
Own Kudos:
Given Kudos: 38
Posts: 65
Kudos: 39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This is a Classic Probability question.
Number of possibilities is 2, its either a boy or a girl
It is mentioned that 3 girls and 2 boys should be the set.
When the first born is a B: B*B*G*G*G (This can be done in 4C3 ways by rearranging the B and G)=1/2*1/2*1/2*1/2*1/2*4= 4 ways
When first born is a G: G*B*B*G*G (This can be done in 4C2 ways) 1/2*1/2*1/2*1/2*1/2*6= 6 ways
Add the two scenarios: 4/32+6/32=10/32 =5/16 (D)
avatar
Himaanshu007
Joined: 18 May 2020
Last visit: 20 May 2024
Posts: 8
Own Kudos:
2
 [1]
Given Kudos: 29
Location: India
Schools: ISB '23
Schools: ISB '23
Posts: 8
Kudos: 2
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In my opinion the probability will be (1/2^5)=1/32

Can someone please explain why the order is considered ?
avatar
iakashp
Joined: 19 Sep 2021
Last visit: 02 Jul 2025
Posts: 5
Own Kudos:
Given Kudos: 19
Location: India
Posts: 5
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
"Post subject: Re: A couple decides to have 5 children. Each child is equally likely to b Post Posted: 21 Sep 2021, 21:10
In my opinion the probability will be (1/2^5)=1/32

Can someone please explain why the order is considered ?"


- You need to consider the arrangement of 3B and 2G, they can be arranged in 5!/(3!*2!) or 10 times. So 10*(1/32)=5/16.
avatar
Himaanshu007
Joined: 18 May 2020
Last visit: 20 May 2024
Posts: 8
Own Kudos:
Given Kudos: 29
Location: India
Schools: ISB '23
Schools: ISB '23
Posts: 8
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
iakashp
"Post subject: Re: A couple decides to have 5 children. Each child is equally likely to b Post Posted: 21 Sep 2021, 21:10
In my opinion the probability will be (1/2^5)=1/32

Can someone please explain why the order is considered ?"


- You need to consider the arrangement of 3B and 2G, they can be arranged in 5!/(3!*2!) or 10 times. So 10*(1/32)=5/16.


Thanks for the reply but still could not understand the fact that why the arrangement matters ? In either of the case probability of 3G and 2B remains same. I'm still not able to understand why arrangement is considered when it has no impact on the probability. Request you to please explain, as I might be missing some critical point here.
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 18 Jul 2025
Posts: 8,351
Own Kudos:
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy)
GMAT Focus 1: 545 Q79 V79 DI73
Posts: 8,351
Kudos: 4,833
Kudos
Add Kudos
Bookmarks
Bookmark this Post
number of ways to have 3 girls and 2 boys ; 5!/3!*2! ; 5
and likelyhood of boy or girl ; 1/2
(1/2)^3 * ( 1/2)^2 ; 1/16
5/16
option D

Kushchokhani
A couple decides to have 5 children. Each child is equally likely to be a boy or a girl. What is the probability that the couple has exactly 3 girls and 2 boys?

A. 1/32
B. 1/16
C. 5/32
D. 5/16
E. 1/6
User avatar
Kushchokhani
Joined: 05 Jan 2020
Last visit: 03 Apr 2024
Posts: 513
Own Kudos:
Given Kudos: 692
Status:Admitted to IIM Shillong (PGPEx 2023-24)
Affiliations: CFA Institute; ICAI; BCAS
Location: India
WE 2: EA to CFO (Consumer Products)
GPA: 3.78
WE:Corporate Finance (Commercial Banking)
Products:
Posts: 513
Kudos: 631
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sujoykrdatta
In each case, probability = (1/2)^2 = 1/32 (since boy or girl has probability 1/2)
sujoykrdatta

It should be (1/2)^2*(1/2)^3 = 1/32
Please rectify your post.
User avatar
ThatDudeKnows
Joined: 11 May 2022
Last visit: 27 Jun 2024
Posts: 1,070
Own Kudos:
911
 [4]
Given Kudos: 79
Expert
Expert reply
Posts: 1,070
Kudos: 911
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Kushchokhani
A couple decides to have 5 children. Each child is equally likely to be a boy or a girl. What is the probability that the couple has exactly 3 girls and 2 boys?

A. 1/32
B. 1/16
C. 5/32
D. 5/16
E. 1/6

If you struggle with when and how to use which of nCr and nPr, I just want to show that you don't need either at all here (and [I'm pretty sure] never really do on official questions...prove me wrong).

Since we have two possibilities for each child, there are 2^5 total possibilities. That's 32.

It's equal probability that we have more girls or more boys, so half of the total possibilities have more girls. That's 16.

There's 1 way to get five girls and no boys: GGGGG.

There are 5 ways to get four girls and one boy: substitute a B for any of the five Gs above.

That means there must be 16-1-5=10 ways to get three girls and two boys.

10/32 = 5/16

Answer choice D.
User avatar
thr3at
Joined: 05 Mar 2024
Last visit: 17 Jan 2025
Posts: 30
Own Kudos:
Given Kudos: 42
Posts: 30
Kudos: 12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
iakashp
"Post subject: Re: A couple decides to have 5 children. Each child is equally likely to b Post Posted: 21 Sep 2021, 21:10
In my opinion the probability will be (1/2^5)=1/32

Can someone please explain why the order is considered ?"


- You need to consider the arrangement of 3B and 2G, they can be arranged in 5!/(3!*2!) or 10 times. So 10*(1/32)=5/16.
­Hello sujoykrdatta

IMO 5/16. But my reasoning is flawed. 

My Process was like:

1st select 3 girls out of 5 students (remaining will be boys)  then arrange them in 5!/3!*2!
but my answer is coming around 100/32 (>1 which is not possible)
So I marked E ( assumed I must have taken 10 extra, but I am not able to figure out how)

Can you please explain? 
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Jul 2025
Posts: 102,619
Own Kudos:
Given Kudos: 98,235
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,619
Kudos: 742,543
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thr3at
iakashp
A couple decides to have 5 children. Each child is equally likely to be a boy or a girl. What is the probability that the couple has exactly 3 girls and 2 boys?

A. 1/32
B. 1/16
C. 5/32
D. 5/16
E. 1/6

"Post subject: Re: A couple decides to have 5 children. Each child is equally likely to b Post Posted: 21 Sep 2021, 21:10
In my opinion the probability will be (1/2^5)=1/32

Can someone please explain why the order is considered ?"


- You need to consider the arrangement of 3B and 2G, they can be arranged in 5!/(3!*2!) or 10 times. So 10*(1/32)=5/16.
­Hello sujoykrdatta

IMO 5/16. But my reasoning is flawed. 

My Process was like:

1st select 3 girls out of 5 students (remaining will be boys)  then arrange them in 5!/3!*2!
but my answer is coming around 100/32 (>1 which is not possible)
So I marked E ( assumed I must have taken 10 extra, but I am not able to figure out how)

Can you please explain? 
\(P(­GGGBB) = \frac{1}{2^5}*\frac{5!}{3!*2!} = \frac{10}{32} = \frac{5}{16}\). We multiply by 5!/(3!*2!) becasue the GGGBB can be arranged in 5!/(3!*2!) ways (GGGBB, GBGGB, BGGGB, ...).

Answer: D.­
User avatar
andreabho99
Joined: 22 Nov 2023
Last visit: 04 Jun 2024
Posts: 2
Given Kudos: 1
Posts: 2
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
\(P(­GGGBB) = \frac{1}{2^5}*\frac{5!}{3!*2!} = \frac{10}{32} = \frac{5}{16}\). We multiply by 5!/(3!*2!) becasue the GGGBB can be arranged in 5!/(3!*2!) ways (GGGBB, GBGGB, BGGGB, ...).

Answer: D.­
­Hi Bunuel, could you please explain why we use that formula to arrange the children? Is that permutation with repetition or another formula?­
User avatar
DmitryFarber
User avatar
Manhattan Prep Instructor
Joined: 22 Mar 2011
Last visit: 16 Jul 2025
Posts: 2,950
Own Kudos:
Given Kudos: 57
GMAT 2: 780  Q50  V50
Expert
Expert reply
GMAT Focus 1: 745 Q86 V90 DI85
Posts: 2,950
Kudos: 8,401
Kudos
Add Kudos
Bookmarks
Bookmark this Post
andreabho99
Bunuel
\(P(­GGGBB) = \frac{1}{2^5}*\frac{5!}{3!*2!} = \frac{10}{32} = \frac{5}{16}\). We multiply by 5!/(3!*2!) becasue the GGGBB can be arranged in 5!/(3!*2!) ways (GGGBB, GBGGB, BGGGB, ...).

Answer: D.­
­Hi Bunuel, could you please explain why we use that formula to arrange the children? Is that permutation with repetition or another formula?­
­The 5!/(3!*2!) part is a combination, which we use since we're just assigning each of the 5 kids to one group or the other (B or G) and not subdividing further. This tells us that there are 10 possible orders that would have 3 G's and 2 B's. The 2^5 part is the total # of possible orders: since each boy can be a boy or a girl (assuming gender binaries, which the GMAT tends to do), we have 2 choices each time, so 32 total possible combos. 
User avatar
andreabho99
Joined: 22 Nov 2023
Last visit: 04 Jun 2024
Posts: 2
Given Kudos: 1
Posts: 2
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
 
DmitryFarber
­The 5!/(3!*2!) part is a combination, which we use since we're just assigning each of the 5 kids to one group or the other (B or G) and not subdividing further. This tells us that there are 10 possible orders that would have 3 G's and 2 B's. The 2^5 part is the total # of possible orders: since each boy can be a boy or a girl (assuming gender binaries, which the GMAT tends to do), we have 2 choices each time, so 32 total possible combos. 
­Thank you for your help Dmitry!
From what I've understood, we are choosing n=5 children to arrange in 3 position (boys) and hence the use of the combination formula. But what about the remaining two girls? Can't they be arranged in 2 different ways?
User avatar
DmitryFarber
User avatar
Manhattan Prep Instructor
Joined: 22 Mar 2011
Last visit: 16 Jul 2025
Posts: 2,950
Own Kudos:
8,401
 [1]
Given Kudos: 57
GMAT 2: 780  Q50  V50
Expert
Expert reply
GMAT Focus 1: 745 Q86 V90 DI85
Posts: 2,950
Kudos: 8,401
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
andreabho99
DmitryFarber
­The 5!/(3!*2!) part is a combination, which we use since we're just assigning each of the 5 kids to one group or the other (B or G) and not subdividing further. This tells us that there are 10 possible orders that would have 3 G's and 2 B's. The 2^5 part is the total # of possible orders: since each boy can be a boy or a girl (assuming gender binaries, which the GMAT tends to do), we have 2 choices each time, so 32 total possible combos. 
­Thank you for your help Dmitry!
From what I've understood, we are choosing n=5 children to arrange in 3 position (boys) and hence the use of the combination formula. But what about the remaining two girls? Can't they be arranged in 2 different ways?
­Well, we're arranging people into two groups, and the math doesn't care what those groups are. In other words, it could be "hired or not hired" or "boys or not boys." In this case, it's "boys or girls." So the 2! in the denominator is already dividing out the 2 ways we could arrange the girls for each arrangement of boys. The general idea is that we take the factorial of the entire group and divide by the factorials of any groups we're not arranging further. We're not arranging either group further, so we divide by 3! and 2!.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,454
Own Kudos:
Posts: 37,454
Kudos: 1,013
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Math Expert
102619 posts
PS Forum Moderator
698 posts