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A couple decides to have 5 children. Each child is equally likely to b

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Kushchokhani
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A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   11 Sep 2021, 07:57
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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
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D
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   11 Sep 2021, 09:15
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Kushchokhani wrote:
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



The 3 boys and 2 girls can be in any order, that being decided in 5!/(3!2!) = 10 ways.
In each case, probability = (1/2)^2 = 1/32 (since boy or girl has probability 1/2)

So required probability is
10 x 1/32 = 5/16

Answer D

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A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   21 Sep 2021, 23:04
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Kushchokhani wrote:
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


I. Working on the number of cases
\(12345\)
a) Say 1 is G, other 2 Gs can be any out of 2, 3, 4 or 5. => 1*4C2=6
b) Say 1 is not G but 2 is G, other 2 Gs can be any out of 3, 4 or 5. => 1*1*3C2=3
c) Say 1 and 2 are not Gs but 3 is G, other 2 Gs can be any out of 4 or 5. => 1*1*1*2C2=1
Total favorable ways = 6+3+1=10
Total ways = 2*2*2*2*2=32
P = \(\frac{10}{32}=\frac{5}{16}\)

II. Direct calculation of P
\(5C3*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{5}{16}\)


Himaanshu007, the reason why arrangement is important..
If P of 3 G and 2 B is 1/32, then following cases will also have P as 1/32
(1G,4B); (2G,3B); (3G,2B); (4G,1B); (5G,0B) and (0g,5B)............6cases, so combined P = 6*1/32=6/32.

But there is NO other possibility apart from 6 cases, and so they should add up to 1. where do \(1-\frac{6}{32}\) or 26/32 go?

When you calculate P as 1/2 *1/2....., what is 2 in the denominator? OR when you calculate total ways as 2*2*2*2*2 or 32, what does 2 stand for
It is for B and G, so B*B*G*G*G is different from G*B*B*B*B.
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   18 Sep 2021, 07:17
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)
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   21 Sep 2021, 07:40
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In my opinion the probability will be (1/2^5)=1/32

Can someone please explain why the order is considered ?
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   21 Sep 2021, 07:51
"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.
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   21 Sep 2021, 18:05
iakashp wrote:
"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.
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   22 Sep 2021, 00:25
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 wrote:
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
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   25 Jun 2022, 06:52
sujoykrdatta wrote:
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.
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   25 Jun 2022, 10:53
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Kushchokhani wrote:
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.
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   02 May 2024, 07:58
iakashp wrote:
"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? 
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A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   02 May 2024, 10:24
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thr3at wrote:
iakashp wrote:
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.­
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A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   04 May 2024, 15:59
Bunuel wrote:
\(P(­GGGBB) = \frac{1}{2^5}*\frac{5!}{3!*2!} = \frac{10}{32} = \frac{5}{16}\). 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?­
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   04 May 2024, 16:29
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andreabho99 wrote:
Bunuel wrote:
\(P(­GGGBB) = \frac{1}{2^5}*\frac{5!}{3!*2!} = \frac{10}{32} = \frac{5}{16}\). 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. 
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   04 May 2024, 16:42
 
DmitryFarber wrote:
­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?
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Re: A couple decides to have 5 children. Each child is equally likely to b [#permalink] A couple decides to have 5 children. Each child is equally likely to b   10 May 2024, 01:00
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andreabho99 wrote:
DmitryFarber wrote:
­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!.
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