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

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

Posted from my mobile device
_________________

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)

In my opinion the probability will be (1/2^5)=1/32

Can someone please explain why the order is considered ?

"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.

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.
_________________

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

sujoykrdatta

It should be (1/2)^2*(1/2)^3 = 1/32
Please rectify your post.
_________________

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.
_________________