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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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.
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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