Application of binomial distributions A family of 6 children. Find the probabilities of the following compositions:
a) 3 boys and 3 girls
b) Fewer boys than girls
The first part of the question a) was pretty easy I knew how to solve that but for part b) the answer for that question is 11/32 in my book but I'm not sure how to get there would anyone have a solution?
 A: For part (b) you have to work out the chance that there are exactly 3 boys and 3 girls.  Call that chance $c$.  The desired answer is $(1-c)/2$.
One might be tempted to say the chance that $B<G$ is $1/2$ but this does not take the possibility that $B=G$ into account.  Obviously $P(B<G)=P(B>G)$.  I get $c = P(B=G)=\binom 6 3 / 2^6$.
A: Note that there are three possibilities:


*

*There are as many boys as girls ($B = G = 3$)

*There are more boys than girls ($B > G$)

*There are fewer boys than girls ($B < G$)


You already know that the probability of event 1 is, since you calculated it for part (a). Also, events 2 and 3 are equally likely, since they're symmetrical. So $Pr(B < G) = Pr(B > G)$ and $Pr(B = G) + Pr(B < G) + Pr(B > G) = 1$ leads you to $\frac{10}{32} + 2 Pr(B < G) = 1 \rightarrow Pr(B < G) = \frac{11}{32}$.
A: Fewer boys than girls would mean the number of boys is either $0,$ $1,$ or $2.$ So you have
$$
\binom 6 0 \left(\frac 1 2 \right)^6 + \binom 6 1 \left(\frac 1 2 \right)^6 + \binom 6 2 \left(\frac 1 2 \right)^6 
$$
$$
= \left(\frac 1 2 \right)^6 \left( 1 + 6 + 15 \right) = \frac {22}{64} = \frac {11}{32} .
$$
