What is the probability that three are males and two are female? The human sex ratio at birth is commonly thought to be 107 boys to 100 girls. Suppose five infants are chosen at random. 
(A) What is the probability that three are males and two are female?
(B) What is the probability that at least one of them is a male?
My work: (A) $(5C2 * 5C3)$ / $207C5$
(B) $(5C1 * 202C)4$ / $207C5$
I'm not sure if these are correct.
 A: 107:100 implies that the chance of being male is $p = \frac{107}{207}$.
Notice that we have $n = 5$ (presumably independent) trials (children) with probability $p = 107/207$ of success (if I consider selecting a boy being as a success). Then the number of boys selected $N$ follows a binomial distribution with $n = 5, p = 107/207$
a) Notice that by selecting 3 males in 5 tries forces us to have 2 females. Hence we only need to consider getting 3 boys in five trials. Can you find that?

 $P(N = 3) =\binom{5}{3}p^3(1-p)^2 = 0.3223292$

b) In terms of notation, this asks for $P(N\geq 1)$, but it is easier to use the complementary probability. Can you do it?

 $P(N\geq 1) = 1-P(N = 0) = 1-\binom{5}{0}p^0(1-p)^5 =1- 0.02631166=0.9736883$.

A: Let $p=\mbox{probability for a boy}$ and $q=\mbox{probability for a girl}=1-p$. Then, we know $p+q=1$ and $\frac{p}{q}=\frac{107}{100}$. Thus, $p=0.516$ and $q=0.483$
Now, we have a binomial distribution $Bin(5,0.516)$
A) $\binom{5}{3}p^3q^2=0.320$
B) $P(\mbox{at least one boy})=1-P(\mbox{no boys})=1-\binom{5}{0}p^0q^5=1-0.026=0.973$
