# 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.

• What are you counting? The probability of selecting 2 from howmany boy-children and 3 from howmany girl-children out of selecting any 5 from 207 all-children. – Graham Kemp Mar 1 '16 at 2:14
• 2 girls and 3 boys out of the 5 random children selected. – Lisa Mar 1 '16 at 2:14
• You do not select from what you have selected. You must select from the source. – Graham Kemp Mar 1 '16 at 2:17
• So for (A) (207C3 * 207C2)/(207C5)? Or (107C3 * 100C2)/(207C5)? – Lisa Mar 1 '16 at 2:18
• Each child of the five infants you choose are taken from the population at large. Since there are "many" children overall, you can well approximate this by considering it as independent events with replacement. In reality, there may only be 200 million infants alive at any given time, and technically we should be doing it as without replacement with those 200 million as our source, but it does not affect the probability in any noticeable way if we don't. – JMoravitz Mar 1 '16 at 2:22

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?

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

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) $P(\mbox{at least one boy})=1-P(\mbox{no boys})=1-\binom{5}{0}p^0q^5=1-0.026=0.973$