Q:

Cauchy–Schwarz inequality probability

(5) Suppose males carry a certain gene with probability $p$ and females carry the same gene with probability $q$. We wish to know the probability that a couple are both carriers.
(a) What is the answer if we assume independence — unreasonable if high risk sub-populations have a statistical bias toward or against in-group marriages.
My answer:
If independent this means that probabilities of joint equal to marginal. So,
$$P(M)P(F)=P(M\cap F)=pq$$
(b) What does the Cauchy–Schwarz inequality tell us about the question.
My answer: In statistics, Cauchy-Schwarz means that correlation is between $-1$ and $1$. Using the formula 
$$E(XY)\le \sqrt{E(X^2)E(Y^2)}$$
rearrange the formula
$$| \rho(X,Y)|= \frac {E(XY)}{\sqrt{E(X^2)E(Y^2)}}\le1$$
the Cauchy–Schwarz inequality tells us that the correlation of a couple who are both carriers is less than $1$?
I don't think my answers are correct as they are too primitive, please help/correct my mistakes. 

A:

Let $X$ be the indicator function for the male being a carrier (i.e. $X=1$ if the male is a carrier, otherwise $X=0$), and let $Y$ be the analogous indicator for the female.
Then,
$$P(M \cap F) = E[XY] \le \sqrt{E[X^2] E[Y^2]} = \sqrt{pq}.$$

