I am having difficulty understanding the solution to this problem.
Since the solution is in the form of Bayes theorem I expected something along the lines that looked similar to Bayes theorem.
Suppose that in each individual of a large population there is a pair of genes that determine eye color. Each of these genes can be x or X and they follow these rules:
- xx – blue eyes
- Xx, xX (heterozygote) – brown eyes
- XX – brown eyes
- The proportion of blue-eyed individuals is $p^2$ and of heterozygotes
it’s $2p(1-p)$, where $0<p<1$.
Each parent transmits one if its own genes to the child. If a parent is a heterozygote, the probability that it transmits the genes of type X is $0.5$.
Assuming random mating, show that among brown-eyed children of brown-eyed parents, the expected proportion of heterozygotes is $\frac{2p}{1+2p}$
the solution is set up as follows:
$Pr$(child is heterozygote | child has brown eyes and parents have brown eyes)
=
Best Answer
It's not Bayes theorem; it's just the definition of conditional probability. The desired prob, by definition of conditional probability, is $${P(\mbox{child is heterozygote & child has brown eyes & parents have brown eyes})\over P(\mbox{child has brown eyes & parents have brown eyes})}.$$ But the numerator simplifies to $$P(\mbox{child is heterozygote & parents have brown eyes}),$$
since the event 'child is heterozygote' implies (is a subset of) the event 'child has brown eyes'.
To evaluate numerator and denominator, there are three cases for the parents to have brown eyes:
(A) Both parents are XX. This has prob $(1-p)^2\cdot (1-p)^2=(1-p)^4$. Given this event, the prob that the child has brown eyes is $1$, while the prob the child is heterozygote is $0$.
(B) One parent is XX and the other is heterozygote. This has prob $(1-p)^2\cdot2p(1-p) + 2p(1-p)\cdot(1-p)^2=4p(1-p)^3$. Given this event, the prob the child has brown eyes is $1$, while the prob the child is heterozygote is $\frac12$.
(C) Both parents are heterozygote. This has prob $2p(1-p)\cdot 2p(1-p)=4p^2(1-p)^2$. Given this event, the prob the child has brown eyes is $\frac34$, while the prob the child is heterozygote is $\frac12$.
Putting these all together should yield the desired forms for the numerator and denominator.