This is a question from the book 'All of Statistics' by Larry Wasserman.
A simple example to illustrate the concept...
Say you record the outcome of the toss of two fair coins. The sample space is {HH, HT, TH, TT}.
If it is known that the first coin is a head, then the probability of getting two heads will be 1/2. The sample space is reduced to {HT, HH} and only {HH} gives the desired outcome.
If it is known that as least one of the coins is a head, then the probability of getting two heads is 1/3. The sample space is reduced to {HH, HT, TH} and only {HH} gives the desired outcome.
Now, say it is known that at least one of the coins is a head. If someone else looks at the coins, picks out a head coin because it is a head, and tells you 'this coin is a head', the probability of two heads is still 1/3.
If, on the other hand, someone picks a coin without regard to whether it is a head or a tail, and tells you 'this coin is a head', then the sample space is reduced to {HH, HT} and the probability of two heads is 1/2.
The answer to the original question hinges on whether the child was chosen because she was the youngest child, or because she had blue eyes.
Apologies for not writing as a comment, I do not have any rep.
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.
Best Answer
Yes, this is correct.
The general form of Bayes' rule is
$$P(A_i|B)= \frac{P(B|A_i)P(A_i)}{\sum_jP(B|A_j)P(A_j)}$$
as you've used above.