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.
Either Smith is BB (Brown-Brown) or Bb (Brown-blue); as you correctly calculated, $P(BB) = \frac13$.
The fact that the first child has brown eyes lessens the likelihood that Smith is Bb. Let $Brown$ represent the event that Smith's first child has brown eyes (given only the information knwon before the birth).
Then
$$
P(Brown\wedge BB) = 1\cdot \frac13 = \frac13 \\ P(Brown|\wedge Bb) = \frac12 \cdot \frac23 = \frac13\\
P(Brown) = \frac13 + \frac13 = \frac23\\
B(BB | Brown) = \frac13/\frac23=\frac12
$$
So the probability of BB, now that we know that $Brown$ is true, is $\frac12$ and the probability of the next child having brown eyes is $$
\frac12\cdot 1 + \frac12 \cdot \frac12 = \frac34$$
Best Answer
If woman's parents are $Bb$, then she may be $BB$, $Bb$, $bB$ or $bb$ with $25\%$ chance for each genotype. As we know that her phenotype is "brown eyes", then the last one is not possible and we have equal chances of $BB$, $bB$ and $Bb$ for this woman. As $bB$ and $Bb$ are the same, we can say that she's $1/3$ homozygous and $2/3$ heterozygous. Blue-eyed man is definitely $bb$, while brown-eyed child can be $Bb$ or $BB$. As the child can't be $BB$ with $bb$ father, that child is definitely $Bb$ (or $bB$, let's consider them as the same).
So we have the following:
- mother is $1/3$ chance $BB$ and $2/3$ chance $Bb$
- father is $100\%$ $bb$
- child is $100\%$ $Bb$
$BB$ mother with $bb$ father will always have $Bb$ child, while $Bb$ mother with $bb$ father only in half of cases. So, $1/3$ for $BB$ and $2/3 * 1/2 = 1/3$ for $Bb$. Normalizing the weights of the events to make the sum $= 1$, we have that chance the woman is $Bb$ is $1/2$.