problem colour of eyes The color of a person’s eyes is determined by a single
pair of genes. If they are both blue-eyed genes,
then the person will have blue eyes; if they are
both brown-eyed genes, then the person will have
brown eyes; and if one of them is a blue-eyed gene
and the other a brown-eyed gene, then the person
will have brown eyes. (Because of the latter
fact, we say that the brown-eyed gene is dominant
over the blue-eyed one.) A newborn child independently
receives one eye gene from each of its
parents, and the gene it receives from a parent is
equally likely to be either of the two eye genes of that parent.
Suppose that Smith and both of his parents have brown eyes, but Smith’s sister has blue eyes.
(a) What is the probability that Smith possesses a
blue-eyed gene?
if Smith's sister has blue eyes , Smith's parents have both the genes Blue-Brown.
The possible cases for Smith are: Brown-Brown, Blue-Brown, Brown-Blue
and the positive cases: Blue-Brown, Brown-Blue
$$p=2/3$$
(b) Suppose that Smith’s wife has blue eyes. What is the probability that their first child will have blue eyes?
$$p(\text{the first child has blue eyes})=p(\mathrm{B|Brown-Brown})*p(\mathrm{Brown-Brown})+p(\mathrm{B|Blue-Blue})*p(\mathrm{Blue-Blue})+p(\mathrm{B|Blue-Brown})*p(\mathrm{Blue-Brown})+p(\mathrm{B|Brown-Blue})*p(\mathrm{Brown-Blue})=0*\frac{1}3+1*0+\frac{1}2*\frac{1}3+\frac{1}2*\frac{1}3=\frac{1}3$$
(c) If their first child has brown eyes, what is the probability that their next child will also have
brown eyes?
C1=event "the first child has brown eyes", $p(C1)=2/3$
C2=event "the second child has brown eyes"
could someone give me a suggestion?
 A: 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 $$
\frac12\cdot 1 + \frac12 \cdot \frac12 = \frac34$$
A: Both answers, by Mark and foo, seem correct, but there's a slightly different way of explaining it (although not fundamentally different).
Like Mark, I let B and b denote the brown and blue alleles, respectively. I also let Brown and Blue denote the resulting eye colour: so bb results in Blue, while BB and Bb (same as bB if we don't care about the order) result in Brown.
As is already known, $\Pr[\text{Smith}=BB]=1/3$ and $\Pr[\text{Smith}=Bb]=2/3$. His wife is bb, so what determines the children's eye colour is which allele they inherit from Smith.
As has already been shown, the chance that the first child inherits the B allele from Smith is $1/3$:
$$
\Pr[C_1\sim\textit{Brown}]
= \Pr[\text{Smith}=BB]\cdot\frac{2}{2}
 + \Pr[\text{Smith}=Bb]\cdot\frac{1}{2}
= \frac{1}{3} + \frac{2}{3}\cdot\frac{1}{2}
= \frac{2}{3}
$$
where the first $1/2$ is the likelihood that the child inherits the B allele from Smith if he has Bb; and the $2/2$ just indicates that when Smith has both B alleles, the child will always inherit B.
We can make the same calculation asking for the likelihood that the first two children both have brown eyes:
$$
\Pr[C_1,C_2\sim\textit{Brown}]
= \Pr[\text{Smith}=BB]
 + \Pr[\text{Smith}=Bb]\cdot\frac{1}{2^2}
= \frac{1}{3} + \frac{2}{3}\cdot\frac{1}{4}
= \frac{1}{2}.
$$
This makes the conditional likelihood of the second child having brown eyes given that the first has brown eyes
$$
\Pr[C_2\sim \textit{Brown}\mid C_1\sim\textit{Brown}]
= \frac{\Pr[C_1,C_2\sim\textit{Brown}]}{\Pr[C_1\sim\textit{Brown}]}
= \frac{1/2}{2/3} = \frac{3}{4}.
$$
