Tree diagram for Conditional probability problem of two sons being disease carriers The question : A woman has a 50% chance of carrying hemophilia. She also has two sons If she is a carrier, each son independently has 0.5 probability of having the disease. If she is not a carrier, her sons will independently be normal (i.e., will not have the disease).
(c) If 1st son is normal, what is the probability that the second son is also normal?
(d) If both sons are normal, what is the probability that she is a carrier?
This is the link to the tree diagram i drew enter image description here
Would the above tree diagram be a correct conceptualization of the given problem?
My understanding of (c) is that since both sons being normal are independent events then 
P (2nd son normal | 1st son normal) = P(2nd son normal) = 0.75 using Partition Rule. 
Also would it be correct to say part (d) can be solved via Bayes Rule?
 A: Let $C$ denote the event that the woman is a carrier.
Let $N_{1}$ denote the event that the first son is normal.
Let $N_{2}$ denote the event that the second son is normal.

$\begin{aligned}P\left(N_{1}\right) & =P\left(N_{1}\mid C\right)P\left(C\right)+P\left(N_{1}\mid C^{\complement}\right)P\left(C^{\complement}\right)\\
 & =\frac{1}{2}\cdot\frac{1}{2}+1\cdot\frac{1}{2}\\
 & =\frac{3}{4}
\end{aligned}
\tag1$
and:
$\begin{aligned}P\left(N_{1}\cap N_{2}\mid C\right) & =P\left(N_{1}\mid C\right)P\left(N_{2}\mid C\right)\\
 & =\frac{1}{2}\cdot\frac{1}{2}\\
 & =\frac{1}{4}
\end{aligned}
\tag2$
Applying $(2)$ we find:
$\begin{aligned}P\left(N_{1}\cap N_{2}\right) & =P\left(N_{1}\cap N_{2}\mid C\right)P\left(C\right)+P\left(N_{1}\cap N_{2}\mid C^{\complement}\right)P\left(C^{\complement}\right)\\
 & =\frac{1}{4}\cdot\frac{1}{2}+1\cdot\frac{1}{2}\\
 & =\frac{5}{8}
\end{aligned}
\tag3$
Applying $(1)$ and $(2)$ we find:
$\begin{aligned}P\left(N_{2}\mid N_{1}\right)\frac{3}{4} & =P\left(N_{2}\mid N_{1}\right)P\left(N_{1}\right)\\
 & =P\left(N_{1}\cap N_{2}\right)\\
 & =\frac{5}{8}
\end{aligned}
\tag4$
Applying $(4)$ we find the solution of c): $$P\left(N_{2}\mid N_{1}\right)=\frac{4}{3}\cdot\frac{5}{8}=\frac{5}{6}$$
Applying $(3)$ and $(2)$ we find:
$\begin{aligned}P\left(C\mid N_{1}\cap N_{2}\right)\frac{5}{8} & =P\left(C\mid N_{1}\cap N_{2}\right)P\left(N_{1}\cap N_{2}\right)\\
 & =P\left(C\cap N_{1}\cap N_{2}\right)\\
 & =P\left(N_{1}\cap N_{2}\mid C\right)P\left(C\right)\\
 & =\frac{1}{4}\cdot\frac{1}{2}\\
 & =\frac{1}{8}
\end{aligned}
\tag5$
Applying $(5)$ we find the solution of d): $$P\left(C\mid N_{1}\cap N_{2}\right)=\frac{8}{5}\cdot\frac{1}{8}=\frac{1}{5}$$

Now check yourself (I can't read your link properly).
