Branching process and calculating the probability of extinction Suppose the male descendants of a man follow a branching process.We further suppose that each man has 3 children, and the number of male descendants follows a binomial distribution, Bin(3,0.5).
(a) What is the probability that Guy A line of male descendants will become extinct by the third generation? How about becoming extinct exactly at the third generation?
(b) Suppose Guy B has two sons and a daughter.What is the probability that Guy B line of male descendants will eventually become extinct?
I have just started learning the branching process and not sure whether I did the question correctly.My working for the question is as follows,
(a)
I know that the generation function is of the form:
$$P(s)=\frac18 + \frac38 s + \frac38 s^2 + \frac18 s^3 = \frac18(1+s)^3,$$
Let Un=P(Xn=0) where Xn denotes the number male descendants at time n.
$$U(1)=P(X1=0)=\frac18$$
$$U(2)=P(U(1))=\frac{729}{4096}$$
$$U(3)=P(U(2))\approx 0.204325$$
Let T=min{n $\ge$ 1 : Xn=0)
So I think the question in part (a) is asking P(T$\le$3|X0=1) and U(3),
$$P(T\le3|X0=1)=U(1)+U(2)+U(3)\approx 0.5073$$
(b)
If the initial generation has 1 male descendant, then the probability of it being extinction will be equals to solving P(u)=u where $0<u<1$.
Solving it I get that the roots of u are 1,$-2+\sqrt 5$,$-2-\sqrt5$. Since $0<u<1$, then the probability of it being extinct given that guy B initial generation has 2 male descendants will be: $$(-2+\sqrt 5)^2=9-4\sqrt5$$
 A: Let $Z_n$ be the number of offspring in generation $n$, then $Z_0=1$ with $$Z_n=\sum_{j=1}^{n-1}Z_{n,j} $$ where $Z_{n,j}\stackrel{\mathsf{i.i.d.}}\sim\mathsf{Bin}(3,1/2)$. The probability of extinction at time $n=1$  is $\mathbb P(Z_1=0)=\frac18$. The probability of extinction at time $n=2$ is 
\begin{align}
\sum_{k=1}^3 \mathbb P(Z_2=0\mid Z_1=k) &= \sum_{k=1}^3 \mathbb P(Z_2=0,Z_1=k)\mathbb P(Z_1=k)\\
&= \sum_{k=1}^3 \left(\frac18\right)^k\binom 3k\cdot\frac18\\
&= \frac{217}{4096}
\end{align}
The probability of extinction at time $n=3$ is $$P^{(3)}(0)-P^{(2)}(0)-P(0)= \frac{112329015625}{549755813888} - \frac{217}{4096} - \frac18 = \frac{14484291913}{549755813888} \approx 0.0263468$$
The probability of becoming extinct by the third generation is
$$
\frac{112329015625}{549755813888}\approx 0.204325.
$$
Let $m=\mathbb E[X_1]$, then $$m = \lim_{s\uparrow1}P'(s)=\lim_{s\uparrow1}\frac18\left(3+6s+3s^2\right) = \frac32>1,$$
so the extinction probability $\pi$ is the unique root of $\pi=P(\pi)$ with $0<\pi<1$. Since
$$
\pi=\frac18(1+\pi)^3 \iff 1-5\pi+3\pi^2+\pi^3=0\iff(1-\pi)\left(\sqrt5-2-\pi\right)\left(2-\sqrt5-\pi\right)=0,
$$
we see that $\pi=\sqrt5-2\approx0.236068$.
