 Solution

Let $q_n=\mathbb{P}(X_n=0)$ for a Galton-Watson process starting from one ancestor. Conditional on $X_1=k$, extinction by generation $n+1$ requires $k$ independent descendant processes to be extinct by generation $n$, soBecause zero is absorbing, $q_n\uparrow q$, the eventual extinction probability, and continuity gives $q=F(q)$. If $r\ge 0$ is any fixed point, monotonicity of $F$ and $q_0\le r$ give $q_n\le r$ inductively. Thus $q$ is the smallest nonnegative fixed point.

Here $F(s)=s/4+3s^3/4$. Every individual has at least one child, so $q=0$. The offspring mean is $F^{\prime }(1)=5/2$, hence