Consider a branching process where the number of offspring of an individual is a binomial random variable with parameters $(2, p),$ with $p \in (0, 1).$ Initially there is one individual.
(a) Calculate the probability, as a function of $p,$ that the branching process survives forever.
(b) Suppose we observe that the initial individual has $2$ offspring, and each of these then has $2$ offspring of its own. What is the probability that the branching process survives forever?
I have calculated for (a), the answer is $1-\min\left\{1,\left(\frac{p-1}{p}\right)^2\right\}$.
However, I don't know how to solve for (b). My instinct tells me it might be $4 \cdot \left(1-\min\left\{1,\left(\frac{p-1}{p}\right)^2\right\}\right)$. Am I correct?
Let's call the probability of the branching process surviving forever $a$.
There is a $(1-p)^2$ probability of zero offspring, $2p(1-p)$ probability of one offspring, and a $p^2$ probability of two offspring.
Then, assuming there is one offspring, the probability of the branching process continuing forever is $a$. If there is two offspring, both branches must not die off, so the probability is $1-(1-a)^2$. Therefore $$a = 0*(1-p)^2 + a*(2p(1-p)) + p^2 * (1-(1-a)^2) = 2pa-p^2*a^2$$Simplifying and setting equal to $0$, $$p^2*a^2+(1-2p)a = 0 \rightarrow a=(2p-1)/p^2$$
Therefore, the probability of the branching process surviving forever is $$a = \frac{2p-1}{p^2}$$. Since this is less than $0$ if $p < 0.5$, $a$ is instead $$\max(0, \frac{2p-1}{p^2})$$ which is the same formula you arrived at.
If there are four branches, then all must not fail. $$b=1-(1-a)^4$$