How to find the probability that the male line continue forever for the problem given below.

Question

Suppose that every man in a certain society has exactly three children, which independently have probability one-half of being a boy and one-half of being a girl. Suppose also that the number of males in the n th generation forms a branching process. (a) Find the probability that the male line of a given man eventually becomes extinct. (b) If a given man has two boys and one girl, what is the probability that his male line will continue forever? Ans: I can calculate part a easily using the formula g(t) =t, where g is the probability generating function and t is extinction probability. After calculating I got probability of extinction is root(5)-2 Help me for the 2nd part

Answer 1

Presumably you set the probability of male-line extinction to $t$ and then tried to solve $$t=\frac18 +\frac38 t +\frac38 t^2 +\frac18 t^3$$ getting the solutions $1$ and $\pm\sqrt 5 -2$, then choosing the sensible one.

What you actually did was say that the probability of male-line extinction overall was

• the probability of male-line extinction immediately due to having no sons, plus

• the probability of one son times the conditional probability he faced male-line extinction, plus

• the probability of two sons times the conditional probability both faced male-line extinction, plus

• the probability of three sons times the conditional probability all three faced male-line extinction

So the conditional probability two sons both faced male-line extinction is $t^2$ and thus the conditional probability two sons did not both faced male-line extinction (i.e. his male-line will continue forever) is $1-t^2$