Final Question - Part 2 (Branching Process $Z_0 = 8$. Find $P(Z_2 =0)$)

Question

Consider a branching process where the number of offspring of an individual is a binomial random variable with parameters (3, p), with $p ∈ (0, 1)$. $Z_0 = 8$. What is $P(Z_2 =0)$

My attempt:

Let $s = P(Z_2 = 0 | Z_0=1)$

s= $P(Z_1=0) + P(Z_1 = 1, Z_2 = 0) + P(Z_1 = 2, Z_2 = 0) + P(Z_1 = 3, Z_2 = 0)$

= $(1-p)^3 + 3(p)(1-p)^2(1-p) + 3(p)^2(1-p)(1-p)^2 + 3(p)^3(1-p)^3$

Therefore, if $Z_0 = 8$, $P(Z_2 =0) = s^8 = ((1-p)^3 + 3(p)(1-p)^2(1-p) + 3(p)^2(1-p)(1-p)^2 + 3(p)^3(1-p)^3)^8$

Answer 1

Let $s = P(Z_2 = 0 | Z_0=1)$, then we have \begin{align} s &= P(Z_1=0| Z_0=1) + P(Z_1 = 1, Z_2 = 0|Z_0=1) \\&+ P(Z_1 = 2, Z_2 = 0|Z_0=1) + P(Z_1 = 3, Z_2 = 0|Z_0=1) \\ &= (1-p)^3 + 3(p)(1-p)^2P(Z_2=0|Z_1=1) \\&+ 3(p)^2(1-p)P(Z_2=0|Z_1=2) + (p)^3P(Z_2=0|Z_1=3) \\ &=(1-p)^3 + 3(p)(1-p)^2(1-p)^3 + 3(p)^2(1-p)((1-p)^3)^2 + (p)^3((1-p)^3)^3 \\ &=(1-p)^3+3p(1-p)^5+3p^2(1-p)^7+p^3(1-p)^9 \\ &= (1-p)^3 (1+3p(1-p)^2+3p^2(1-p)^4 + p^3(1-p)^6) \\ &= (1-p)^3 (1+p(1-p)^2)^3 \\ &= [(1-p)(1+p(1-p)^2)]^3 \end{align}

If $Z_0=8$, just raise to the $8$-th power.

$$[(1-p)(1+p(1-p)^2)]^{24}$$

Answer 2

Starting with 8 individuals, the generating function for the number, $X$, in the next generation is $$\sum_ip(X=i)t^i=(pt+q)^{24}.$$ The probability there are no offspring in the following generation is$$\sum_ip(X=i)q^{3i}=(pq^3+q)^{24}.$$