A unique Binomial Distribution A single cell can either die, with probability $0.1$, or split into two cells, with probability $0.9$, producing a new generation of cells. Each cell in the new generation dies or splits into two cells independently with the same probabilities as the initial cell.
Find the probability distribution for the number of cells in any generation? 
My attempt:
Let $X_i$ be the number of cells in generation $i$ where $i$ is any positive integer and we assume we are starting with one cell. 
Noting that the problem have has to do with the binomial distribution (there are two possible outcomes for each cell going from one generation to the next and such transitions are independent) we define : 
$p = P($ a cell lives$)  = 0.9$ and $q = P($ a cell dies$)  = 0.1$ 
From my working the following pattern emerged;
$$P(X_i = x ) = {2^{i-1} \choose \frac{x}{2}}p^{\frac{x}{2}}q^{2^{i-1}-\frac{x}{2} } $$  where $0 \leq x \leq 2^{i}$ and $x$ is an even integer number. 
Have I answered my own question?
Thanks 
 A: Reading the original question in the context of all of the other exercises, it is clear that the intent of the question is to ask for the probability distribution of the number of cells in the third generation, if the first generation consists of the single initial cell.
To find this, it is a simple matter of creating a probability tree:  after the first generation, there are $0$ cells with probability $0.1$, and $2$ cells with probability $0.9$.  In the second generation, if there are no cells, then the third generation will have no cells either.  If there are two cells, then both need to die out, each with probability $0.1$, in order for the third generation to be devoid of cells.  So in the third generation, the probability of $0$ cells is $$\Pr[X_3 = 0] = 0.1 + (0.9)(0.1)^2.$$
Note that because cells either die or split in two, it is not possible for there to be an odd number of cells apart from the first generation.
I leave it to you to figure out the remaining probabilities; note that we have positive probabilities only for $X_3 = 2$ and $X_3 = 4$.  We can't have any other outcomes.
