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In an hour, a bacterium dies with probability $p$ or else splits into two. What is the probability that a single bacterium produces a population that will never become extinct?

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Let $x$ (for "extinction") be the probability that the colony dies out. This can happen in two ways: either the first bacterium dies immediately (with probability $p$) or it splits and then neither of its children has infinite descendants.

This means we can write $x$ as

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

which is a quadratic with two real roots. One solution is $x=1$ and the other is $x=\frac{p}{1-p}$. The latter is a valid probability (ie, it's between 0 and 1) only when $p \in [0,1/2]$ when the graph looks like this:

Graph of p/(1-p) from 0 to 1/2

So extinction is guaranteed for $p \geq 1/2$ and then decreases as shown in the graph above as $p$ decreases. Survival is guaranteed, of course, if $p=0.$

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\begin{align} r & =\Pr(\text{posterity lives forever}) \\[10pt] & = \Pr(\text{split})\cdot\Pr(\text{at least one offspring lives forever}) \\[10pt] & = (1-p)\cdot(1-\Pr(\text{both lines eventually die out})) \\[10pt] & = (1-p)\cdot(1-(1-r)^2). \end{align} So $$ r = (1-p)(1-(1-r)^2). $$ That's a quadratic equation in the variable $r$. One solution is obviously $r=0$, so the polynomial $(1-p)(1-(1-r)^2)-r$ can be factored as $$ (1-p)(1-(1-r)^2)-r = r\cdot\text{something}. $$ Then write $\text{something}=0$ and it's not even quadratic.

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