# What is the probability of extinction?

In every second, the probability that a certain bacterium will die is p and the probability that it will divide into two identical bacteria is 1-p. (The offsprings will die or divide independently of each other.) What is the probability of extinction?

### 7 Answers

- Scythian1950Lv 71 decade agoFavourite answer
This has already been answered correctly here, but I'll offer the summation for the exact probability:

p( ∑(n = 0 to ∞) (p(1-p))^n) ( (2n)!/(n!(n+1)!) ) ) = p/(1-p)

so that for p = 1/2, the probability is 1. For p < 1/2, the probability is < 1, so that for p = 0, the probability is 0.

We can look at this as an asymmetrical random walk, where increasing the bacteria population by 1 has probability p, and decreasing the same by 1 has probability (1-p). For every increase there has to be a decrease, hence the (p(1-p))^n term. The (2n)!/(n!(n+1)!) term counts the different ways a 2n step path can be done.

Edit: Fixed summation to show that n starts at 0, not 1, excuse me.

Edit 2: I'm still uncomfortable about this result, since my first guess was the same as First Grade Rocks!'s answer, "certain exinction, if given enough time". I keep thinking maybe I have made a mistake or overlooked something.

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- simplicitusLv 71 decade ago
The previous answer had the write idea but the wrong algebra.

If the probability of the line associated with a single bacterium will die out is E, then

E = p + (1-p) probability both offspring will die

E = p + (1-p) E^2

Thus: (1-p)E^2 - E + p = 0

Given p, you can solve for E. (Note that E=1 is always one root. Use the other if it is in [0,1].)

For p = 1/2, E = 1, which is what you'd expect:

50% chance of extinction in the the first generation,

25% chance of extinction in the second generation,

etc.

In fact, whenever p>= 1/2, the probability of extinction should be 1.

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- steppenwolfLv 51 decade ago
This is a branching process. The way to get the extinction probability is to first find the PGF (probability generating function) as: ((1-p) + pt^2) and solve for the smallest positive root in the equation:

(p + (1-p)t^2) = t

⇒ (1-p)t^2 - t + p = 0

[1 +/- (1-2p)]/[2(1-p)]

i.e., p/(1-p) or 1

If p > 0.5, then p/(1-p) > 1 i.e., prob of extinction = 1.

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- Anonymous1 decade ago
If at time t0 the population is N0, then after i seconds the expected population is Ni = N0*(2^i)*(1-p)^i.

After a lot of seconds i→∝, the population will be close to Ni:

1) If p<=1/2 (and N0 big enough), then the probability of extinction is a little bigger than p^N0

2) If p>1/2 then the probability of extinction is 100%

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- 1 decade ago
I really like simplicitus's answer and think it's spot on. I just wanted to add some additional comments . . .

Using simplicitus's notation, where E is the probability of a bacterium's lineage dying out, then

E = min( 1, p / (1-p) ).

If you started with some initial population of bactiera A0, then

E(A0) = E^A0 = probability of all A0 lineages dying out

= min( 1, (p / (1-p))^A0 )

where E(A0) is the probability of extinction starting with an initial population of A0.

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- 1 decade ago
Given an infinite amount of time, the probability is 100%.

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- 1 decade ago
This is a binomial distribution. Look it up online. And do your own homework.

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