Amoeba's Survival

A population starts with a single amoeba. For this one and for the generations thereafter, there is a probability of 3/4 that an individual amoeba will split to create two amoebas, and a 1/4 probability that it will die out without producing offspring. What is the probability that the family tree of the original amoeba will go on forever?

Answer

Solution

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Copyright © 1996-2017 Alexander Bogomolny

A population starts with a single amoeba. For this one and for the generations thereafter, there is a probability of 3/4 that an individual amoeba will split to create two amoebas, and a 1/4 probability that it will die out without producing offspring. What is the probability that the family tree of the original amoeba will go on forever?

The probability in question is 2/3. You can check the solution.

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Copyright © 1996-2017 Alexander Bogomolny

A population starts with a single amoeba. For this one and for the generations thereafter, there is a probability of 3/4 that an individual amoeba will split to create two amoebas, and a 1/4 probability that it will die out without producing offspring. What is the probability that the family tree of the original amoeba will go on forever?

Good and general notations help solve the problem. Let p be the probability of a successful split for a single amoeba, and P the probability in question, the probability that an amoeba's family tree is infinite.

With the probability p we have a second generation of two amoebas. The probability that at least one of them will have an infinite family tree is 1 - (1 - P)2, because (1 - P)2 is the probability that both of them will perish undivided. Therefore,

P = p(1 - (1 - P)2)

because both sides of the equation represent the probability of the long term survival to the original amoeba.

Simipification yields

pP2 + (1 - 2p)P = 0,

or

P·(pP + (1 - 2p)) = 0,

and since P ≠ 0,

pP + (1 - 2p) = 0,

or

P = (2p - 1)/p.

We see that if a generic amoeba divides with the probability not exceeding 1/2, it stands no chance to survive for ever. However, for the specific case where p = 3/4, the probability of survival P = 2/3.

References

  1. R. Blum et al., Mathemagic, Main Street, 2002

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Copyright © 1996-2017 Alexander Bogomolny

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