Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Learn to enjoy mathematics.
Google
Web CTK
Best sites for teachers
Sites for teachers
Sites for parents
Terms of use
Awards

Interactive Activities
CTK Exchange
CTK Insights - a blog

Games & Puzzles
What Is What
Arithmetic/Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Reciprocal links
Privacy Policy

Guest book
News sites

Recommend this site

Best sites for teachers
Sites for teachers
Sites for parents

Education & Parenting

Manifesto: what CTK is about Search CTK Buying a book is a commitment to learning Table of content Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

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 for ever?

Answer

Solution

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

2/3. You can check the solution.

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

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

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

28699867Page copy protected against web site content infringement by Copyscape


Search:
Keywords:


Latest on CTK Exchange
Math
Posted by Laura
2 messages
06:56 AM, Apr-15-08

Divisibility rules - Jargon buste ...
Posted by Carolyn
2 messages
08:35 AM, Apr-04-08

drawing puzzle
Posted by martin gran
31 messages
06:53 PM, May-09-08

Distance to the horizon
Posted by Monty
3 messages
04:38 PM, May-08-08

Mistake on the page (an aside, Be ...
Posted by Max
4 messages
10:28 AM, Feb-28-08

Deriving functions based on diffe ...
Posted by ke_45
1 messages
12:47 PM, May-10-08

A typo in
Posted by alexwajn
1 messages
11:36 PM, Apr-19-08