S for p075 pe is expected to be 13 but for ple05 pe is


The aim of this exercise is to simulate bacteria growth.

Suppose that a certain type of bacteria divides or dies according to the following assumptions:

(a) During a fixed time interval, called a generation, a single bacterium divides into two identical replicas with probability p.

(b) If it does not divide during that interval, it dies.

(c) The offspring (called daughters) will divide or die during the next generation, independently of the past history (there may well be no offspring, in which case the colony becomes extinct).

Start with a single individual and write a script which simulates a number of generations. Take p=0.75. The number of generations which you can simulate will depend on your computer system. Carry out a large number (e.g., 100) of such simulations. The probability of ultimate extinction, p(E), may be estimated as the proportion of simulations that end in extinction. You can also estimate the mean size of the nth generation from a large number of simulations. Compare your estimate with the theoretical mean of (2p)n. Statistical theory shows that the expected value of the extinction probability p(E) is the smaller of 1, and (1-p)/p. So for p=0.75, p(E) is expected to be 1/3. But for p≤0.5, p(E) is expected to be 1, which means that extinction is certain (a rather unexpected result). You can use your script to test this theory by running it for different values of p, and estimating p(E) in each case.

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Basic Computer Science: S for p075 pe is expected to be 13 but for ple05 pe is
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