Q. A jar has a single cell of a bacteria. After every fixed interval of time the bacteria either splits into two with probability 2/5, does nothing with probability 2/5 or dies out with probability 1/5. What is the probability that the bacteria would continue to make a large family tree over an extended period of time?
A. The situation can be described by the following visual.
The total of each of the above must add up to the probability that the bacteria eventually die out, which is 'x'. So you can phrase the problem recursively as
This simplifies to
which is a quadratic equation, yielding a solution as
This further yields x = 1/2 or x = 1. It's easy to see why x = 1 isn't a solution because scenario 1 would rule that out immediately. This yields a probability of 50%.
In order to test this out, it's also relatively easy to code it up in Python. The following code shows how to simulate this scenario.
import sys import numpy as np import matplotlib.pyplot as plt import seaborn as sns dist_estimates =  for index in range(400): counter = 0 ''' Repeating 100 times for estimating the probability ''' for epoch in range(100): num_bacteria = 1 ''' Looking ahead 10 steps ''' for iter in range(20): bacteria = np.zeros(int(num_bacteria)) ''' Step through each bacteria ''' for nb_iter in range(int(num_bacteria)): ''' Decide if each of the bacteria will die/stay/split in this time slice ''' prob = np.random.uniform(0,1) if prob < 0.2: ''' This one dies ''' bacteria[nb_iter] = 0 elif prob >= 0.2 and prob <= 0.6: ''' This one stays as is ''' bacteria[nb_iter] = 1 else: ''' This one splits into two ''' bacteria[nb_iter] = 2 num_bacteria = np.sum(bacteria) if np.sum(bacteria) > 0: counter += 1 dist_estimates.append(float(counter)/100) ''' Visualization is easy Use matplotlib and seaborn ''' p = sns.distplot(dist_estimates,kde=False,rug=True).get_figure() p.savefig('t.png')
The above code generates a plot of estimates for the said probability. This came out the following way which validates our method.
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