Bacteria Growth

Solution

Answer: 62.5%

This situation is a classic branching process. Each bacterium lives for one day, produces some number of children, and then dies. The children repeat the same process independently. Let \(X\) be the number of children produced by a bacterium. The distribution is \[ P(X=0)=0.25, \quad P(X=1)=0.35, \quad P(X=2)=0.40. \] The probability generating function of the offspring distribution is \[ f(s) = E[s^X] = 0.25 + 0.35s + 0.40s^2. \] Let \(q\) be the probability that the population eventually dies out. A standard result from branching processes is that \(q\) satisfies \[ q = f(q). \] The intuition is that extinction occurs if all children lineages go extinct. Since each child independently dies out with probability \(q\), we substitute \(s=q\) into the generating function. So we solve \[ q = 0.25 + 0.35q + 0.40q^2. \] \[ 8q^2 - 13q + 5 = 0. \] Using the quadratic formula, \[ q = \frac{13 \pm \sqrt{13^2 - 4(8)(5)}}{16}. \] \[ q = \frac{13 \pm 3}{16}. \] This gives the two solutions \[ q = 1 \quad \text{or} \quad q = \frac{10}{16} = 0.625. \] In a branching process, the extinction probability is the smallest solution in the interval \([0,1]\). Therefore \[ q = 0.625. \] So the probability that the bacteria population eventually dies out is \( 62.5\% \).