# Bacteria Evolution

A wonderful property of modern life is that we have anti-biotics to kill bacterial infections. However, we only have a fixed number of anti-biotic medicines, and bacteria are evolving to become resistent to our anti-biotics. In this example we are going to use probability to understand evolution of anti-biotic resistence in bacteria.

Imagine you have a population of 1 million infectious bacteria in your gut, 10% of which have a mutation that makes them slightly more resistant to anti-biotics. You take a course of anti-biotics. The probability that bacteria with the mutation survives is 20%. The probability that bacteria without the mutation survives is 1%.

What is the probability that a randomly chosen bacterium survives the anti-biotics?

Let $E$ be the event that our bacterium survives. Let $M$ be the event that a bacteria has the mutataion. By the By Law of Total Probability (LOTP): \begin{align} \p(E) &= \p(E \and M) + \p(E \and M\c) && \text{LOTP} \\ &= \p(E | M)\p(M) + \p(E | M\c)\p(M\c) && \text{Chain Rule} \\ &= 0.20 \cdot 0.10 + 0.01 \cdot 0.90 && \text{Substituting} \\ &= 0.029 \end{align}

What is the probability that a surviving bacterium has the mutation?

Using the same events in the last section, this question is asking for $\p(M | E)$. We aren't givin the conditional probability in that direction, instead we know $P(E|M)$. Such situations call for Bayes' Theorem: \begin{align} \p(M | E) &= \frac{\p(E|M)\p(M)}{\p(E)} && \text{Bayes} \\ &= \frac{0.20 \cdot 0.10}{\p(E)} && \text{Given} \\ &= \frac{0.20 \cdot 0.10}{0.029} && \text{Calculated} \\ &\approx 0.69 \end{align}

After the course of anti-biotics, 69% of bacteria have the mutation, up from 10% before. If this population is allowed to reproduce you will have a much more resistent set of bacteria!