This has already been answered correctly here, but I'll offer the summation for the exact probability:
p( ∑(n = 0 to ∞) (p(1-p))^n) ( (2n)!/(n!(n+1)!) ) ) = p/(1-p)
so that for p = 1/2, the probability is 1. For p < 1/2, the probability is < 1, so that for p = 0, the probability is 0.
We can look at this as an asymmetrical random walk, where increasing the bacteria population by 1 has probability p, and decreasing the same by 1 has probability (1-p). For every increase there has to be a decrease, hence the (p(1-p))^n term. The (2n)!/(n!(n+1)!) term counts the different ways a 2n step path can be done.
Edit: Fixed summation to show that n starts at 0, not 1, excuse me.
Edit 2: I'm still uncomfortable about this result, since my first guess was the same as First Grade Rocks!'s answer, "certain exinction, if given enough time". I keep thinking maybe I have made a mistake or overlooked something.