First red light run on the fourth light

At each traffic light, a careless driver runs it independently with the same probability $p$, where $p<\tfrac12$. The chance that the second light is the first one she runs equals $\tfrac{9}{100}$. What is the probability that the fourth light is the first one she runs?

$\tfrac{729}{10000}$. We need $(1-p)p=0.09$, whose roots are $0.1$ and $0.9$; the condition $p<\tfrac12$ picks $p=0.1$. Then the probability of three passes followed by a run is $(0.9)^3(0.1)=0.0729$.