A car on a quiet road

Idea. “At least one car” does not split nicely over time, but “no car” does: no car in 30 minutes means no car in each of three separate 10-minute windows, and those are independent.

Step 1. Let $q$ be the probability of no car in 10 minutes. The three 10-minute pieces of a 30-minute window are independent and have the same distribution, so the probability of no car in 30 minutes is $$q^3=1-0.95=\mathrm{0.05.}$$

Step 2. $$q=0.05^{1/3}\approx 0.368,$$ so the probability of at least one car in 10 minutes is $$1-0.05^{1/3}\approx \mathrm{0.632.}$$

With the rate. The same answer comes out of the Poisson formula: no car in time $t$ has probability $e^{-\lambda t}$, so $e^{-30\lambda }=0.05$ and $1-e^{-10\lambda }=1-0.05^{1/3}$.

The trap. Dividing by three, $0.95/3\approx 0.317$, treats the probability as if it grew in proportion to time. That is only close to the truth for small probabilities.