Rescaling a Poisson Time Window

Model the crashes as a Poisson process of rate $\lambda$ per hour. The number of crashes in a window of length $t$ hours is $\mathrm{Poisson}(\lambda t)$ , so the probability of no crash in that window is $e^{-\lambda t}$ .

The hour has no crash with probability $1-\frac{8}{9}=\frac19$ , so

e^{-\lambda}=\tfrac19 .

For a $30$ -minute window, $t=\tfrac12$ , and

\mathbb{P}(\text{no crash in 30 minutes}) = e^{-\lambda/2} = \bigl(e^{-\lambda}\bigr)^{1/2} = \tfrac13 .

Hence the probability of at least one crash in $30$ minutes is $1-\tfrac13=\tfrac23$ .

\boxed{\tfrac{2}{3}}

Alternative. The value of $\lambda$ is never needed. Split the hour into its two halves. Let $p$ be the probability of at least one crash in a half-hour; both halves share this $p$ because the rate is constant. The hour is crash-free exactly when both halves are. Counts in disjoint windows are independent, so

(1-p)^2=\tfrac19 .

Taking the root in $[0,1]$ gives $1-p=\tfrac13$ , so $p=\tfrac23$ .

Key idea. No-event probabilities multiply over disjoint windows. If the long window is $k$ times the short one, the short window is event-free with probability equal to the $k$ -th root of the long window’s.