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Seeing a car in 5 minutes from the 20-minute odds

Question

On a quiet road, the probability of seeing at least one car during any 2020-minute window is 609625\tfrac{609}{625}. Assuming cars pass at a constant average rate (so non-overlapping windows are independent), what is the probability of seeing at least one car in a 55-minute window?

Short answer

3/53/5. No car for 2020 minutes means no car in four independent 55-minute windows: q4=1−609625=16625q^4=1-\tfrac{609}{625}=\tfrac{16}{625}, so q=25q=\tfrac25 and the answer is 1−251-\tfrac25.

Adapted from Zhou, A Practical Guide to Quantitative Finance Interviews, sec. 4.3 Conditional Probability and Bayes' Formula, 'Cars on road', p.84; question and answer rewritten by SelfMadeQuant.com.

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