Spotting cars on a highway

You're standing by a highway. The probability of observing at least one car during any 20-minute window is $609/625 = 97.44\%$. Assume the probability of seeing a car is constant (uniform) over time.

What's the probability of observing at least one car in a 5-minute window?

The decomposition trick

The 20-minute interval splits into four non-overlapping 5-minute intervals. By the constant-rate assumption, these intervals are independent and identical.

The four independent "no car" probabilities multiply together. Drag the slider to change $p$ and watch both the individual intervals and the combined result:

Work with the complement — the probability of seeing no cars:

$P(\text{no car in 20 min}) = 1 - \frac{609}{625} = \frac{16}{625}$

Since the four 5-minute intervals are independent:

$P(\text{no car in 20 min}) = P(\text{no car in 5 min})^4$

Verify with the simulation — set $p = 0.6$ and run:

Why complements matter

It's tempting to try: $P(\text{car in 5 min}) = P(\text{car in 20 min}) / 4$. But this gives $609/2500 \approx 0.244$, which is wrong. Why?

"At least one car" is not an additive quantity. If you see a car in the first 5-minute window, that guarantees you see a car in the 20-minute window regardless of what happens next. The events overlap.

The complement "no cars" avoids this overlap:

$P(\text{no car in all four windows}) = P(\text{no car in window 1}) \times P(\text{no car in window 2}) \times \cdots$

This works because "no car in window 1" and "no car in window 2" are independent (by the constant-rate assumption), and their intersection is exactly "no car in the full 20 minutes."

Connection to the Poisson process

This problem is closely related to the Poisson process. If cars arrive according to a Poisson process with rate $\lambda$, then:

$P(\text{no car in time } t) = e^{-\lambda t}$

For our problem: $e^{-20\lambda} = 16/625$, giving $\lambda = \ln(625/16)/20 \approx 0.183$ cars per minute.

And $P(\text{at least one car in 5 min}) = 1 - e^{-5\lambda} = 1 - (e^{-20\lambda})^{1/4} = 1 - (16/625)^{1/4} = 3/5$.

Generalizing

If the probability of at least one event in time $T$ is $P_T$, then the probability in time $T/n$ is:

$p = 1 - (1 - P_T)^{1/n}$

Drag the time slider to see how the probability grows. The green curve (correct) diverges from the red dashed line (wrong linear scaling) — probability growth is decidedly non-linear:

Competition appearances

This problem appears in quant interviews (testing independence and complements), actuarial exams (claim arrival rates and time-interval decomposition), AMC/AIME contests (decomposing probabilities over large intervals), and Poisson process problems from insurance modeling to network traffic.

This technique applies when converting between probabilities at different time scales under a stationarity assumption.

The takeaway

Work with complements when decomposing time intervals. The probability of "at least one" event doesn't split multiplicatively — it's not additive over time. But the probability of "zero events" does split cleanly when intervals are independent.