Cars on the Highway: From a 20-Minute Window to a 5-Minute Window
Cars pass a fixed point on a highway at a constant average rate, and arrivals in disjoint time intervals are independent (a Poisson process). The probability of seeing at least one car in any 20-minute window is $\dfrac{609}{625}$.
What is the probability of seeing at least one car in a 5-minute window?
Hints
- Work with the complement: the probability of seeing no car at all in the window.
- A 20-minute window is four consecutive 5-minute windows, and with a constant rate and independence the no-car events in these four windows are independent and equally likely.
- $(1 - p)^4 = 1 - \tfrac{609}{625} = \tfrac{16}{625}$, so $1 - p = \tfrac{2}{5}$.
Worked Solution
How to Think About It: The trap is to divide by four: "at least one car in 5 minutes" is not one quarter of the 20-minute probability. What does scale cleanly is the complement. With a constant rate and independent increments, "no car in 20 minutes" is the intersection of four independent "no car in 5 minutes" events.
Quick Estimate: $609/625 \approx 0.974$, so no car in 20 minutes has probability about $0.026$. The fourth root of $0.026$ is about $0.4$, so a 5-minute window is car-free about $40\%$ of the time and contains a car about $60\%$ of the time.
Formal Solution:
*Step 1 -- Complement over 20 minutes.*
$$P(\text{no car in 20 min}) = 1 - \frac{609}{625} = \frac{16}{625}.$$
*Step 2 -- Split into four 5-minute windows.* Let $p$ be the probability of at least one car in a 5-minute window. By stationarity each 5-minute window is car-free with probability $1 - p$, and by independent increments the four windows are independent:
$$(1 - p)^4 = \frac{16}{625} = \left(\frac{2}{5}\right)^4 \quad\Longrightarrow\quad 1 - p = \frac{2}{5}, \qquad p = \frac{3}{5}.$$
*Step 3 -- Poisson-rate check.* For a Poisson process with rate $\lambda$ per minute, $e^{-20\lambda} = 16/625$ gives $e^{-5\lambda} = 2/5$, so $P(\text{at least one in 5 min}) = 1 - e^{-5\lambda} = 3/5$. Same answer.
Answer: $P(\text{at least one car in 5 minutes}) = \dfrac{3}{5}$.
Intuition
"No cars in 20 minutes" means "no cars in each of four independent 5-minute pieces," so the no-car probability over 20 minutes is the fourth power of the no-car probability over 5 minutes. The number $609/625$ is chosen so that $16/625 = (2/5)^4$ comes out exactly. This is the Poisson-process scaling rule $P(\text{no event in } t) = e^{-\lambda t}$ in its most elementary form, and it is the same logic used to convert a default probability or arrival probability from one horizon to another.