Poisson Process: From 20 Minutes to 5 Minutes

Given the probability of seeing at least one car in 20 minutes, find the probability of seeing at least one car in 5 minutes.

Given
P(at least 1 car in 20 min) = 609 / 625
Use the complement
P(0 cars in 20 min) = 1 - 609/625 = 16/625
Key Poisson idea

In a Poisson process, disjoint time intervals are independent, and equal-length intervals behave the same. Since 20 minutes consists of four 5-minute intervals:

0–5 min P(0 cars) = q
5–10 min P(0 cars) = q
10–15 min P(0 cars) = q
15–20 min P(0 cars) = q
  1. Let q = P(0 cars in 5 min).
  2. Then by independence,
    P(0 cars in 20 min) = q⁴
  3. Substitute the known value:
    q⁴ = 16/625 = (2/5)⁴
  4. So:
    q = 2/5
  5. Therefore:
    P(at least 1 car in 5 min) = 1 - q = 1 - 2/5 = 3/5
Final Answer
P(at least 1 car in 5 min) = 3/5 = 0.6 = 60%
Equivalent exponential view
P(0 arrivals in t) = e-λt

So from e-20λ = 16/625, we get e-5λ = (16/625)1/4 = 2/5, hence the same answer: 1 - 2/5 = 3/5.