Cars on the Road: Poisson Probability Over Time
The Puzzle
You’re told that the probability of seeing at least one car on a highway during any 20-minute interval is:
\[ p_{20} = \frac{609}{625} \]
Assuming car arrivals are uniform and follow a Poisson process, what is the probability of seeing at least one car in a 5-minute interval?
Step 1: Use Poisson Properties
In a Poisson process, the number of arrivals in time \( t \) follows the distribution:
\[ P(\text{no arrivals in } t) = e^{-\lambda t} \]
So:
\[ P(\geq 1 \text{ in } t) = 1 - e^{-\lambda t} \]
Given:
\[ 1 - e^{-20\lambda} = \frac{609}{625} \]
Solve for \( \lambda \):
\[ e^{-20\lambda} = 1 - \frac{609}{625} = \frac{16}{625} \]
Take logs:
\[ -20\lambda = \ln\left(\frac{16}{625}\right) \Rightarrow \lambda = -\frac{1}{20} \ln\left(\frac{16}{625}\right) \]
Step 2: Compute \( p_5 \)
\[ p_5 = 1 - e^{-5\lambda} = 1 - \left(e^{-20\lambda}\right)^{1/4} = 1 - \left(\frac{16}{625}\right)^{1/4} \]
Calculate:
\[ \left(\frac{16}{625}\right)^{1/4} = \frac{2}{5} \]
So:
\[ p_5 = 1 - \frac{2}{5} = \boxed{\frac{3}{5}} \]
Final Answer
\[ \boxed{\frac{3}{5}} \]