uppose that cars arrive at a traffic light according to a Poisson process with a rate

of one car per minute starting from time t = 0 minutes. Determine
a. The probability that no cars arrive in the rst two minutes.
b. The probability that no cars arrive ever.
c. The probability that exactly one car arrives in the rst minute.
d. The probability that the second car arrives within 10 seconds of the
rst car.
e. The probability that the second car takes more than two minutes to
arrive.

To solve these questions, we need to rely on the properties of a Poisson process.

A Poisson process is a mathematical model used to describe the arrival of events in a continuous time interval, where these events occur at a constant average rate and independently of one another. In our case, the events are cars arriving at a traffic light, and the rate is one car per minute.

a. The probability that no cars arrive in the first two minutes:
Since the cars arrive according to a Poisson process, the number of arrivals in any given time interval follows a Poisson distribution. The probability of no arrivals in a given time interval is given by the Poisson distribution formula: P(X = 0) = λ^k * e^(-λ) / k!, where λ is the average rate and k is the number of arrivals.

In this case, the average rate is one car per minute, and the time interval is two minutes. Therefore, the probability of no cars arriving in the first two minutes is:
P(X = 0) = (1^0 * e^(-1)) / 0! = e^(-1) ≈ 0.3679

b. The probability that no cars arrive ever:
Since the Poisson process continues indefinitely, the probability of no arrivals ever is given by the limit as time approaches infinity. The probability of no arrivals in a given time interval is e^(-λ), where λ is the average rate.

In this case, λ is still one car per minute. Therefore, the probability of no cars arriving ever is:
P(X = 0) = e^(-1) ≈ 0.3679

c. The probability that exactly one car arrives in the first minute:
Similarly to part a, we can use the Poisson distribution formula to determine the probability of exactly one arrival in a given time interval. The average rate λ is one car per minute, and k = 1.

P(X = 1) = (1^1 * e^(-1)) / 1! = e^(-1) ≈ 0.3679

d. The probability that the second car arrives within 10 seconds of the first car:
Since the cars arrive independently of one another, the interarrival times between cars follow an exponential distribution with a rate equal to the average rate λ.

The probability of the second car arriving within 10 seconds of the first car can be calculated as the probability that an exponential random variable with rate λ = 1/minute is less than or equal to 10 seconds (which is 10/60 minutes).

P(T ≤ 10/60) = 1 - e^(-λ * t) = 1 - e^(-1/6) ≈ 0.0512

e. The probability that the second car takes more than two minutes to arrive:
Similarly to part d, we need to calculate the probability that an exponential random variable with rate λ = 1/minute takes more than two minutes (which is 2 minutes in this case).

P(T > 2) = 1 - P(T ≤ 2) = 1 - (1 - e^(-λ * t)) = e^(-2) ≈ 0.1353

In summary:
a. The probability that no cars arrive in the first two minutes is approximately 0.3679.
b. The probability that no cars arrive ever is approximately 0.3679.
c. The probability that exactly one car arrives in the first minute is approximately 0.3679.
d. The probability that the second car arrives within 10 seconds of the first car is approximately 0.0512.
e. The probability that the second car takes more than two minutes to arrive is approximately 0.1353.