Question

Cars arrive at a toll booth according to a poisson process with mean 80 cars per hour. a. if the attendant makes a one-minute phone call, what is the probability that at least 1 car arrives during the call? b. how long can the attendant's phone call last if the probability is at least 0.4 that no cars arrive during the call?

Answer 1

a. The probability that at least 1 car arrives during a one-minute phone call is 1 - e^(-4/3). b. The longest duration the phone call can last for a probability of no cars arriving of at least 0.4 is -ln(0.4) hours.

Step-by-step explanation:

a. To find the probability that at least 1 car arrives during a one-minute phone call, we need to calculate the probability that 0 cars arrive and subtract it from 1. The average number of cars arriving per minute is 80/60 = 4/3 cars. Using the Poisson distribution, the probability of 0 cars arriving is given by e^(-λ) where λ is the average number of cars. Therefore, the probability that at least 1 car arrives during the call is 1 - e^(-4/3).

b. To find the longest duration of the phone call where the probability of no cars arriving is at least 0.4, we need to find the value of λ that satisfies P(0) ≥ 0.4. Rearranging the equation P(0) = e^(-λ) gives e^(-λ) ≥ 0.4. Taking the natural logarithm of both sides gives -λ ≥ ln(0.4), which simplifies to λ ≤ -ln(0.4). Therefore, the longest duration the phone call can last is -ln(0.4) hours.