Question

Given an arrival process with λ (lambda) = 2.0, what is the probability that an arrival occurs in the first t = 5 time units? Round to four decimal places as needed.

Answer 1

To find the probability of an arrival within 5 time units in a Poisson process with λ = 2.0, calculate the complement of the probability of no arrivals, which is 1 - e^(-10), where e is the base of the natural logarithm.

Step-by-step explanation:

The question concerns the calculation of the probability of the arrival of an event in a Poisson process within a certain time frame. In a Poisson process with a rate parameter λ (lambda), the probability of exactly k events in an interval of time is given by the formula P(X = k) = (λt)^k * e^(-λt) / k!, where λ is the average number of events per time unit, t is the time, and e is the base of the natural logarithm.

For this case, with λ = 2.0 and t = 5, we want to find the probability of at least one arrival, which is easier to calculate by finding the complement of the probability of having no arrivals (k=0) and subtracting it from 1. The probability of no arrivals is calculated as P(X = 0) = (2.0*5)^0 * e^(-2.0*5) / 0! = e^(-10). Thus, the probability of at least one arrival is 1 - e^(-10).