Question

Suppose that the number of calls per hour arriving at an answ service follows a Poisson process with λ = 4. What is the probability that fewer than two calls come in the first hour?

Answer 1

To find the probability that fewer than two calls come in the first hour, we can use the Poisson distribution formula with an arrival rate (λ) of 4 calls per hour. The probability is found to be approximately 0.6766.

Step-by-step explanation:

To find the probability that fewer than two calls come in the first hour, we can use the Poisson distribution formula. The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space.

For this problem, we have a Poisson process with an arrival rate (λ) of 4 calls per hour. We can use the formula P(X < 2) = e^(-λ) * ((λ^0)/0!) + e^(-λ) * ((λ^1)/1!). By plugging in the value of λ=4, we can calculate the probability as follows:

P(X < 2) = e^(-4) * ((4^0)/0!) + e^(-4) * ((4^1)/1!)

P(X < 2) = e^(-4) * 1 + e^(-4) * 4

P(X < 2) = e^(-4) * 1 + e^(-4) * 4 = 0.1353 + 0.5413 = 0.6766