Question

The number of requests for assistance received by a towing service is a Poisson process with a mean rate of 5 calls per hour.

a. What is the probability that 3 or more requests are received during a one hour period?

b. If the operator of the towing service takes a 30-minute break for lunch, what is the probability that they do not miss any requests for assistance?

Answer 1

a)
Here, λ = 5 and x = 2
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X > 2) = 1 - P(X <= 2).
P(X > 2) = 1 - (5^0 * e^-5/0!) + (5^1 * e^-5/1!) + (5^2 * e^-5/2!)
P(X > 2) = 1 - (0.0067 + 0.0337 + 0.0842)
P(X > 2) = 1 - 0.1246 = 0.8754
P(X>=3) = 0.8754

b)
Here, λ = 5/2 = 2.5 (for 30 mins) and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 0)
P(X = 0) = 2.5^0 * e^-2.5/0!
P(X = 0) = 0.0821
Ans: 0.0821