A customer service firm receives an average of three calls per hour on its toll-free number. For any given hour,find the probability that it will receive at least three calls. Use the Poisson distribution.
A) 0.5768
B) 0.1891
C) 0.4232
D) 0.6138

Question

Quizplus · Accepted Answer

The probability of receiving at least three calls in an hour is given by the complement of the probability of receiving less than three calls. The probability of receiving less than three calls is given by the sum of the probabilities of receiving 0, 1, or 2 calls. The probability of receiving 0 calls is e^-3, the probability of receiving 1 call is 3e^-3/1!, and the probability of receiving 2 calls is 9e^-3/2!. Therefore, the probability of receiving less than three calls is e^-3 + 3e^-3 + 9e^-3/2! = 0.4232. The probability of receiving at least three calls is therefore 1 - 0.4232 = 0.5768.

A Customer Service Firm Receives an Average of Three Calls