The number of requests for assistance received by a towing service is a Poisson process with ? = 4 rate per hour. a. ?Compute the probability that exactly ten requests are received during a particular 2-hour period. b. ?If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance? c.? ?How many calls would you expect during their break?

Answer : Step 1 of 3 : Given, The number of requests for assistance received by a towing service is a Poisson process with = 4 rate per hour. a) The claim is to compute the probability that exactly ten requests are received during a particular 2-hour period. Let X be the number of requests are received during a particular 2-hour period. x Therefore, P(X = 10) = e x! Where, = 2 = 2×4 = 8 e8810 P(X = 10) = 10! = 0.09926 Therefore, the probability that exactly ten requests are received during a particular 2-hour period is 0.09926 Step 2 of 3 : b) If the operators of the towing service take a 30-min break for lunch. The claim is to find the probability that they do not miss any calls for assistance. Let Y be the number of requests received when the operators of the towing service take a 30-min break for lunch e x Therefore, P(Y = 0) = x! Where, = /2 = 2 2 0 P(Y = 0) = e 0! = 0.1353 0.1353 be the probability that they do not miss any calls for assistance