25 Telephone calls to a single operator office is modeled as

(25%) Telephone calls to a single operator office is modeled as a Poisson process of 10 calls per hour. The operator wants to take a coffee break of 20 minutes. What is the probability of missing no phone calls? What is the probability of missing at most one phone call? (you need a calculator to compute e^x, for calculation purpose, you may use e = 2.73).

Solution

Note that the probability of k events on a certain interval T is

P(k) = e^(-lambda T) (lambda T)^k / k!

As

lambda = 10/hr

T = 1/3 hr = 0.33333 hr

k = 0

Then

P(0) = 0.03567 [ANSWER]

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Getting the probability of 1 phone call,

P(1) = 0.23884

Thus, the probability of at most 1 is

P(at most 1) = P(0) + P(1) = 0.2745 [ANSWER]

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Note: I used the exact value for e = 2.71828. I didn\'t use 2.73 because it was not a correct rounding off of e. I may have been 2.72 instead.