At a telephone enquiry system, the number of phone calls regarding relevant enquiry follow. Poisson distribution with an average of 5 phone calls during 10 min time intervals. The probability that there is almost one phone call during a 10 min time period is

Solution :

Required Probability = P(X = 0) + P(X = 1)

= \(e^{-5}\over 0!\).\(5^0\) + \(e^{-5}\over 1!\).\(5^1\)

= \(e^{-5}\) + 5\(e^{-5}\) = \(6\over {e^5}\)


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