At a telephone enquiry system, the number of phone calls regarding relevant enquiry follows a Poisson distribution with an average of 5 phone calls du - Mathematics MCQ AIEEE 2006 2026 | ExamTest.live

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At a telephone enquiry system, the number of phone calls regarding relevant enquiry follows a Poisson distribution with an average of 5 phone calls during 10 -minute time intervals. The probability that there is at most one phone call during a 10 -minute time period is:

Solution & Step-by-step Explanation

For a Poisson distribution, P (X = r) = \frac{e ^{- m} m ^{r}}{r !}, where m is the mean.Here, m = 5 .We need to find P (X \leq 1) = P (X = 0) + P (X = 1) . P (X = 0) = \frac{e ^{- 5} 5 ^{0}}{0 !} = e^{- 5} P (X = 1) = \frac{e ^{- 5} 5 ^{1}}{1 !} = 5 e^{- 5} Sum = e^{- 5} + 5 e^{- 5} = 6 e^{- 5} .This can also be written as \frac{6}{e ^{5}} .

Try it yourself before checking the explanation above.

At a telephone enquiry system, the number of phone calls regarding relevant enquiry follows a Poisson distribution with an average of 5 phone calls during 10 -minute time intervals. The probability that there is at most one phone call during a 10 -minute time period is:

\frac{6}{5} e

\frac{5}{6}

\frac{6}{55}

6 e^{- 5}

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