Q. 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 at the most one phone cell during a 10 min time period, is
A) $\frac{5}{6}$
B) $\frac{6}{55}$
C) $\frac{6}{e^{5}}$
D) $\frac{6}{5^{e}}$
Explanation
(c)
Required probability $= P(X = 0) + P(X = 1)$
$= \frac{e^{- 5}}{0!}.5^{0} + \frac{e^{- 5}}{1!}.5^{1}$
$= e^{- 5} + 5e^{- 5} = \frac{6}{e_{5}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.