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 minute time intervals. The probability that there is at the most one phone call during a 10 minute time period, is :

Question

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 minute time intervals. The probability that there is at the most one phone call during a 10 minute time period, is : (A) (5/6) (B) (6/55) (C) (6/e5) (D) (6/5e)

Tardigrade Admin · Accepted Answer

Required probability = P (X = 0)+ P (X = 1) =(e-5/0 !)50+(e-5/1 !)⋅51 =e-5+5e-5 =(6/e5)

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 minute time intervals. The probability that there is at the most one phone call during a 10 minute time period, is :

$\frac{5}{6}$

$\frac{6}{55}$

$\frac{6}{e ^{5}}$

$\frac{6}{5 ^{e}}$

Required probability $= P (X = 0) + P (X = 1)$
$= \frac{e ^{- 5}}{0 !} 5^{0} + \frac{e ^{- 5}}{1 !} \cdot 5^{1}$
$= e^{- 5} + 5 e^{- 5}$
$= \frac{6}{e ^{5}}$

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 minute time intervals. The probability that there is at the most one phone call during a 10 minute time period, is :

65​

556​

e56​

5e6​

Required probability =P(X=0)+P(X=1) =0!e−5​50+1!e−5​⋅51 =e−5+5e−5 =e56​

$\frac{5}{6}$

$\frac{6}{55}$

$\frac{6}{e ^{5}}$

$\frac{6}{5 ^{e}}$

Required probability $= P (X = 0) + P (X = 1)$
$= \frac{e ^{- 5}}{0 !} 5^{0} + \frac{e ^{- 5}}{1 !} \cdot 5^{1}$
$= e^{- 5} + 5 e^{- 5}$
$= \frac{6}{e ^{5}}$