Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • At a telephone enquiry system, the number of phone calls regarding relevant equiry follows the Poisson distribution with average 5 phone calls during <img alt="\mathrm{10\: min}" src="https://

At a telephone enquiry system, the number of phone calls regarding relevant equiry follows the Poisson distribution with average 5 phone calls during \mathrm{10\: min} time intervals. The probability that there is Most one phone call during a 10 min time period, is
 

Option: 1

\frac{5}{6}
 


Option: 2

\frac{6}{55}
 


Option: 3

\frac{6}{e^2}
 


Option: 4

\mathrm{\frac{6}{5^{\prime}}}


Answers (1)

best_answer

We know that, by Poisson distribution,

\mathrm{ P(X=r)=\frac{e^{-m} m^r}{r !} }

\mathrm{ \therefore P(X \leq 1)=P(X=0)+P(X=1)=e^{-m}+\frac{e^{-m} m}{1 !} }

\mathrm{ \text { Given, } m=\text { mean }=5 }

\mathrm{ \therefore P(X \leq 1)=e^{-5}+5 \times e^{-5} }

\mathrm{ =e^{-5}(1+5)=\frac{6}{e^5} }
Hence option 4 is correct.



 

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question