Thread: Poisson process probabilites

A shop opens at 9am and shuts at 5pm. While the shop is open customers arrive as a Poisson process of rate 20 per hour.
a) Calculate the following probabilities:

i) P(no customers enter the shop before 9:03am)
I am confused whether to convert the hours to minutes or not but this is what i have tried.
P(X(1/3)=0)=P(X(1/3)-X(0)=0)=

ii) P(exactly 20 customers enter the shop between 2pm and 3pm)
I can work this out if know how to write the above in the form P(X(s+t)=k) etc..
iii) P(exactly 20 customers enter the shop before 11am given that 150 customers

enter during the whole day)
I can work this out if know how to write the above in the form P(X(s+t)=k) etc..

Thanks for any help.

Originally Posted by charikaar

A shop opens at 9am and shuts at 5pm. While the shop is open customers arrive as a Poisson process of rate 20 per hour.
a) Calculate the following probabilities:

i) P(no customers enter the shop before 9:03am)
I am confused whether to convert the hours to minutes or not but this is what i have tried.
P(X(1/3)=0)=P(X(1/3)-X(0)=0)=

ii) P(exactly 20 customers enter the shop between 2pm and 3pm)
I can work this out if know how to write the above in the form P(X(s+t)=k) etc..
iii) P(exactly 20 customers enter the shop before 11am given that 150 customers

enter during the whole day)
I can work this out if know how to write the above in the form P(X(s+t)=k) etc..

Thanks for any help.

i) The number of customers entering between 9:00 and 9:03 has a Poisson distribution with mean 1. You want P(X=1), where X has that distribution.

ii) The number of customers entering between 2 and 3 has a Poisson distribution with mean 20. You want P(X=20), where X has that distribution.

iii) Let X be the number of customers entering before 11 am and Y be the number of customers entering between 11 am and 5 pm, so X has a Poisson distribution with mean 40, Y has a Poisson distribution with mean 120, and X+Y has a Poisson distribution with mean 160. Then