Simple Poisson problem I wanted to ensure that I solve the following problem correctly:
There is a door to the building and I do not have a key. On average, every 5 minutes a person enters or exits the building (i.e., eventually opening the door). What is probability that I will enter the door within 3 minutes.

Solution:
1. Lambda: on average 0.6 people entering/exiting the building in 3 minutes


The probability to enter the building in 3 minutes is the probability 1-the probability that no one enters / exits the building.
The probability that no one enters / exits the building: P(X=0) = e^-0.6 = 0.55
Probability to enter the building within 3 minutes is 1-0.55 = 0.45


Thanks!
Since average number of people who enter or exit in 5min is 1 we can conclude probability that at least one person enters or exits in 1min is 0.2. So, probability that at least one person does it in 3min is =1-(1-0.2)^3=0.488
I tried this in matlab simulations and its checks.
Since average number of people who enter or exit in 5min is 1 we can conclude probability that at least one person enters or exits in 1min is 0.2. So, probability that at least one person does it in 3min is =1-(1-0.2)^3=0.488
I tried this in matlab simulations and its checks.