Question:
Suppose there are number of accidents at location \(A\) and location \(B\) each day with Poisson Distribution with rate \(0.001\) and the number of patients during each accident is Poisson distributed with rate \(0.1\). Find the probability that the number of patients less than \(3\) in one day.
Attempt:
So I did \(E(N)=0.001\times 0.1=0.0001\) and then using this mean to find
P(X<3)=\(\sum_{0}^{2}Poi(0.0001)=1\)
and I got a value of \(1\). I am sure this is the wrong answer. Where did it go wrong? I need some help. Thank you!
Suppose there are number of accidents at location \(A\) and location \(B\) each day with Poisson Distribution with rate \(0.001\) and the number of patients during each accident is Poisson distributed with rate \(0.1\). Find the probability that the number of patients less than \(3\) in one day.
Attempt:
So I did \(E(N)=0.001\times 0.1=0.0001\) and then using this mean to find
P(X<3)=\(\sum_{0}^{2}Poi(0.0001)=1\)
and I got a value of \(1\). I am sure this is the wrong answer. Where did it go wrong? I need some help. Thank you!