I’m stuck with a problem for my Statistics class. The problem says that there is a hospital where patients arrive at a constant rate of 2 patients per hour, and there is a doctor that works 12 hours, from 6 am to 6 pm. If the doctor has treated 6 patients by 8 am, what is the probability that he treats 9 patients by 10 am?
So, I was planning to solve this problem using Poisson distribution and conditional probability. I identified two events: A, where after 2 hours the doctor treats 6 patients, and B, where after 4 hours the doctor treats 9 patients.
For A, m=lambda*hours=2*2hrs=4, and P(X=6)= (e^-4 * 4^6)/6! = 0.104
For B, m=lambda*hours=2*4hrs=8, and P(X=9)= (e^-8 * 8^9)/9! = 0.124
So, given that events are independent in Poisson distributions, I try to calculate P(B|A) as P(B)/P(A) but I get a number > 1.
Also, I tried working the answer considering the time interval from 8 am to 10 pm, where the number of patients increased by (9-6)=3
In that scenario, m=lambda*hours=2*2hrs=4, and P(X=3)= (e^-4 * 4^3)/3! = 0.53 , but I think it must be wrong too, since it does not consider the event A.
Could someone help me with this?