Customers arrive in single server line to be served according to Poisson process with intensity 5 customers an hour. (there is a single server who services each customer in the order they arrive while all the customers wait in line).
If the customers begin to arrive at 8am, find the probability that at least 4 customers arrived between 9am and 10am.
my answer:
rate = 5 customers per hour
t = 1
k = 4
Thus we find the answer by Poisson approximation:
1 – ( (e^-5) + ((e^-5) (4)/1!) + ((e^-5) (4)^2/2!) + ((e^-5) (4)^3/3!)))
Can someone please confirm my answer?
Best Answer
Denote by $N_t$ the number of customers arrived in the time-interval $[0,t]$. We identify 8am with $t=0$. From the definition of a Poisson process you know that the number of customers arrived between 9am and 10am is independent on the number of customers arrived between 8am and 9am. Moreover we have:
$P(N_{t+s}-N_s=k)=P(N_t=k)=e^{-\lambda t}\displaystyle\frac{(\lambda t)^k}{k!}$, where $\lambda=5$.
Hence the probability that between 9am and 10am arrive at least 4 customers is $P(N_{10}-N_9\geq 4)=P(N_1\geq 4)=1-P(N_1<4)$.
But $P(N_1<4)=P(N_1=0)+P(N_1=1)+P(N_1=2)+P(N_1=3)$. Now you just compute each probability $P(N_1=k)$, $k=0,1,2,3$ and replace them into $1-P(N_1<4)$.