Suppose that an average of 45 people arrive per hour to a restaurant according to a Poisson Process.
What is the probability that more than 5 minutes pass before two clients arrive to the restaurant?
I know that = .75 per minute and that 0 or 1 client can arrive from 0 to 5 minutes.
My set up for the problem is:
$$P(N(5)<=1) = \frac{e^{-3.75}*3.75^5}{5!}$$
Is it right?
Let $\{N(t):t\geqslant0\}$ be a Poisson process with rate $\lambda>0$, $T>0$, and $n$ a positive integer. Then $$ \mathbb P(N(t)\leqslant n-1) = \sum_{k=0}^{n-1}e^{-\lambda t}\frac{(\lambda t)^k}{k!}. $$ Here $\lambda=\frac34$, $T=5$, and $n=2$, so we have $$ \mathbb P(N(5)\leqslant 1) = e^{-\frac34\cdot5}\left(1 + \frac34\cdot 5\right)= \frac{19}4e^{-\frac{15}4}\approx0.1117093. $$