A Poisson process has a mean rate of $10$ orders per second. The orders could be from type $A$ with a probability of $0.7$ or from type $B$ with a probability of $0.3$. Orders are independent of each other. Consider a $3$ second time interval.
- What is the probability that we have $5$ orders of type $A$?
- What is the probability that we have $5$ orders of type $B$?
Should I use a different $\lambda$ when I'm calculating each kind of order?
Let $\{N(t):t\geqslant0\}$ be a Poisson process with rate $\lambda>0$. Let $0<p<1$ and consider the split processes $\{N_p(t):t\geqslant0\}$ and $\{N_{1-p}(t):t\geqslant0\}$ corresponding to orders from type A and type B, respectively. Let $T>0$ be fixed.
We can think of $N_p$ as $N_p(t) = N(t)B$ where $B\sim\mathsf{Ber}(p)$ is independent of $\{N(t):t\geqslant 0\}$, and similarly $N_{1-p}(t) = N(t)(1-B)$.
Because the split processes are independent with rates $\lambda p$ and $\lambda (1-p)$ (this is a nontrivial fact that you should prove on your own), we have that the probability of $k$ type $A$ orders in $[0,T]$ is
$$ \mathbb P(N_p(T) = k) = e^{-\lambda pT}\frac{(\lambda p T)^k}{k!} $$
and similarly the probability of $k$ type $B$ orders in $[0,T]$ is $$ \mathbb P(N_{1-p}(T) = k) = e^{-\lambda (1-p)T}\frac{(\lambda (1-p) T)^k}{k!} $$
Plug in $p=\frac7{10}$, $k=5$, $\lambda = 10$, and $T=3$ for your specific problem.