Proof of $ \text{Poisson}(\lambda p) $-arrivals. We have a queue where people pass out of it with $ \text{Poisson}(\lambda) $ and they come in with probability $ p $.

I understand that the arrivals follow $ \text{Poisson}(\lambda p) $, but how can I prove it?
Let X be the number of arrivals.
\begin{eqnarray*}
&amp;&amp;P(X=r)=\sum\limits_{k=r}^{\infty}\dfrac{\lambda ^k}{k!}\exp{(-\lambda)}C_k^rp^r(1-p)^{k-r}\\
&amp;&amp;=\sum\limits_{k=r}^{\infty}\exp{(-\lambda)}\dfrac{\lambda ^k k!}{r!(k-r)!k!}p^r(1-p)^{k-r}\\
&amp;&amp;=\dfrac{\lambda^r p^r}{r!}\sum\limits_{k=r}^{\infty}\exp{(-\lambda)}\dfrac{\lambda ^{k-r} }{(k-r)!}(1-p)^{k-r}\\
&amp;&amp;=\dfrac{\lambda^r p^r}{r!}\exp{(-\lambda)}\exp{(\lambda(1-p))}=\dfrac{(\lambda p)^r}{r!}e^{-\lambda p}
\end{eqnarray*}
So $X\sim P(\lambda p)$
Here is a cool proof I thought about.

I am going to split the original process into two parts, one that can arise with probability p and the other with probability 1-p. Here in your example, the original process is passing out, with proba p going in, proba 1-p not going in. 

Now what we want to prove is how the splitted process is still a poisson process, and this will give us the proof that your process is indeed poisson with parameter $\lambda p$.

Also, since it is the case that interested me, I prove it for poisson proceses, not poisson random variables (replace $\lambda$ by $\lambda t $.


  We have the poisson process
  
  $$X(t) = X_1(t) + X_2(t) $$
  
  First we calculate the joint probability 
  
  $$P[X_1(t) = k, X_2(t) = m] = \sum_{n=0}^{\infty} P[X_1(t) = k, X_2(t)
&gt; = m \mid X(t) = n]P[X(t) = n]$$
  
  Note that
  
  $$P[X_1(t) = k, X_2(t) = m \mid X(t) = n] = 0 \:\:\: \text{when}\:\: n
&gt; \neq k+m$$
  
  Now
  
  $$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k, X_2(t) = m \mid X(t) =
&gt; k+m]P[X(t) = k+m]$$
  
  $$= P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m]e^{-\lambda
&gt; t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$
  
  Now, given that $k+m$ events occurred, since each event has
  probability $p$ of being a type $1$ event and probability $1-p$ of
  being a type $2$ event, it follows that
  
  $$P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m] = \binom{k+m}{k}
&gt; p^k(1-p)^m$$
  
  Thus, 
  
  $$P[X_1(t) = k, X_2(t) = m] = \binom{k+m}{k} p^k(1-p)^m e^{-\lambda
&gt; t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$
  
  $$= \frac{(k+m)!}{k!m!} p^k(1-p)^m e^{-\lambda t}\frac{(\lambda
&gt; t)^{k+m}}{(k+m)!}$$
  
  $$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
&gt; \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\:\:\:\:(1)$$
  
  Then 
  
  $$P(X_1 = k) = \sum_{m=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$
  
  $$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
&gt; \sum_{m=1}^{\infty} \frac{[\lambda (1 - p)t]^m}{m!}$$
  
  $$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
&gt; e^{\lambda(1-p) t}$$
  
  $$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} \:\:\:\:\:\: (2)$$
  
  which indicates that $X_1(t)$ is a poisson process with rate $\lambda
&gt; p$.
  
  Similarly, we can obtain
  
  $$P(X_2(t) = m) = \sum_{k=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$
  
  $$= e^{-\lambda(1-p) t} \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\:
&gt; (3)$$
  
  and so $X_2(t)$ is a poisson process with rate $\lambda (1-p)$.  
  
  Finally, from equations $(1)$, $(2)$ and $(3)$
  
  $$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k]P[X_2(t) = m]$$
  
  Hence, $X_1(t)$ and $X_2(t)$ are independent.


Then you have the answer to your problem ! I am not sure that there is a simplier way to prove it, other than intuition.

The original part of the proof comes from here : Splitting Poisson process formal proof and the reward of the work should be attributed to JKnecht. I wanted to use his proof for another interesting similar case, which is yours, because in the end, it is the same problem.