How can I correct the expectation formula?

Question

Let $p_i$ be probability a packet sends from a client $i^{th}$ to a server will be lost. I have two clients which will send packet to the server. The first client will send $n_1$ packets, with loss probability $p_1$ and the second client will send $n_2$ packet, with loss probability $p_2$

It is easy to compute the expected number of packet can reaches to the server as $(1-p_1)n_1+(1-p_2)n_2$

However, I do not know how can I represent the expectation number of packets at the server by using mathematic form. is it right if I write as follows:

$E(X)=E(X_1)+E(X_2)=(1-p_1)n_1+(1-p_2)n_2$

where $X_i$ is event that sends packet from client $i^{th}$ to the server

Answer 1

Indeed, exactly so.

We may let $X_i$ be a random variable counting the number of packets that arrive among the $n_i$ sent from server$\#~i$.   For two servers the total is $X=X_1+X_2$.

Since each has a binomial distribution, $X_i\sim\mathcal {Bin}(n_i, 1-p_i)$, then the expectation of each is $\mathsf E(X_i)=n_i(1-p_i)$.

Finally, for the expectation of the total we use the law of total expectation, so:

$$\begin{align}\mathsf E(X)~=~&\mathsf E(X_1)+\mathsf E(X_2) \\ =~& n_1(1-p_1)+n_2(1-p_2) \end{align}$$