Problem
Information packets arrive at a server with a poisson process having rate $\lambda = 2$ per hour.
The server processing time for a packet follows the distribution : $f(x) = 1, 0\leq x\leq1$
The status of the server is busy if it is processing a packet, otherwise it is waiting. If a packet arrives while the server is busy, that packet is lost.
Let $t_B$ denote the length of one busy period, $t_I$ denote the length of one idle period. Find the distrubutions of $t_B$ & $t_I$.
Let $N$ denote the total number of lost packets, $T_B$ denote the total busy time of the server up until time $T=10$ hours.
Show that : $$ \begin{align} E[N] = \lambda E[T_B] \end{align} $$
My attempt :
$t_B$ is simply the time taken for the server to process a packet, so its distribution is also $f(x) = 1, 0 \leq x \leq 1$
$t_I$ is the inter-arrival time between two packets which follows an $Exp(\lambda = 2)$ distribution, using the relation between poisson & exponential rv's.
In trying to prove the equation, this is what I have so far :
N = total no. of lost packets = no. of packets arriving during $T_B$(while the server is busy)
$$ \begin{align} E[N] &= E[\textrm{No. of arrivals during }T_B] \\ &=E[\textrm{No. of arrivals per unit time}]\cdot E[T_B] \\ &=\lambda E[T_B]~\textrm{(Shown)} \end{align} $$
Am I going about this the right way? If so is there a proper way to express this in math notation? Thank you!