Given that, total load-in request = $50 \hspace{0.1cm} requests/sec$

& frame transmission time = $40 \hspace{0.1cm} msec$

∴ Channel traffic load rate (G) = $50 \times (40 \times 10 ^{-3})$

$\qquad \qquad \qquad \qquad \qquad = 50 \times 0.04$

$\qquad \qquad \qquad \qquad \qquad = 2.00$

$\qquad \qquad \qquad \qquad \qquad = 2 \hspace{0.1cm}requests/slot$

Now, we know that the probability that K frames are generated during a given time frame in which G frames are expected , is given by Poisson Distribution

$$P[k] = \dfrac{G ^k e^{-G}}{K!}$$

a) Now, chances of success on the first attempt =

$P(0) = \dfrac{G^0e^{-G}}{0!} \qquad∵ [ \text{P(0) because no other requests occurs within the first slot }]\\ \qquad = \dfrac{1 \times e^{-G}}{1}\\\qquad = e^{-G} = e^{-2} \qquad ∵ [ \text{We already derive G = 2}] \\ \qquad = 0.135$

b) We know that,

The probability that it will avoid a collision is $e^{-G}$, which is the probability that all the other stations are silent in that slot.The probability of a collision is then just $(1-e^{-G})$. The probability that a transmission requiring exactly k attempts (i.e k-1 collisions followed by one success ) is

$$P_k = e^{-G}(1-e^{-G})^{k-1} $$

∴Probability of exactly k collisions and then a success i.e. k collisions and success in (k+1) attempts = $P_k = e^{-G}(1-e^{-G})^{k}$

$= P(\text{success in k+1 attempts}) \times P(\text{collisions in k attempts})$

$= e^{-G}(1-e^{-G})^{k}$

$= 0.135 \times (1-0.135)^k$

$= 0.135 \times (0.865)^k$

c) The expected number of transmission attempts needed = $\dfrac{\text{traffic load(G)}}{throughput(S)} = \dfrac{G}{Ge^{-G}} = e^{G} = e^2 = 7.3890 \approx 7.39 \approx 7.4$ .