We know , here Poisson distribution is followed .So

P(k) = e^{-}^{G} . (G)^{k} / k! where G is the number of frames generated in one frame transmission time(or frame slot time)

Here to find G we have to find number of frames that are generated in 1 slot time which is given as 40 ms.

Given , in 1000 ms no of frames generated = 50

So , no of frames generated in 40 ms = (50 / 1000) * 40

= 2

Hence the value of G obtained = 2

Now we solve the given parts one by one.

**Solution to part a) : **

P(success at first attempt) = P(0)

= e^{-G}

** = e**^{-2}

**Solution to part b) : **

P(success after k collisions) = [ P(failure) ]^{k} P(success)

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

^{ }= ** (1 - e**^{-2})^{k} e^{-2}

**Solution to part c) :**

Expected Number of attempts = 1 . P(1) + 2 . P(2) ............... infinite terms

= Σ k . P(k) [ Where P(k) is the probability of success at kth attempt ]

** **= Σ k . e^{-G} (1 - e^{-G})^{k-1}

This is an infinite arithmetico geometric progression.So finding this sum , we get :

Number of attempts(expected or mean number) = e^{G}

= e^{2}

= 7.39

**= 8 [As number of attempts should be an integer and hence we need to round it to higher integer in this case ] **