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Explain please

in Computer Networks 214 views

4 Answers

2 votes
There are "n" stations in LAN. According to question, only one station want to transmit , in the sense among "n" stations, "n-1" should not be allowed to transmit the data  when one station is already is tranmitting the data.

In our question suppose "n" leveled station is transmitting the data with probability "p" then remaining '"n-1" should not be allowed to transmit the data at given slot.

So, np.(1-p)^ n-1.

(B)
0
(Building on what @akash.dinkar12 pointed)

it is a simple probability+PnC question, of type:

given n events E, P(E) = p and P(E') = 1-p,

Find the P(Success), where Success = exactly 1 event occurs.
2 votes

Out of all stations only 1 allowed to transmit so to select  1 station out of N stations there are N ways possible (nC1 ) then 
it must transmit and other n-1 stations should not transmit 
therefore    n * p * (1-p)*(1-p)*.............(1-p)
= np(1-p)n-1

0 votes
as u can do it with the help of  probability distribution

n is the no of station out of  which 1 sation send the data then the probability of transmitting of1 station is (p)^1

and the probabilty of transmitting "n-1" staion is (1-p)^n-1 which is unsuccess probability then u got the answer.
0 votes

Event 1: Only one station should transmit

We can select one station out of 1 station in C(n,1 ) ways = n

This station has to transmit with the probability 'p'

Therefore, P(E1) = n * p

Event 2: Remaining (n-1) stations should not transmit

Then remaining (n-1) stations should not transmit.

So, for all of them probability will be (1-P)

therefore, P(E2) = (1-p)^(n-1).

Now, both the events should occur simultaneously

Hence, n * p * (1 - p )^(n-1)

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