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In a aloha implemented shared channel probability of transmission of a station in a time span of T is p. Given, probability such that NO station transmit in a time duration of $2T$ is $50\%$ , where T = one frame transmission time. What is the value of p is if total no of station = $100$
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is aloha in gate 2017 syllabus ?
Yes both pure and slotted aloha as they are collision detection techniques in MAC sublayer.

"Given, probability such that NO station transmit in a time duration of 2T" implies vulnerable time = 2G,

as T = frame transmission time which is for pure aloha , then why question says slotted aloha .??

What I think, In poisson distribution we can consider any time span and calculate some required probability irrespective of how small the probability comes out to be. Whether it is slotted or not in this question , concerned time span of time is 2T. (it can be anything like 3.2T etc. for this question)
okk, so wht is P , prob. of success ?
probability of transmission of a station in a time span of T is p.

I did in the following way,

For any time span probability of there being x transmission attempts is

$$\frac{A^xe^{-A}}{x!}$$

Where A is the average no of transmission attempt in that time span.

In that question time duration is $2T$ . And we know that channel load G is the average transmission attempt in T time duration. So, in 2T time the average transmission attempts will be 2G.

SO, plugging A = 2G in the above formula,

$$\frac{(2G)^xe^{-2G}}{x!}$$

For the given condition , No transmission attempts is 2T  => x = 0

Probability = $\frac{(2G)^0e^{-2G}}{0!}$ = 0.5

=> G = 0.3465

Now for poisson Mean of average is  = Np

=> p = 0.003465 (very low)

I also did in the same way, and its only what i m getting and even i am not getting any other approach to solve this .
same !

+1 vote