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Here the probability of frame being lost is $P.$ So the probability of frame reaching safely would be $(1-P).$

Now lets consider that the frame will reach safely in $k$-th transmission. That means that the frame being lost $k-1$ times and reached in $k$-th time with probability $(1-P).$ Now a frame requires $k$-transmissions exactly when the first $k-1$ attempts fail .... this happens with probability $P^{ \text{k-1} }$ and the $k$-th transmission succeeds , this happens with probability $1-P.$

For $k=1,$ the probability $= (1-P)$

For $k=2,$ the probability $= P(1-P)$

For $k=3,$ the probability $=$ $P^2$$(1-P)$ $.............. ............. {\infty}$

So the mean number of transmission will be $= (1-P) + P(1-P) +$ $P^2$$(1-P)$$.......... $Which gives me $1.$

But solution saying,

$\sum_{k=1}^{\infty} kP_k$

$=$ $\sum_{k=1}^{\infty} k(1-P)P^{k-1}$

$=$$(1-P)$$\sum_{k=1}^{\infty}kP^{k-1}$

$= (1-P).$$\frac1{{(1-P)}^2}$ $=$$\frac1{{(1-P)}}$

I don't understand how they multiply $P^{ \text{k-1} }$$(P-1)$ by $k.$

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You have to compute the "average of the number of transmissions" and not the "average of probabilities". Think of it as you are rolling a dice and a number $k$ comes with probability $P^{k-1} \cdot (1-P)$. So, the average value that you get on dice is $\sum_{k = 1}^{\infty} k \cdot (\textrm{Probabaility of $k$}) = \sum_{k = 1}^{\infty} k \cdot P^{k-1} \cdot (1-P)$

In other words, you have to compute the expected value of "the number of transmissions". Therefore, we multiply by $k$.

Read more about geometric distribution and its expected analysis here.

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  • $\begingroup$ still not understood... Wikipedia also don't saying why multiply by k... Please elaborate little bit more.. $\endgroup$
    – S. M.
    Commented Sep 19, 2021 at 5:28
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Commented Sep 19, 2021 at 5:52
  • $\begingroup$ @yagami please approve $\endgroup$
    – S. M.
    Commented Sep 19, 2021 at 11:59
  • $\begingroup$ @Saslok Please avoid asking specific users to help you with another question. Be patient, and wait to let others help you on their own terms. (remember, it is currently Sunday in many places in world, so not many people are available to help you) $\endgroup$
    – Discrete lizard
    Commented Sep 19, 2021 at 12:49
  • $\begingroup$ @yagami one thing tell $k$ isn't equally likely or probability isn't equally likely? $\endgroup$
    – S. M.
    Commented Sep 19, 2021 at 20:43

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