# If the probability of frame being lost is $P.$ Then, calculate the mean no. of transmission for the frame to make it success$.$ [closed]

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 $$k.$$

• Possibly helps en.wikipedia.org/wiki/Expected_value#Definition Commented Sep 19, 2021 at 7:07
• Because this is the Probability that the no of transmission times=K, so to get the expected time u multiply each time with its probability. It just slipped out of ur mind that E(X)= SUM(Xi*Prob(Xi))
– ShAr
Commented Sep 19, 2021 at 15:18
• – D.W.
Commented Oct 4, 2021 at 7:13
• I’m voting to close this question because it was cross-posted.
– D.W.
Commented Oct 4, 2021 at 7:13

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 $$k$$.
• @yagami one thing tell $k$ isn't equally likely or probability isn't equally likely? Commented Sep 19, 2021 at 20:43