If the probability of a bit is received corrupted is P and there are n bits in a packet. Then what is the probability of a failed transmission?
My attempt- Probability of a successful transmission == probability that all bits in a frame was received successfully == (1-P)^n. So, the probability of a successful transmission = 1 - (1-P)^n now the issue is, how 1 - (1-P)^n approximates to n*P (I saw it from a slide)? Or is my calculation wrong?
Thanks.
On
Your calculation is valid.
This assumption ($1-(1-p)^n\approx np$) is only valid when $p$ is very small and $n$ is not very large.
In such case, the probability of two errors or more is so small that is approximately zero. So only consider one error patterns. It is also assumed the errors occur independently from each other and also independent of the bits in the packets.
The packet decoding is failed if either bit is in error. There are $n$ bits, hence the probability of failed reception is approximately $np$:
$$\begin{align}\Pr(\text{packet error})&=\Pr(\text{first bit in error, or second bit in error or..., or n-th bit in error})\\ &=p+p+\cdots+p\\ &=np \end{align}$$
Hint: Use the first order (taylor/binomial) approximation of $(1-p)^n$. When $p$ is small, the other terms can be ignored for practical purposes.
$$(1-p)^n=1-np+\frac{n(n-1)}{2}p^2-...\approx 1-np$$