Probability of at least two bits

Question

I have a frame of N bits with per-bit error rate as p sent over wire. Considering this as a binomial distribution and assuming bit error independence, I am finding the probability of at least 2 bits being corrupted as this :

Probability of none bits being corrupt = $ P(0) = (1-p)^N$

Probability of 1 bit being corrupt = $ P(1) = N \cdot p \cdot (1-p)^{N-1}$

Probability of at least 2 bits being corrupt = $ P = 1 - P(0) - P(1) = 1-(1-p)^N-N \cdot p \cdot (1-p)^{N-1}$

Am doing this correctly or missing something?

Thank you

Answer 1

Yup, you are correct.

$$P=1-(1-p)^N-Np(1-p)^{N-1}=1-(1-p)^{N-1}(1-p+Np)$$