Consider a bit (say $0$) duplicate $n$ times in some device:
$$[0]_n = (0,0,\dots, 0).$$
During some procedure, let $p$ be the probability for a bit to change by error. For the error correcting code to work, we need the bit $0$ to be still in a strict majority after the procedure. Here is the probability for that to happen:
$$ P(n,p)= \sum_{k=0}^{\left\lceil n/2 - 1 \right\rceil} \binom{n}{k} p^k (1-p)^{n-k}.$$
Question: Is it possible to calculate* this sum in general?
*By calculate I mean something like the following examples:
$$\sum_{k=0}^n \binom{n}{k} = 2^n,$$
$$\sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k}=1,$$
$$\sum_{k=0}^n p^k = \frac{1-p^{n+1}}{1-p}.$$
The precise form of the cumulative distribution function of binomial distribution is a bit complicated, but handle-able. There are some bounds you can use.
Depending on your $p$ and $n$, you might be able to get away with Normal Approximation to the Binomial or De Moivre–Laplace theorem.
There are a number of other approximations on Wikipedia as well