What is an approximate closed formula for this probability, with a derivation:
$p(k,n)$ is the probability, that among $n$ PC disks and $k$ errors in sum on them, there will be at least $1$ disk with at least $3$ errors? The errors are uniformly and independently distributed on the disks.
E.g. for $k=3$ errors and $n=7$ disks, $p(3,7)=1-77/\binom96=0.0833333$, if I'm right. Here, $9=n+k-1$ and $6=n-1$, so $\binom96$ is the number of distribution of $k$ errors on $n$ disks. The distributions themselves are uniform. There are precisely $k$ errors altogether.
Here, $77$ were obtained via recursive equation below with appropriate border conditions:
$M(k,0)=1$, $M(k,1)=1$ for $k<3$,
$M(0,n)=1$, $M(1,n)=n$, $M(2,n)=n+\frac{n}{2}$.
The number $M(k,n)$ means that there are $M(k,n)$ possibilities of $k$ errors on $n$ disks with the property that there are $0$, $1$, or $2$ errors on each disk.
The problem is that $M(k,n)=M(k, n-1)+M(k-1, n-1)+M(k-2, n-1)$ is infeasible to compute for larger $k$'s and $n$'s. We need a precision of about dozen digits for $k$ and $n$ being hundreds of thousands together with justification of the approximation formula. Perhaps the gamma function will be involved?