The $k$-of-$r$-out-of- $n$circular reliability system, $k\le r\le n$, consists of $n$components that are arranged in a circular fashion. Each component is either functional or failed, and the system functions if there is no block of $r$consecutive components of which at least $k$are failed. Show that there is no way to arrange $47$components, $8$of which are failed, to make a functional $3$-of-$12$-out-of-$47$circular system.

The $k$-of-$r$-out-of-$n$circular reliability system, localid="1647321919939" $k\le r\le n$consists of $n$components that are arranged in a circular fashion.

Define random variables $N_i$that marks the number of failed components in $i$th block of $12$components. Observe that $N_i$has Hypergeometric distribution, i.e.

$P\left(N_i=k\right)=\frac{\left(\begin{array}{c}12\\ k\end{array}\right)\left(\begin{array}{c}35\\ 8-k\end{array}\right)}{\left(\begin{array}{c}47\\ 8\end{array}\right)}$

So we have that

$E\left(N_i\right)=8·\frac{12}{47}\approx 2.0425$

Now, let's prove that there exists a block that does not work, i.e., that has three or more failed components. If that would not be the case, we would have that all blocks have two or fewer failed components. But, that would be in contradiction with a fact that the expected number of failed components is $2.0425$. So, it is impossible to arrange these components to obtain a functional $3$-of-$12$-out-of-$47$circular system.