Question

There are n components lined up in a linear arrangement. Suppose that each component independently functions with probability p. What is the probability that no 2 neighboring components are both nonfunctional?

Short answer

The answer is$P\left(E\right)$localid="1646910197664" $=\sum _{0⩽m\le (n+1)/2}\left(\begin{array}{c}n+m-1\\ m\end{array}\right)p^{n-m}(1-p{)}^m$

Step-by-step solution

Step 1:Given Information

Let $X$denotes the number of non functional components and let $E$denotes the event. if no two nonfunctional components are to be constructive, then the space between the functional components must each contain at most one non functional components.

Step 2:Calculation

$P\left(E\right)$$=\sum _{m=0}^{n}P(E\mid X=m)P(X=m)$,(From Bayes theorem)

$=\sum _{0\le m\le n+1)/2}P(E\mid X=m)P(X=m)$

localid="1646910154485" $=\sum _{0\le m\le (n+1)/2}\frac{\left(\begin{array}{c}n+m-1\\ m\end{array}\right)}{\left(\begin{array}{l}n\\ m\end{array}\right)}\left(\begin{array}{l}n\\ m\end{array}\right)p^{n-m}(1-p{)}^m$

localid="1646910170371" $=\sum _{0\le m\le (n+1)/2}\left(\begin{array}{c}n+m-1\\ m\end{array}\right)p^{n-m}(1-p{)}^m$

Step 3:Final Answer

The answer is$P\left(E\right)$localid="1646910184882" $=\sum _{0\le m\le (n+1)/2}\left(\begin{array}{c}n+m-1\\ m\end{array}\right)p^{n-m}(1-p{)}^m$