Q.4.12 - A First Course in Probability

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? 

Step-by-Step Solution

Answer

The answer is$P\left(E\right)$$=\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 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)

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

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

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

Step 3:Final Answer

The answer is  $P\left(E\right)$$=\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$