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Q.4.12

Expert-verifiedFound in: Page 170

Book edition
9th

Author(s)
Sheldon M. Ross

Pages
432 pages

ISBN
9780321794772

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?

The answer is$P\left(E\right)$localid="1646910197664" $=\sum _{0\u2a7dm\le (n+1)/2}\left(\begin{array}{c}n+m-1\\ m\end{array}\right){p}^{n-m}(1-p{)}^{m}$

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.

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

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

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