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Q.4.12

Expert-verified
Found in: Page 170

### A First Course in Probability

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

See the 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\left(E\mid X=m\right)P\left(X=m\right)$,(From Bayes theorem)

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

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

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

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