Q. 3.67 An engineering system consisting... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 3: Q. 3.67 (page 103)

An engineering system consisting of $n$ components is said to be a $k$ -out-of- $n$ system role="math" localid="1649415337837" $(k 鈮� n)$ if the system functions if and only if at least $k$ of the $n$ components function. Suppose that all components function independently of one another. (a) If the component functions with probability $p_{i}, i = 1, 2, 3, 4,$ , compute the probability that a $2$ -out-of- $4$ system functions. (b) Repeat part (a) for a $3$ -out-of- $5$ system. (c) Repeat for a k-out-of-n system when all the Pi equal p (that is, $p_{i}, i = 1, 2, . . . . . n)$

Short Answer

Expert verified

a) The probability of $2 -$ out ofrole="math" localid="1649415686148" $- 4$ system function is

$1 鈭� Q_{1} Q_{2} Q_{3} Q_{4} 鈭� P_{1} Q_{2} Q_{3} Q_{4} 鈭� Q_{1} P_{2} Q_{3} Q_{4} 鈭� Q_{1} Q_{2} P_{3} Q_{4} 鈭� Q_{1} Q_{2} Q_{3} P_{4}$

b) The Probability of $3 -$ out of5 system function $A + B + C$ where

$\begin{aligned} A = Q_{1} Q_{2} P_{3} P_{4} P_{5} + Q_{1} P_{2} Q_{3} P_{4} P_{5} + Q_{1} P_{2} P_{3} Q_{4} P_{5} + Q_{1} P_{2} P_{3} P_{4} Q_{5} + P_{1} Q_{2} Q_{3} P_{4} P_{5} + \\ + P_{1} Q_{2} P_{3} Q_{4} P_{5} + P_{1} Q_{2} P_{3} P_{4} Q_{5} + P_{1} P_{2} Q_{3} Q_{4} P_{5} + P_{1} P_{2} Q_{3} P_{4} Q_{5} + P_{1} P_{2} P_{3} Q_{4} Q_{5} \end{aligned}$

$B = Q_{1} P_{2} P_{3} P_{4} P_{5} + P_{1} Q_{2} P_{3} P_{4} P_{5} + P_{1} P_{2} Q_{3} P_{4} P_{5} + P_{1} P_{2} P_{3} Q_{4} P_{5} + P_{1} P_{2} P_{3} P_{4} Q_{5}$

$C = P_{1} P_{2} P_{3} P_{4} P_{5}$

Step by step solution

Component Function

a) The events name was given below,

$S = \{t h e 2 - o u t o f - 4 s y s t e m f u n c t i o n s\}$

$A_{i} = {the i -th component functions}, i = 1, 2, 3, 4 .$

$B_{k} = {exactly k components function}, k = 0, 1, 2, 3, 4,$

$S = {at least 2 components functions} = B_{2} 鈭� B_{3} 鈭� B_{4} .$

The disjoint sets arelocalid="1649682218389" $i 鈮� j,$ we have $S^{c} = B_{0} 鈭� B_{1}$

System Function

we mentioned the equations, $Q_{i} = 1 鈭� P_{i} = P (A_{i}^{c})$

$P (B_{0}) = P (A_{1}^{c} 鈭� A_{2}^{c} 鈭� A_{3}^{c} 鈭� A_{4}^{c}) = [independence of A_{i}]$

$\begin{aligned} = P (A_{1}^{c}) P (A_{2}^{c}) P (A_{3}^{c}) P (A_{4}^{c}) \end{aligned}$

$= Q_{1} Q_{2} Q_{3} Q_{4} .$

$P (B_{1}) = P ({鈭�}_{i = 1}^{4} 鈥� \underset{j 鈮� i}{鈭�} 鈥� A_{i} A_{j}^{c}) = [aditivity]$

$= {鈭�}_{i = 1}^{4} 鈥� P (\underset{j 鈮� i}{鈭�} 鈥� A_{i} A_{j}^{c}) = [independence of A_{i}]$

$= {鈭�}_{i = 1}^{4} 鈥� P (A_{i}) \underset{j 鈮� i}{鈭�} 鈥� P (A_{j}^{c}) = {鈭�}_{i = 1}^{4} 鈥� P_{i} \underset{j 鈮� i}{鈭�} 鈥� Q_{j}$

$= P_{1} Q_{2} Q_{3} Q_{4} + Q_{1} P_{2} Q_{3} Q_{4} + Q_{1} Q_{2} P_{3} Q_{4} + Q_{1} Q_{2} Q_{3} P_{4}$

The chances of $2 - o u t o f - 4$ System function is

$P (S) = 1 鈭� P (S^{c})$

$= 1 鈭� P (B_{0} 鈭� B_{1})$

$= 1 鈭� [P (B_{0}) + P (B_{1})]$

$= 1 鈭� Q_{1} Q_{2} Q_{3} Q_{4} 鈭� P_{1} Q_{2} Q_{3} Q_{4} 鈭� Q_{1} P_{2} Q_{3} Q_{4} 鈭� Q_{1} Q_{2} P_{3} Q_{4} 鈭� Q_{1} Q_{2} Q_{3} P_{4}$

Independent Event

we mentioned the event name,

$S = \{t h e 3 - o u t o f - 5 s y s t e m f u n c t i o n s\}$

$A_{i} = {the i -th component functions}, i = 1, 2, 3, 4 .5$

$B_{k} = {exactly k components function}, k = 0, 1, 2, 3, 4, 5$

role="math" localid="1649417563600" $S = {at least3components functions} = B_{3} 鈭� B_{4} 鈭� B_{5} .$

Twice the disjoint sets are, $B_{k}$

probability of system function

we mentioned $Q_{i} = 1 鈭� P_{i} = P (A_{i}^{c})$

$P (B_{3}) = P (\underset{1 鈮� i, j, k 鈮� 5}{鈭�} 鈥� \underset{m, n 鈮� i, j, k}{鈭�} 鈥� A_{i} A_{j} A_{k} A_{m}^{c} A_{n}^{c})$

$P (B_{4}) = P ({鈭�}_{i = 1}^{5} 鈥� \underset{j 鈮� i}{鈭�} 鈥� A_{i}^{c} A_{j})$

$\begin{aligned} = Q_{1} P_{2} P_{3} P_{4} P_{5} + P_{1} Q_{2} P_{3} P_{4} P_{5} + P_{1} P_{2} Q_{3} P_{4} P_{5} + \\ + P_{1} P_{2} P_{3} Q_{4} P_{5} + P_{1} P_{2} P_{3} P_{4} Q_{5} \end{aligned}$

$P (B_{5}) = P (A_{1} A_{2} A_{3} A_{4} A_{5}) = P_{1} P_{2} P_{3} P_{4} P_{5}$

The probability that a $3 - o u t o f - 5$ system function is,

$P (S) = P (B_{3} 鈭� B_{4} 鈭� B_{5})$

$= P (B_{5}) + P (B_{4}) + P (B_{5})$

$\begin{aligned} = Q_{1} Q_{2} P_{3} P_{4} P_{5} + Q_{1} P_{2} Q_{3} P_{4} P_{5} + Q_{1} P_{2} P_{3} Q_{4} P_{5} + Q_{1} P_{2} P_{3} P_{4} Q_{5} + \\ + P_{1} Q_{2} Q_{3} P_{4} P_{5} + P_{1} Q_{2} P_{3} Q_{4} P_{5} + P_{1} Q_{2} P_{3} P_{4} Q_{5} + P_{1} P_{2} Q_{3} Q_{4} P_{5} + \\ + P_{1} P_{2} Q_{3} P_{4} Q_{5} + P_{1} P_{2} P_{3} Q_{4} Q_{5} + \\ + Q_{1} P_{2} P_{3} P_{4} P_{5} + P_{1} Q_{2} P_{3} P_{4} P_{5} + P_{1} P_{2} Q_{3} P_{4} P_{5} + \\ + P_{1} P_{2} P_{3} Q_{4} P_{5} + P_{1} P_{2} P_{3} P_{4} Q_{5} + \\ + P_{1} P_{2} P_{3} P_{4} P_{5} . \end{aligned}$

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91影视

Short Answer

Step by step solution

Component Function

System Function

Independent Event

probability of system function

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