Q. 3.67

Question

An engineering system consisting of n components is said to be a k-out-of-n system (kn) 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 pi,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,  pi,i=1,2,.....n)

Step-by-Step Solution

Verified
Answer

a) The probability of 2-out of -4 system function is

1Q1Q2Q3Q4P1Q2Q3Q4Q1P2Q3Q4Q1Q2P3Q4Q1Q2Q3P4


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

A=Q1Q2P3P4P5+Q1P2Q3P4P5+Q1P2P3Q4P5+Q1P2P3P4Q5+P1Q2Q3P4P5++P1Q2P3Q4P5+P1Q2P3P4Q5+P1P2Q3Q4P5+P1P2Q3P4Q5+P1P2P3Q4Q5

B=Q1P2P3P4P5+P1Q2P3P4P5+P1P2Q3P4P5+P1P2P3Q4P5+P1P2P3P4Q5

C=P1P2P3P4P5

1Step 1: Component Function

a) The events name was given below,

S=the 2-out of-4 system functions

Ai={ the i-th component functions },i=1,2,3,4

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

S={ at least 2 components functions }=B2B3B4

The disjoint sets are ijwe have Sc=B0B1

2Step 2: System Function

we mentioned the equations, Qi=1Pi=PAic

PB0=PA1cA2cA3cA4c= independence of Ai

=PA1cPA2cPA3cPA4c

=Q1Q2Q3Q4. 

PB1=Pi=14jiAiAjc=[ aditivity ]


=i=14PjiAiAjc= independence of Ai

=i=14PAijiPAjc=i=14PijiQj

=P1Q2Q3Q4+Q1P2Q3Q4+Q1Q2P3Q4+Q1Q2Q3P4

 The chances of 2-out of-4 System function is

P(S)=1PSc

=1PB0B1

=1PB0+PB1

=1Q1Q2Q3Q4P1Q2Q3Q4Q1P2Q3Q4Q1Q2P3Q4Q1Q2Q3P4

3Step 3: Independent Event

we mentioned the event name,

S=the 3-out of-5 system functions

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

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

S={ at least 3 components functions }=B3B4B5

Twice the disjoint sets are, Bk 

4Step 4: probability of system function

we mentioned Qi=1Pi=PAic

PB3=P1i,j,k5m,ni,j,kAiAjAkAmcAnc

=Q1Q2P3P4P5+Q1P2Q3P4P5+Q1P2P3Q4P5+Q1P2P3P4Q5++P1Q2Q3P4P5+P1Q2P3Q4P5+P1Q2P3P4Q5+P1P2Q3Q4P5++P1P2Q3P4Q5+P1P2P3Q4Q5

PB4=Pi=15jiAicAj

=Q1P2P3P4P5+P1Q2P3P4P5+P1P2Q3P4P5++P1P2P3Q4P5+P1P2P3P4Q5

PB5=PA1A2A3A4A5=P1P2P3P4P5

The probability that a 3-out of-5 system function is,

P(S)=PB3B4B5

=PB5+PB4+PB5

=Q1Q2P3P4P5+Q1P2Q3P4P5+Q1P2P3Q4P5+Q1P2P3P4Q5++P1Q2Q3P4P5+P1Q2P3Q4P5+P1Q2P3P4Q5+P1P2Q3Q4P5++P1P2Q3P4Q5+P1P2P3Q4Q5++Q1P2P3P4P5+P1Q2P3P4P5+P1P2Q3P4P5++P1P2P3Q4P5+P1P2P3P4Q5++P1P2P3P4P5.