This is one related question from number 23-26. please, write the number of the
ID: 388194 • Letter: T
Question
This is one related question from number 23-26. please, write the number of the question next to each answer
For questions 23-26: a product is being used at a given point in time; the product has s components (A,B & C) that all must function for the product to function. Component A fails 1 per 50 trials; B fails 1 per 100 trials; C fails 1 per 1000 trials What is the probability component C will fail? 23 A. 0.001 B. 0.01 C. 0.02 D. 0.969 E. 0.999 24 What is the probability this product will perform reliably? A. 0.969 B. 0.972 C. 0.978 D. 0.981 E. 0.999 25 What is the reliability of this product if component A is backed up with an identical component? A. 0.922 B. 0.969 C. 0.989 D. 0.992 E. 0.998 26 What is the reliability of this product if all 3 components are backed up with identical compone and switches that are .95 reliable? A. 0.9695 B. 0.9997 C. 0.9971 D. 0.9980 E. 0.9999Explanation / Answer
Reliability of component A = 49/50 = 0.98
Reliability of component B = 99/100 = 0.99
Reliability of component C = 999/1000 = 0.999
Since product must function if all the components function, it should be assumed that component A , B and C are connected in series.
23. probability that component C will fail = 1/1000 = 0.001
24.Probability that product will perform reliably ( where A, B and C are connected in series )
= Reliability of A x reliability of B X Reliability c
= 0.98 X 0.99 X 0.999
= 0.9692 ( 0.969 rounded to 3 decimal places )
Answer : 0.969
25. If component A is backed up by identical component , two components of A will be considered connected in parallel. Let us call this subsystem as AA .
Reliability of subsystem AA = 1 – ( 1 – 0.98 ) x ( 1 – 0.98 ) = 1 – 0.02 x 0.02 = 1 – 0.0004 = 0.9996
With this, we consider AA , B and C are connected in series .
The reliability of this system
= Reliability of AA x Reliability of B x Reliability of C
= 0.9996 x 0.99 x 0.999
= 0.9886 ( 0.989 rounded to 3 decimal places )
ANSWER : C ) 0.989
26. reliability of the subsystem AA consisting of A backed up with identical component of reliability 0.98 and switch that is 0.95 reliable
= 1 – ( 1 – 0.98 ) x ( 1 – 0.95 x 0.98)
= 1 - 0.02 x 0.069
= 1 – 0.00138
= 0.99862
Reliability of subsystem BB consisting of B backed up with identical component of reliability 0.99 and switch that is 0.95 reliable
= 1 – ( 1 – 0.99 ) x ( 1 – 0.99 x 0.95)
= 1 – 0.01 x 0.0595
= 1 – 0.000595
= 0.9994
Reliability of subsystem CC consisting of C backed up with identical component of reliability 0.999 and switch that is 0.95 reliable
= 1 – ( 1 – 0.999) x ( 1 – 0.95 x 0.999)
= 1 – 0.001 x 0.05095
= 1 – 0.00005095
= 0.99994
The system now consists of subsystems AA, BB and CC connected in series
The reliability of the system will be
= Reliability of AA x Reliability of BB x Reliability of CC
= 0.99862 X 0.9994 X 0.99994
= 0.99796 ( 0.9980 rounded to 4 decimal places )
Answer : D) 0.9980
Remarks: the key assumption made here is that the identical component and the switch are connected in series and the wholething is connected in parallel to another identical component
Answer : 0.969
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