For questions 23-26: a product is being used at a given point in tit me; the pro

Question

For questions 23-26: a product is being used at a given point in tit me; the product has 3 components (A,B & C) that all must function for the product to function. Component A fails 1 per S0 trials; B fails 1 per 100 trials; C fails 1 per 1000 trials What is the probability component C will fail? A. 0.001 B. 0.01 C. 0.02 D. 0.969 E. 0.999 23 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 What is the reliability of this product if all 3 components are backed up with identical componen and switches that are .95 reliable? A 0.9695 B. 0.9997 C. 0.9971 D. 0.9980 E. 0.9999 26

Explanation / Answer

23)P(compoennt C will fail) =1/1000 =0.001

option A

24)

P( function relaibly) =P(all function work) =(1-1/50)*(1-1/100)*(1-1/1000)=0.969

option A

25)

P( function properly) =P(at least one of A works)*P(B works)*P(C works)

=(1-P(none of A works))*P(B works)*P(C works) =(1-1/50*1/50)*(1-1/100)*(1-1/1000) =0.989

option C

26)

here probability that identical component A works =P(switch works and component works)

=0.95*49/50

hence  identical component A does not works =(1-0.95*49/50)

similarly component B does not works =(1-0.95*99/100)

component C does not works =(1-0.95*999/1000)

P( function properly) =P(at least one of A works)*P(at least one of B works)*P(at least one of C works)

=(1-(1/50)*(1-0.95*49/50))*(1-(1/100)*(1-0.95*99/100))*(1-(1/1000)*(1-0.95*999/1000))=0.9980

option D

( please revert for any clarificaiton required)_

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