Answer (a) The probability that this product will be reliable = p(A)*p(B)*p(C)*p(D)
= (0.90) *(0.75) *(0.80) *(0.60) = 0.32
Answer (b) Component D should be selected as it has the lowest reliability probability value and most likely to be unsuccessful.
Answer (c) =>Finding the new reliability of component C:
For Component C to be reliable then the probability = 1 – (probability of component C getting unsuccessful or fail) = 1 – (probability of component C and failed redundant backup)
= 1 – [(1- 0.8) *(1- 0.6)] = 1 – [(0.2) *(0.4)] = 1 – 0.08 = 0.92
Hence, probability for the reliable product = (0.9) *(0.75) *(0.92) *(0.6) = 0.37
Therefore, the failure probability = 1 – 0.37 = 0.63
Answer (d) => Finding the new reliabilities for each component:
For component A => 1- [(1-0.90) * 0.5] = 1 – (0.1) *(0.5) = 0.95
For component B => 1- [(1-0.75) * 0.5] = 1 – (0.25) *(0.5) = 0.875
For component C => 1- [(1-0.8) * 0.5] = 1 – (0.2) *(0.5) = 0.9
For component D => 1- [(1- 0.6) * 0.5] = 1 – (0.4) *(0.5) = 0.8
Hence, the reliability of product = (0.95) * (0.875) * (0.9) * (0.8) = 0.599
Therefore, the failure probability = 1 – 0.599 = 0.401
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For the questions mentioned on labour productivity, Input values are not available in the question.
However, Labor productivity should be evaluated using output and input ( = Output / Input)
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