Suppose that a process is comprised of 12 components, each having a success rate of 71%. By how much would reliability change if the processswitched from having the components in parallel to having the components in series? (Round intermediate calculations to three decimal places.Express final answer as a percentage rounded to one decimal place.)OA. 98.4% increaseOB. 1.6% decreaseOC. 5.9% increaseD. 98.4% decreaseOE. no change

Given data: Total components = 12 Success rate = 71% or 0.71Now, the reliability in parallel is calculated as follows: = 1-(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)×(1-0.71)=1−0.29(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.29)(0.29)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−0.0841(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.0841)(0.29)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−0.024389(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.024389)(0.29)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−0.007073(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.007073)(0.29)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−0.002051(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.002051)(0.29)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−0.000595(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.000595)(0.29)(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−0.000172(1−0.71)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.000172)(0.29)(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−0.00005(1−0.71)(1−0.71)(1−0.71)(1−0.71) =1−(0.00005)(0.29)(1−0.71)(1−0.71)(1−0.71) =1−0.000015(1−0.71)(1−0.71)(1−0.71) =1−(0.000015)(0.29)(1−0.71)(1−0.71) = 1 or 0.9999 Now, the reliability in series connections is calculated as follows: =0.71×0.71×0.71×0.71×0.71×0.71×0.71×0.71×0.71×0.71×0.71×0.71=(0.5041)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71) =(0.357911)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71) =(0.254117)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71) =(0.180423)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71) =(0.1281)(0.71)(0.71)(0.71)(0.71)(0.71)(0.71) =(0.090951)(0.71)(0.71)(0.71)(0.71)(0.71) =(0.064575)(0.71)(0.71)(0.71)(0.71) =(0.045849)(0.71)(0.71)(0.71) =(0.032552)(0.71)(0.71) =(0.023112)(0.71) =0.01641 The reliability is decreased from 0.9999 to 0.01641, the percentage is:Decrease % = (0.9999-0.01641)/0.9999 = 0.98349/0.9999 = 0.983588 or 98.4% Therefore the correct option is (d) 98.4% decrease