Correct Answer - 0.2647
Let `E_(1), E_(2), E_(3)` be the respective events that the 1st, 2nd and 3rd components function. Then,
`P(bar(E)_(1))=0.14, P(bar(E)_(2))=0.10` and `P(bar(E)_(3))=0.05`
`implies P(E_(1))=(1-0.14)=0.86, P(E_(2))=(1-0.10)=0.90` and `P(E_(3))=(1-0.05)=0.95`
`implies` P(machine fails) `=1 -P` (machine functions)
`=1-P [(E_(1) and E_(2) and E_(3))]`
`=1-[P(E_(1))xxP(E_(2))xxP(E_(3))]`.