P(failing)=1−P(notfailing)P(failing)=1-P(not failing)P(failing)=1−P(notfailing)
=1-P(A not failing and B not failing and C not failing)
=1−P(AˉBˉCˉ)=1-P(\bar{A}\bar{B}\bar{C})=1−P(AˉBˉCˉ)
=1−(0.85×0.95×0.9)=0.27325=1-(0.85\times0.95\times0.9)=0.27325=1−(0.85×0.95×0.9)=0.27325
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