the probablity of A failing =p(a)=0.15, not failing p(!a)=0.85
the probablity of b failing =p(b)=0.05, not failing p(!b)=0.95
the probablity of c failing =p(c)=0.10, not failing p(!c)=0.90
the probability that the equipment will fail in one year is (0.15*0.95*0.90)+(0.85*0.05*0.90)+(0.85*0.95*0.10)=0.247
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