system a:
P(subsystem 1-2-3 works) =(1-p1)*(1-p2)*(1-p3)=0.9*0.95*0.95=0.81225
P(subsystem 4-5 works)=(1-p4)*(1-p5)=0.85*0.85=0.7225
hence system (a) fails =P(subsystem 1-2-3 fails and subsystem 4-5 fails)
=(1-0.81225)*(1-0.7225)=0.0521
system b.
P(subsystem 4-5 works)=1-p4*p5 =1-0.15*0.15=0.9775
hence P(subsystem 1-4-5) works =P(1 works)*P(subsystem 4-5 works)=0.9*0.9775=0.87975
P(subsystem 6-7 works)=1-p6*p7 =1-0.2*0.2=0.96
P(subsystem 2-3-6-7 works)=0.95*0.95*0.96=0.8664
hence P(system (b) fails)=P(subsystem 6-7 fails and subsystem 2-3-6-7 fails)
=(1-0.87975)*(1-0.8664)=0.016065
Problem 4. Two communication systems (shown below) are composed from several links, where, for proper operation,...
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