a)probability that a device rejected =1-P(it has none of defect) =1-(1-0.3)*(1-0.4)=0.58
b)P(reject for either E1 or C1 or both) =P(E1)+P(C)-P(E1nC) =P(E1)+P(C)-P(E1)*P(C)=0.1+0.4-0.1*0.4 =0.46
c)
P(reject for either E1;E2 or C or all)= P(E1 U E2 UC)
=P(E1)+P(E2)+P(C)-P(E1)P(E2)-P(E1)P(C)-P(E2)P(C)+P(E1)P(E2)P(C)=
=0.1+0.2+0.4-0.1*0.2-0.2*0.4-0.1*0.4+0.1*0.2*0.4=0.568
Hh Electrical devices, housed ih a metl 2h 슨re-produced lag A Process in whicl 307. are...