Here, B and C are connected in Serial, so Reliability would be = 0.80* 0.87 = 0.696,
again, D and E are connected in Serial, so Reliability would be = 0.75*0.79 = 0.5925,
System Reliability would be = 0.98*(1-((1-0.696)*(1-0.5925)))*0.99 = 0.98*0.876*0.99 = 0.850,
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