Question

(9) A device is composed of two components, A and B, subject to random failures. The components are connected in parallel and, consequently, the device is down only if both components are down. The two components are not independent. We estimate that the probability of a failure of component A is equal to 0.2; a failure of component B is equal to 0.8 if component A is down; a failure of component B is equal to 0.4 if component A is active. (a) Calculate the probability of a failure (i) of component A if component B is down; (i) of exactly one component

I want help to solve this question

I want help to solve this question

Answer 1

$(a)~Let~A~be~an~event~that~component~A~is~failed~and\\ B~be~an~event~that~component~B~is~failed,~then\\ P(A)=0.2,~P(A^c)=1-P(A)=0.8,~P(B|A)=0.8,~P(B|A^c)=0.4.\\\\ (i)~P(A|B^c)=\frac{P(B^c|A)P(A)}{P(B^c)}=\frac{P(B^c|A)P(A)}{P(B^c|A)P(A)+P(B^c|A^c)P(A^c)}\\\\ =\frac{(1-P(B|A))P(A)}{(1-P(B|A))P(A)+(1-P(B|A^c))P(A^c)}\\\\ =(1-0.8)*0.2/((1-0.8)*0.2+(1-0.4)*0.8)=0.0769\\\\ (ii)~P(Exactly~one~component~is~failed)=P(A\cap B^c)+P(A^c\cap B)\\\\ =P(B^c|A)P(A)+P(B|A^c)P(A^c)=(1-P(B|A))P(A)+0.4*0.8\\ =(1-0.8)*0.2+0.4*0.8=0.36\\\\ (c)~P(C)=0.2\\\\ P(C\cap (A^c\cup B^c))=P(C)P(A^c\cup B^c)\\ (Since~C~is~independent~of~the~state~of~components~A~and~B)\\ =P(C)(1-P(A\cap B))~~(DeMorgan's~law)\\ =P(C)(1-P(B|A)P(A))=0.2*(1-0.8*0.2)=0.168\\\\ P(Device~is~active)=P(At~least~one~component~is~active)\\ =P(A^c\cup B^c\cup C^c)=1-P(A\cap B\cap C)\\ =1-P(A\cap B)P(C)=1-P(B|A)P(A)P(C)=1-0.8*0.2*0.2=0.968\\\\ P(C|Device~is~active)=\frac{P(C\cap (A^c\cup B^c))}{P(Device~is~active)}=0.168/0.968=0.1736.$