An electronic assembly consists of two sub-systems say A and B. From previous testing procedures, the following probabilities are assumed to be known. `P(Afa i l s)=0. 2 ,P(Bfa i l sa lon e)=0. 15 ,P(Aa n dF a i l)=0. 15.` Evaluate the following probabilities: `P(Afa i l s//Bh a sfa i l e d)` (ii) `P(Afa i l sa lon e)`

Answer

N/a

Explanation

Let the events that A and B fails are represented by A' and B' respectively.
`P(A')=0.2,P(A'nnB')=0.15`
`:' P` (B fails alone)`=P(B')-P(A'nnB')`
`implies 0.15=P(B')-0.15impliesP(B')=0.30`
(i) `P` (A fails /B fails)
`=P(A'//B')=(P(A'nnB'))/(P(B'))=0.15/0.30=0.5`
(ii) `P` (A fails alone)
`=P(A')-P(A'nnB')=0.2-0.15=0.05`