Misc 15

Transcript

Misc 15 An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15 Evaluate the following probabilities (i) P(A fails |B has failed) Finding P(B fails) P(B fails alone) = P(B fails) – P(A & B fail) 0.15 = P(B fails) – 0.15 0.15 + 0.15 = P(B fails) 0.30 = P(B fails) P(B fails) = 0.30 P(A fails |B has failed) = 𝑃(𝐴 & 𝐵 𝐹𝑎𝑖𝑙)﷮𝑃(𝐵 𝐹𝑎𝑖𝑙)﷯ = 0. 15﷮0. 30﷯ = 1﷮2﷯ = 0.5 ∴ P(A fails |B has failed) = 0.5 Misc 15 (ii) P(A fails alone) P(A fails alone) = P(A fails) – P(A & B fail) = 0.20 – 0.15 = 0.05