

Last updated at Dec. 8, 2016 by Teachoo
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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. 150. 30 = 12 = 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
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