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Misc 9 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 Now, 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 9 (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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo