Misc 15 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by
Last updated at Feb. 15, 2020 by
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) 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 15 (ii) P(A fails alone) P(A fails alone) = P(A fails) – P(A & B fail) = 0.20 – 0.15 = 0.05
Conditional Probability - Values given
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