Q. 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\mid B\; has\; failed)

Answers (1)
S seema garhwal

Let event in which A fails and B fails be E_A,E_B

           P(E_A)=0.2

P(E_A\, and \, E_B)=0.15

P(B\, fails\, alone)=P(E_B)-P(E_A\, and\, E_B)

\Rightarrow \, \, \, 0.15=P(E_B)-0.15

\Rightarrow \, \, \, P(E_B)=0.3

   P(E_A|E_B)=\frac{P(E_A\cap E_B)}{P(E_B)}

                      =\frac{0.15}{0.3}=0.5

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