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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

                (ii) P(A\; fails \; alone)

Answers (1)

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

                                                    =0.2-0.15=0.05                      

Posted by

seema garhwal

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