Question

# 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) (ii) P(A fails alone)

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Solution

## (i) Let E A be the event that subsystem A fails and E B be the event that subsystem B fails. Now the probabilities are defined as, P( E A )=0.2 P( E A  and  E B )=0.15 The probability of occurrence of event E B is calculated as, P( E B )=P( B fails alone )+P( E A  and  E B ) P( E B )=0.15+0.15 P( E B )=0.3 The probability of failure of subsystem A after failure of subsystem B is, P( E A | E B )= P( E A  and  E B ) P( E B ) = 0.15 0.3 =0.5 (ii) Let E A be the event that subsystem A fails and E B be the event that subsystem B fails. Now the probabilities are defined as, P( E A )=0.2 P( E A  and  E B )=0.15 The probability of occurrence of event E B is calculated as, P( E B )=P( B fails alone )+P( E A  and  E B ) P( E B )=0.15+0.15 P( E B )=0.3 The probability that subsystem A fails alone is, P( A fails alone )=P( E A )−P( E A  and  E B ) =0.2−0.15 =0.05

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