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)
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)

Solution:

(i) Take, for example, the event that is failed by A, which is symbolised by the symbol E A.

Furthermore, an event that is failed by B is marked by the symbol E B.

The following information is provided in the question:

Event failed by A, P \left(E_{A}\right)=0.2

Event failed by both, P\left(E_{A} \cap E_{B}\right)=0.15

And, event failed by B alone =P\left(E_{B}\right)-P\left(E_{A} \cap E_{B}\right)

0.15=\mathrm{P}\left(\mathrm{E}_{8}\right)-0.15

\therefore \mathrm{P}\left(\mathrm{E}_{\mathrm{B}}\right)=0.30

Hence, P\left(E_{A} \mid E_{B}\right)=\frac{P\left(E_{A} \cap E_{B}\right)}{P\left(E_{4}\right)}

=\frac{0.15}{0.3}

=0.5

(ii) We have, probability where A fails alone =P\left(E_{A}\right)-P\left(E_{A} \cap E_{B}\right)
=0.2-0.15

=0.05