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)
Let the event in which $A$ fails and $B$ fails be denoted by $E_{A}$ and $E_{B}$.
$P\left(E_{A}\right)=0.2$
$P\left(E_{A} \cap E_{B}\right)=0.15$
$P$ (B fails alone) $=P\left(E_{B}\right)-P\left(E_{A} \cap E_{B}\right)$
$\Rightarrow 0.15=P\left(E_{B}\right)-0.15$
$\Rightarrow P\left(E_{B}\right)=0.3$
(i) $\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \mid \mathrm{E}_{\mathrm{B}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \cap \mathrm{E}_{\mathrm{B}}\right)}{\mathrm{P}\left(\mathrm{E}_{\mathrm{B}}\right)}=\frac{0.15}{0.3}=0.5$
(ii) $P$ (A fails alone) $=P\left(E_{A}\right)-P\left(E_{A} \cap E_{B}\right)$
$=0.2-0.15$
$=0.05$