An electronic assembly consists of two subsystems, say, $A$ and $B$. From previous testing procedures, the following probabilities are assumed to be known :

$\mathrm{P}$ $( A$ fails $)=0.2$

$P(B$ fails alone $)=0.15$

$P(A$ and $ B $ fail $)=0.15$

Evaluate the following probabilities $\mathrm{P}(\mathrm{A}$ fails alone $)$

Let the event in which $A$ fails and $B $ fails be denote by $E_{A}$ and $E_{B}$.

$P\left(E_{A}\right)=0.2$

$\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \text { and } \mathrm{E}_{\mathrm{B}}\right)=0.15$

$\mathrm{P}(\mathrm{B} \text { fails alone })=\mathrm{P}\left(\mathrm{E}_{\mathrm{B}}\right)-\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \text { and } \mathrm{E}_{\mathrm{B}}\right)$

$\therefore $  $ 0.15=P\left(E_{B}\right)-0.15$

$\therefore $ $ \mathrm{P}\left(\mathrm{E}_{\mathrm{B}}\right)=0.3$

$\mathrm{P}$ $(A$ fails alone $)$ $=\mathrm{P}\left(\mathrm{E}_{\mathrm{A}}\right)-\mathrm{P}\left(\mathrm{E}_{\mathrm{A}} \text { and } \mathrm{E}_{\mathrm{B}}\right)$

$=0.2-0.15$

$=0.05$