# An electronic assembly consists of two subsystems,

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)

Solution:

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) $=P\left(E_{A}\right)-P\left(E_{A} \cap E_{B}\right)$

$=0.2-0.15$

$=0.05$