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

P (A fails/B has failed )

P(A fails alone)

Answer 1

Let the event in which A fails and B fails be denoted by A' and B'

P(A')=0.2, $P(A^{\prime }\cap B^{\prime })=0.15$

P(B fails alone)= P(B')- $P(A^{\prime }\cap B^{\prime })$

$\therefore$ 0.15= P(B')-0.15

$\Rightarrow$ P(B')=0.30

P(A fails/B has failed)

$=P\left(A^{\prime }/B^{\prime }\right)=\frac{P(A^{\prime }\cap B^{\prime })}{P\left(B^{\prime }\right)}=\frac{0.15}{0.30}=0.5$

Let the event in which A fails and B fails be denoted by A' and B'

P(A')=0.2, $P(A^{\prime }\cap B^{\prime })=0.15$

P(B fails alone)= P(B')- $P(A^{\prime }\cap B^{\prime })$

$\therefore$ 0.15= P(B')-0.15

$\Rightarrow$ P(B')=0.30

P(A fails alone) =P(A fails) -P(Both A and B fails)

=$P\left(A^{\prime }\right)-P(A^{\prime }\cap B^{\prime })=0.2-0.15=0.05$