ACE Practice Test question

Question

An electronic assembly consists of two sub systems say A and B from previous testing procedures , the following probability are assumed to be known,

P(A fails) = 0.2

P(B fails alone) = 0.15

P (A and B fail) = 0.15

then the conditional probability P(A fails| B has failed) =

Comments

P(A/B) = P(B/A) * P(A)/ P(B/A) * P(A) + P(B|A') * P(A')

=0.15*0.2/0.15 *0.2 + 0.15 * 0.8

where am i wrong?

Answer 1

P(A/B)=P(A∩B) / P(B)

here given P(B-A)=.15

P(A)=.20

P(A∩B)=.15

as P(B)=P(B-A)+P(A∩B)= .15+.15=.30

so P(A/B)= .15/.30 = .50