Two reinforced concrete buildings A and B are located in a seismic region. It is estimated that an impending earthquake in the region might be strong (S), moderate (M), or weak (W) with probabilities.
P (S) = 0.02, P (M) = 0.2, and P (W) = 0.78.
The probabilities of failure of each building if these earthquakes occur are 0.20, 0.05, and 0.01, respectively.Due to similar procedures used in the design and construction of the two buildings, it is estimated that if building A fails the probability that building B will also fail is 0.50, 0.15, and 0.02 for the three types of earthquakes. If building A has failed and building B has survived, what is the probability that the earthquake was not strong?
$$P( \bar{S}|(A\cap\bar{B}))=\frac{P((A\cap\bar{B})|\bar{S})P(\bar{S})}{P(A \cap\bar{B})}$$ I am able to calculate $$ P(A \cap\bar{B}) and P(\bar{S})$$ but am suck on $$P((A\cap\bar{B})|\bar{S})$$ Can someone please help.
$P(AF/S) = P(BF/S) = 0.2\times 0.02 = 0.004$
$P(AF/M) = P(BF/M) = 0.05\times 0.2 = 0.01$
$P(AF/W) = P(BF/W) = 0.01\times 0.78 = .0078$
$P(AS/S) = P(BS/S) = 0.8\times 0.02 = .016$
$P(AS/M) = P(BS/M) = 0.95\times 0.2 = 0.19$
$P(AS/W) = P(BS/W) = 0.99\times 0.78 = 0.7722$
These are independent probabilities of failure of A and B.
$P(BFS/AFS) = 0.5$
$P(BFM/AFM) = 0.15$
$P(BFW/AFW) = 0.02$
This is what is given,
$P(BF) = P(BFS/AFS)*P(AFS)+P(BFM/AFM)*P(AFM)+P(BFW/AFW)*P(AFW)$
$=0.5\times 0.004+.15\times 0.01+0.02\times 0.0078$
$=0.003656$ $P(BF/AF) = P(BSS/AFS)*P(AFS)+P(BSM/AFM)*P(AFM)+P(BSW/AFW)*P(AFW)$
$P(BS/AF) = 0.5\times 0.004+0.85\times 0.01+0.98\times 0.0078=0.002+0.0085+0.007644=0.018144$
Now use Bayes' theorem:
$P(\bar S/(BS/AF)) = \dfrac{(P(BSM/AFM)*P(AFM)+P(BSW/AFW)*P(AFW))}{P(BS/AF)} =\frac{(0.0085+0.007644)}{0.018144} = \frac{0.016144}{0.018144}=0.88977$