Traditional Fire Alarm detects smoke not fire , so if there are other reason of smoke like someone smoking, it can increase the chance of alarm though it is not related to fire broken out.

He forgot!

P(y|x)P(x) = P(x|y)P(y) because both are equal to P(x,y). See below: P(x|y) = P(x,y)/P(y) P(x,y)P(y) = P(x,y) P(y|x) = P(x,y)/P(x) P(y|x)P(x) = P(x,y) P(y|x)P(x) = P(x|y)P(y)

@@tubesteaknyouri And he did that so you get Bayes' Rule out of it. It wasn't just for the heck of it

After 5 years, its still the only one on whole internet

this is a good thought, but the given observation value of +z is a constant, not a variable, so although it is contained in f2(U, V, W, +z) the only variables of f2 are U, V, W, hence 2^3 = 8 .

+Zs Sj assume f5 gives you a vector with 2 entries for +y and -y, say [1/5, 3/5]. to normalize this vector simply divide each coordinate by the sum of all coordinates [1/5 * 5/4 , 3/5 * 5/4] = [1/4, 3/4]

Thanks

hansen1101 do you know why we should normalize this and how this became non-normalized one?

+Zs Sj in this particular case you are calculating a distribution of the form P(Q|e) where e is an instantiation of some evidence variables. By definition this form has to sum to 1 over all instances of the query variable Q (i.e. P(q1|e) + P(q2|e) = 1 in the binary case). Be careful, there are queries of other forms that need not sum to 1 and therefore normalization is not necessary (i.e P(Q,e) or P(e|Q)). This became non normalized after applying Baye's rule and only working with the term in the numerator, leavin out the joint prob. over the instantiated evidence vars in the denominator. Therefore you'll have to rescale in the end.

+sahdeV ok I got it... Observation we have is +u and not -u. So there are 4 ways in which +u is possible. Rain, Rain, Umbrella or TT-U Sun Rain Umbrella or FT-U Sun Sun Umbrella or FF-U Rain Sun Umbrella or TF-U Probability of each is respectively: 0.5*0.7*0.9 0.5*0.3*0.9 0.5*0.7*0.2 0.5*0.3*0.2 T-U probability is therefore 63+27/(63+27+14+6)=0.818 F-U probability is 14+6/(63+27+14+6)-0.182 *************** For the next stage, time based update alone gives us probabilities as B'(T)=0.818*0.7+0.182*0.3=0.6272 B'(F)=0.818*0.3+0.182*0.7=0.3728 observation (u+) based updates give us B(T)=0.6272*0.9/(0.6272*0.9+0.3728*0.2)=0.883 B(F)=0.3728*0.2/(0.6272*0.9+0.3728*0.2)=0.117

@@vedhasp Thank you very much.