Accomplished_Milk514B to Machine Learning@academy.gardenEnglish · 10 months ago[D] Naive bayesmessage-squaremessage-square12fedilinkarrow-up11arrow-down10file-text
arrow-up11arrow-down1message-square[D] Naive bayesAccomplished_Milk514B to Machine Learning@academy.gardenEnglish · 10 months agomessage-square12fedilinkfile-text
hi everyone, I could not understand answer. I found 1/4 from table. answer is 1/2. thanks everyone. https://preview.redd.it/tht330mp7h2c1.png?width=939&format=png&auto=webp&s=2084932c8e79929563ddc757f012fce0c040453c
minus-squaremofossBlinkfedilinkEnglisharrow-up1·10 months agoP(K=1) = 1/2 P(a=1|K=1) = P(a=1,K=1)/P(K=1) = (1/4)/(1/2)=1/2 P(b=1|K=1) = P(b=1,K=1)/P(K=1) = (1/8)/(1/2)=1/4 P(c=0|K=1) = P(c=0, K=1)/P(K=1) = (1/4)/(1/2)=1/2 P(a=1, b=1, c=0, K=1) = 0 P(a=1, b=1, c=0, K=0) = 1/8 [0.5 * 0.25 * 0.5] / (0 + 1/8) = (1/16) / (1/8) = 1/2 For conditionals, convert it into joints and priors first and THEN use the table to count instances out of N samples. P(X|Y) = P(X,Y)/P(Y) :)
P(K=1) = 1/2
P(a=1|K=1) = P(a=1,K=1)/P(K=1) = (1/4)/(1/2)=1/2
P(b=1|K=1) = P(b=1,K=1)/P(K=1) = (1/8)/(1/2)=1/4
P(c=0|K=1) = P(c=0, K=1)/P(K=1) = (1/4)/(1/2)=1/2
P(a=1, b=1, c=0, K=1) = 0
P(a=1, b=1, c=0, K=0) = 1/8
[0.5 * 0.25 * 0.5] / (0 + 1/8) = (1/16) / (1/8) = 1/2
For conditionals, convert it into joints and priors first and THEN use the table to count instances out of N samples.
P(X|Y) = P(X,Y)/P(Y)
:)