3 machines (a1,a2,a3) produce parts
10% of a1 are defective
6% of a2 is defecyive
2% of a3 is defective
a)prob of a random part chosen to not be defective
b)part produced by a1
a) i thought of finding the defective first then subtracting it from 1 to get the not defective so...defective= d p(d|a1)=.1 p(d|a2)=.06 p(d|a3)=.02 p(d)= p(dUa1)+p(dUa2)+p(dUa3)=.1x.45+.06x.35+.02x.2 then 1-(above calculation) should give me the not damaged part right?
for b is it just .45? because it asks for the prob that it was produced by a1, so 45% is produced by a1 so 45/100?