三门问题

先验:
p(y1)=p(y2)=p(y3)=1/3 ,1,2,3门有车的概率
p(k2|y1)=1/2 1门有车,开2门的概率
p(k2|y2)=0 2门有车,开2门的概率
p(k2|y3)=1 3门有车,开2门的概率

推倒:p(y1|k2)=(dfrac{p(k2|y1)p(y1)}{p(k2)})=(frac{1/2 * 1/3}{1/2}) =1/3 =p(y1)
其中:p(k2)=p(k2|y1)p(y1)+p(k2|y2)p(y2)+p(k2|y3)p(y3)
=(frac{1}{2}) * (frac{1}{3}) + 0 * (frac{1}{3}) + 1 * (frac{1}{3})=(frac{1}{2})

原文地址:https://www.cnblogs.com/niulang85/p/10720946.html