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 * \(\frac{1}{3}\)=\(\frac{1}{2}\)

posted @ 2019-04-16 23:31  番茄瓜园  阅读(301)  评论(0编辑  收藏  举报