Exam P/1 里想不通的题

前面的题目环境忘了..
主干是
1/2 children of which two parents are left-handed are left-handed,
1/6 children of which only one parent is left-handede are left-handed,
1/16 children of which no parent is left-handede are left-handed.
1/5 children have two left-handed parents.
1/50 children have one left-handed parent.
find the probability that the left-handed children have no left-handed parent.
我要好好研究下这个题目了...

那天晚上和Kelvin说了下,他说他也有这题还被他小笑了下.呜~其实很简单的，只要用一次Bayes' Rule就可以了~我的英文真够Suck的
正解：
P=[(1/2)(1/5)+(1/6)(1/50)+(1/16)(1-1/5-1/50)]/(1/16)(1-1/5-1/50)
晕~真活该做不出来~如Kelvin所说的我对条件分布还不是很了解=.=