Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 16 / b

Let $Y$ indicate whether the draw from the second bag is green. It is independent of $X$, with $P(Y=1)=1/3$ and $P(Y=0)=2/3$. Exactly one green overall arises from the disjoint alternatives $(X,Y)=(1,0)$ or $(0,1)$. Therefore

The sole green ball comes from the first bag exactly on the event $(X,Y)=(1,0)$. Divide this joint probability by the conditioning event's probability:This is the conditional probability form of Bayes' theorem.