There are two urns each containing red and green balls. Urn A has 60% red balls and 40% green. Urn B has 80% red balls and 20% green. We chose one Urn randomly and draw a ball out of it. It turns out to be red. Now what is the probability that Urn A was chosen.
Let A=Urn A chosen, B=Urn B chosen, R=red ball drawn. Then,
$P(A)=P(B)=0.5$
$P(R\mid A)=0.6$ and $P(R\mid B)=0.8$ and we need $P(A\mid R)$.
Bayes' rule provides us with the solution $P(A\mid R)=\frac{P(A)\,P(R\mid A)}{P(A)\,P(R\mid A)+P(B)\,P(R\mid B)}$. It also says that for the above to be applicable the two hypothesis should be mutually exclusive and jointly exhaustive.
A and B are mutually exclusive, it that only one Urn can be chosen at any instant. How to check if they are jointly exhaustive ?.
Jointly exhaustive is defined as alt least one of the hypothesis must be true.
Yes. Or to put it another way: one of the events must occur. In this case: you must chose one urn from $A$ or $B$. There is no other choice; the union of the events includes all outcomes in the sample space.$$A\cup B = \Omega \qquad\text{exhaustive}\\A\cap B=\emptyset \qquad \text{exclusive}~~$$