Understanding Bayes' theorem through an example

Question

Suppose I have three nodes A,B,C such that A and B are independent and pointed to C as the following:

A --> C <-- B

Also Suppose that each node takes a peobability between (0,1) so that the sum of them will be 1 (for example, A=0.3, B=0.5, C=0.2).

My question is the following:

In this case, is it true that calculating P(C|A,B) is equivalent to calculating P(C|A)*P(C|B)? If not, what is the difference between both?

Thank you.

Answer 1

Suppose $\Omega=\{1,2,3,\ldots,20\}$, with all $20$ outcomes equally probable, and

$$A= \{1,2,3,4,5,6\}\quad \text{and} \quad B= \{1,3,5,7,9,11,13,15,17,19\}.$$

Then $\Pr(A)=0.3$ and $\Pr(B)=0.5$ and $\Pr(A\ \&\ B) = \Pr(\{1,3,5\}) = 0.15 = 0.3\times0.5$, so $A$ and $B$ are independent.

Now suppose $C=\{2,4,17,19\}$. Then

$$\Pr(C\mid A)\Pr(C\mid B) = \dfrac 1 3 \times \dfrac 1 5 = \dfrac 1 {15} = 0.666666\ldots\ne 0 = \Pr(C\mid A\ \&\ B).$$