Bayes’ Theorem for three events

Question

When researching this topic, I found a website that said $$P(A|BC) = \cfrac{P(C|BA)P(A|B)}{P(C|B)}$$ but they did not provide a proof so I was wondering if someone could explain the proof for this formula? I want to understand the formula instead of just plugging numbers in.

Answer 1

Write it as $$P(A|BC)P(C|B) =P(ABC|B)= P(C|BA)P(A|B) $$

Answer 2

$$P(A|BC) = \cfrac{P(ABC)}{P(BC)} \mathit{\tag{By definition}}$$

$$P(C|AB) = \cfrac{P(ABC)}{P(AB)} \therefore P(ABC)=P(C|AB)P(AB)$$

$$P(A|B) = \cfrac{P(AB)}{P(B)},\ P(C|B)=\cfrac{P(BC)}{P(B)} \\ \therefore P(AB) =P(A|B)P(B), \ P(BC) = P(C|B)P(B)$$

$$\therefore P(A|BC) = \cfrac{P(C|AB)P(AB)}{P(BC)}=\cfrac{P(C|AB)P(A|B)P(B)}{P(C|B)P(B)}=\cfrac{P(C|AB)P(A|B)}{P(C|B)}$$

If you want some further intuition, I suggest drawing a three-set venn diagram consisting of overlapping events $A,B,$ and $C$, and seeing what the region $A|BC$ looks like.

Answer 3

I want to understand the formula instead of just plugging numbers in.

Draw a big rectangle. Inside the rectangle, draw 3 overlapping circles in a Venn diagram. Label them $A$, $B$, $C$.

For the left-hand side, $P(A | BC)$ by definition represents the fraction of the area of $BC$ that's occupied by $A$. This is $P(ABC)/P(BC)$.

For the right-hand side, $P(C|BA)$ is the fraction of $C$ in both $B$ and $A$, which is $P(ABC)/P(AB)$. To turn this into the previous expression, we need to multiply by $P(AB)/P(CB)$, which is $P(A|B)/P(C|B)$.