probability involving conditional prob

Question

I just want to know if i am able to prove that $P(A|B,C)P(B|C)=P(A|C)$

I tried to solve but i couldn't get rid of the B,C intersection part. I cancelled a lot of terms but the $P(B,C|A)$ term is bothering me. thanks!

Answer 1

Your statement is not true, so you cannot prove it! On the other hand what would be true is a statement of the following form: (Where I am assuming that by $A$, $B$, $C$ you are denoting events).

$$P(A|C)= P(A|B,C)P(B|C) + P(A|\neg B,C)P(\neg B|C) $$

By $\neg B$ I am denoting the complementary event of $B$, i.e. if $\Omega$ is your sample space, then $\neg B = \Omega \setminus B$.

Answer 2

Note that $\Pr(X\mid Y)=\frac{\Pr(X\cap Y)}{\Pr(Y)}$.

Using that a couple of times, we get that the left-hand side is equal to $\Pr((A\cap B)\mid C)$.

Now we have a clue for producing an example where this not $\Pr(A\mid C)$.

Let $C$ be any event with probability $1$. Toss a fair coin once. Let $A$ be the event we get a head, and let $B$ be the event we get a tail. Then $\Pr(A\mid B,C)=0$ so the product on the left-hand side is $0$, while $\Pr(A|C)=1/2$. (One can also construct less trivial counterexamples!)