Conditional Probability and Independence Let $A,B,C$ be events, and $A,C$ be pairwise independent. Can I say that $$P(A\mid B \cap C)=P(A\mid B)$$
Any help would be appreciated.
 A: A counterexample. Let $Ω=\{a_1,a_2,a_3,a_4\}$ with $P(a_i)=1/4$ (equally likely). Let $$A=\{a_1,a_2\},\quad B=\{a_2,a_3\},\quad C=\{a_1,a_3\}$$ The sets $A,B,C$ are pairwise but not mutually independence. In this case $$P(A\mid B\cap C)=0 \quad \text{while} \quad P(A\mid B)=\frac12$$
A: No, that's only true if $A,C$ are conditionally independent given $B$.
Counter-example: toss a fair die with events:
$$B=\{2,4,6\},\quad A=\{2,4\},\quad C=\{1,2,3\}.$$
Then, $P(A\cap C)=P(\{2\}) = 1/6 = P(A)P(C),\;$ so that $A,C$ are independent.
However,
$$P(A\mid B,C)=P(\{2,4\}\mid \{2\})=1 \\
  P(A\mid B)=P(\{2,4\}\mid \{2,4,6\})=2/3.$$
A: The answer is: "no".
E.g. substitute $B=A\cup C$. If $0<P(A\cup C)<1$ then:
$P(A\mid (A\cup C)\cap C)=P(A\mid C)=P(A)\neq P(A\mid A\cup C)$

You are actually asking wether the following statement holds if $P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$:
$$P\left(A\cap B\cap C\right)P\left(B\right)=P\left(A\cap B\right)P\left(B\cap C\right)$$
and the answer on that is: not in general.
Note that the LHS equals $P\left(A\mid B\cap C\right)P\left(B\cap C\right)P\left(B\right)$
and the RHS equals $P\left(A\mid B\right)P\left(B\cap C\right)P\left(B\right)$.
