Any counter example for $P(A|C)=P(A),P(B|C)=P(B)$ but $P(A\cap B|C)\neq P(A\cap B)$? Let $A,B,C$ be three events, what would be an example that $P(A|C)=P(A)$ and $P(B|C)=P(B)$ do not imply $P(A\cap B|C)= P(A\cap B)$?

Let $\Omega = \{1,2,3,4\}$ be the sample space, events $A = \{1,2\}$, $B = \{1,3\}$, $C = \{1,4\}$.
\begin{align}
P(A|C) &= P(\text{choose 1 or 2 | 1 or 4 chosen}) = \frac12 \\
P(A) &= P(\text{choose 1 or 2}) = \frac12 = P(A|C) \\
P(B|C) &= P(B) = \frac12 \quad \text{similarly} \\
P(A\cap B | C) &= P(\text{choose 1 | 1 or 4 chosen}) = \frac12, \text{ but} \\
P(A\cap B) &= P(\text{choose 1}) = \frac14 \ne P(A\cap B | C).
\end{align}
