The problem was posed in video PP 2.3 of Math Monk's probability primer.
Problem
$\Omega = \{1, 2, 3, 4\}$, $P(k) = \frac{1}{4} \forall k \in \Omega$
$A = \{1, 2\}$, $B = \{1, 3\}$, $C = \{2, 3\}$
Are $A$ and $B$ independant?
Are $A, B, C$ mutually independent? Pairwise independent?
Are $A, B$ conditionally independent given $C$?
Solution
$A$ and $B$ independant?
$P(A) = \frac{1}{2}$, $P(B) = \frac{1}{2}$, $P(A \cap B) = P(\{1\}) = \frac{1}{4} = P(A)*P(B)$
So yes, $A$ and $B$ are independent.
$A, B, C$ mutually independent?
$P(A) = P(B) = P(C) = \frac{1}{2}$
$A \cap B \cap C = \emptyset$
So $P(A \cap B \cap C) = 0 \neq P(A)P(B)P(C) = \frac{1}{8}$
And thus they are not independent.
$A, B, C$ pairwise independent?
Yes, each pair shares one element and are independent for the same reason $A$ and $B$ are independent.
$A, B$ conditionally independent given $C$?
This asks whether $P(A \cap B \mid C) = P(A \mid C)P(B \mid C)$
$P(A \mid C) = \frac{P(A \cap C)}{P(C)} = \frac{1/4}{1/2} = \frac{1}{2}$
Similarly, $P(B \mid C) = \frac{1}{2}$
$P(A \cap B \mid C) = \frac{P(A \cap B)}{P(C)} = \frac{1/4}{1/2} = 1/2$
So $P(A \cap B \mid C) = 1/2 \neq 1/4 = 1/2 * 1/2 = P(A \mid C)P(B \mid C)$
and thus no, $A$ and $B$ are not conditionally independent given $C$.