Pairwise independent events that are not mutually independent

Counterexample

Toss a fair coin two times, so $\Omega = \{HH, HT, TH, TT\}$ with probability $1/4$ for each outcome. Let $A$ be "the first toss is heads", $B$ be "the second toss is heads", and $C$ be "the two tosses are equal". Then $A = \{HH, HT\}$, $B = \{HH, TH\}$ and $C = \{HH, TT\}$, so $P(A) = P(B) = P(C) = 1/2$. Each intersection of two of them is $\{HH\}$, with probability $1/4 = \frac12 \cdot \frac12$. So $A$, $B$ and $C$ are pairwise independent. But $P(A \cap B \cap C) = P(\{HH\}) = 1/4$, and $P(A) \, P(B) \, P(C) = 1/8$. So $A$, $B$ and $C$ are not mutually independent.

Remarks

Here any two of the events determine the third: if $A$ and $B$ both occur, then $C$ occurs. The same idea with $n$ fair bits and their sums modulo $2$ gives large pairwise independent families. Such families are useful in computer science.

Explanations

L1 human

If you know the first toss, you know nothing about $C$, and the same for the second toss. But if you know both tosses, you know $C$ exactly. So the three events are not independent as a group.