Three fair coins are tossed. Let A be the event that the first and second coins show the same face…

Step 1. Each event. $A$ holds for HHH, HHT, TTH and TTT, so $P(A) = \dfrac{4}{8} = \dfrac{1}{2}$. By the same count $P(B) = P(C) = \dfrac{1}{2}$.

Step 2. A pair. $A \cap B$ means the first two agree and the last two agree, so all three coins agree: HHH or TTT. So $P(A \cap B) = \dfrac{2}{8} = 0.25$. This equals $\dfrac{1}{2} \times \dfrac{1}{2}$, so $A$ and $B$ are independent. The pairs $A, C$ and $B, C$ give the same.

Step 3. All three. $A \cap B \cap C$ is again HHH or TTT, so $P(A \cap B \cap C) = 0.25$.

Step 4. The product. $P(A)\,P(B)\,P(C) = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = 0.125$, which is not $0.25$.

Why it happens. If you know that $A$ and $B$ both happened, then $C$ is certain. Two of the events together tell you everything about the third.