Prove that the pairwise independence of events A , B , C…

Prove that the pairwise independence of events $A$, $B$, $C$ does not imply their independence as a whole (mutual independence).

Explanation of Method

To prove that one property does not follow from another, one counterexample suffices. We need three events such that every pair satisfies the product rule, but $P(ABC)\ne P(A)P(B)P(C)$. Bernstein's tetrahedron gives such events; Problem 113 gave an equivalent example with two coin tosses.

Step 1: What must be shown

We need events with

\[ P(AB)=P(A)P(B),\quad P(AC)=P(A)P(C),\quad P(BC)=P(B)P(C), \quad\text{but}\quad P(ABC)\ne P(A)P(B)P(C). \]

Step 2: Bernstein's tetrahedron

A regular tetrahedron is thrown; it lands on each of its four faces with probability $\frac14$. One face is painted red, one blue, one green, and the fourth face has all three colours (red, blue and green stripes). Let

\[ A=\text{"the bottom face has red"},\quad B=\text{"... blue"},\quad C=\text{"... green"}. \]

Face	red	blue	green
1	✓
2		✓
3			✓
4 (all colours)	✓	✓	✓

Step 3: Pairwise independent

Each colour appears on $2$ faces: $P(A)=P(B)=P(C)=\frac24=\frac12$. Each pair of colours appears together only on face $4$:

\[ P(AB)=P(AC)=P(BC)=\frac14=\frac12\cdot\frac12. \ \checkmark \]

Step 4: Not independent as a whole

\[ P(ABC)=P(\text{face }4)=\frac14\ne\frac18=P(A)P(B)P(C). \]

In words: knowing that red and blue are both at the bottom (face $4$), green is certain: $P(C\mid AB)=1\ne\frac12=P(C)$.

\[ \boxed{\text{Pairwise independence does not imply mutual independence.}} \quad\blacksquare \]

Step 5: The same with two coins

Toss two fair coins: $A$ = "first heads", $B$ = "second heads", $C$ = "both show the same side". Again every pair is independent but $P(ABC)=\frac14\ne\frac18$ (see Problem 113).

Result


Example	tetrahedron with faces R, B, G, RBG; $A,B,C$ = colour on the bottom face
Pairs	$P(AB)=P(AC)=P(BC)=\frac14=\frac12\cdot\frac12$
Triple	$P(ABC)=\frac14\ne\frac18$