# Prove 3 coin toss events as pairwise independent but not mutual

Suppose you roll three distinguishable fair dice and call the resulting numbers a, b, and c. Define events X= “a+b is even", ”Y= “b+c is even", and Z= “a+c is even”. Prove that these three events are pairwise independent but not mutually independent.

Right now, this is how I understand the difference between pairwise independence and mutual independence.

𝐴,𝐵,𝐶 are mutually independent if $$𝑃(𝐴∩𝐵∩𝐶)=𝑃(𝐴)𝑃(𝐵)𝑃(𝐶)$$

But this is not the case for pairwise independence.

Any help would be greatly appreciated!

Suppose we've been given a collection of events $$X_1,X_2,...,X_n.$$

The events are mutually independent if $$P\left(\bigcap_{k=1}^n X_k\right)=\prod_{k=1}^n P(X_k),$$ and pairwise independent if for all $$j,k$$ with $$1\le j we have $$P(X_j\cap X_k)=P(X_j)\cdot P(X_k).$$

So, you'll need to show the following: $$P(A\cap B)=P(A)P(B)\\P(A\cap C)=P(A)P(C)\\P(B\cap C)=P(B)P(C)\\P(A\cap B\cap C)\neq P(A)P(B)P(C)$$

Note that $$Z= X\odot Y$$ (coincidence operator).

COnsqeuently $$P X = P Y = PZ = {1 \over 2}$$ and $$P[X \cap Y \cap \bar Z| = 0$$.

The definition of mutual independence for three events (using $$A,B$$ to denote $$A\cap B$$ is: $$P(A,B)=P(A)P(B), \qquad P(B, C)=P(B)P(C),\qquad P(C,A)=P(C)P(A),\\ P(A,B,C)=P(A)P(B)P(C)$$ So there are four conditions to check, not just the three way one you listed. For pairwise, you only need the first three.

For mutual independence of $$n$$ events, there are $$2^n-n-1$$ conditions, one for each subset of the events.

Note that $$a+b,a+c,$$ and $$b+c$$ are all even iff and only if $$a,b,c$$ are all even or $$a,b,c$$ are all odd. Therefore,

$$P(X,Y,Z)=P(a,b,c\text{ all even})+P(a,b,c\text{ all odd})=(1/2)^3+(1/2)^3=1/4.$$ However, $$P(X)=P(a,b\text{ both even})+P(a,b\text{ both odd})=(1/2)^2+(1/2)^2=1/2,$$ so $$P(X)P(Y)P(Z)=(1/2)^3=1/8.$$