# “Pairwise independent” is weaker that “independent”

Can someone please give me a reference to an (simple, realworld, i.e. not constructed) example of a discrete probability space such that there are three events in it that are pairwise independent but all three together are not independent (although I wouldn't mind, if someone would give me the example as an answer).

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another example of not googling first... –  Bombyx mori Dec 28 '12 at 16:45
1) For my defense: I did google it -albeit not that long - but dismissed wikipedia a priori. Nonetheless I think -4 votes in a couple of minutes for a legitimate question is harsh and not entirely justifiable I think!  2) The example in Wikipedia does it's just, but it is very artificial/constructed! I would like a more natural example, that can be taken from a real-world scenario (I know, should specified this from the beginning - but thats still not an excuse for the people who brutally voted down) –  user26698 Dec 28 '12 at 17:05
There are really quite a lot of good math articles in wikipedia, though typesetting is an issue, and some statements does not have proofs. You should take a look before you raise a question like this. –  Bombyx mori Dec 28 '12 at 17:08
@user26698: I've voted up your comment, but in general you'll get better answers if you mention what you've tried (e.g. "I looked at wikipedia and found the example artificial".) –  ShreevatsaR Feb 11 '14 at 5:59

The standard example involves tossing $2$ fair coins. For a more symmetrical example, toss $3$ fair coins. Let $A$ be the event Toss $1$ and Toss $2$ give the same result, $B$ be the event Toss $2$ and Toss $3$ give the same result, and $C$ the event Toss $3$ and Toss $1$ give the same result.
We have $\Pr(A)=\Pr(B)=\Pr(C)=\frac{1}{2}$ and $\Pr(A\cap B)=\Pr(B\cap C)=\Pr(C\cap A)=\frac{1}{4}$.
However, it is clear that $A$, $B$, and $C$ are not mutually independent.