Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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) – MyCatsHat 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
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.