TITLE: For 3 independent events A, B and C, is $\mathrm{P(A \cap B \cap C) = P(A)P(B)P(C)}$?
QUESTION [0 upvotes]: For 3 independent events A, B and C
is $\mathrm{P(A \cap B \cap C) = P(A)P(B)P(C)}$?
Just like for two independent events $\mathrm{P(A\cap B) = P(A)P(B)}$. 
Intuitively I think it is because the denominator gives the total number of possible outcomes and the numerator gives the favourable outcomes. 
Also, if we suppose that A,B and C are not independent and we keep changing the sample space after A is performed and then after B is performed, then would $\mathrm{P(A\cap B\cap C) = P(A)P(B)\text{(with altered sample space)}P(C)\text{(with twice altered sample space)}}$
I think I have used this in many problems and got the right answer but never thought about it rigorously. 
Example problem: 
A bag contains 1 red and 2 blue balls. An experiment consists of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour. 
Probability that at least one blue ball is drawn is? 
So to solve it we use $P(E) = 1- P(RRR) = 1- \dfrac{1}{3}\dfrac{2}{4}\dfrac{3}{5} = 0.9$
Why did we multiply here? Didn't we use that $P(RRR)= P(R)P(R)P(R)$ ?

REPLY [1 votes]: The definition of "{A,B,C} is independent" implies that the equation you wrote is true, as well the pairwise equations being true. As shown in one of the answers below, this is stronger just having the three pairwise independence equations hold, and it's also stronger than just having the equation you wrote hold, although I don't remember a counterexample. In general, independence for a finite set $S$ of events means, by definition, that the intersection/product equation holds for EVERY subset of $S$.
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