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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 $P(RRR)= P(R)P(R)P(R)$ ?

Archer
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3 Answers3

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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 $S$.

Ned
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  • What does pairwise independence mean? – Archer Nov 24 '18 at 04:05
  • It means "taken two at a time" – Ned Nov 24 '18 at 04:05
  • Could you give a counterexample to when three events are independent but not pairwise independent? – Archer Nov 24 '18 at 04:06
  • I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S. – Ned Nov 24 '18 at 04:11
  • https://math.stackexchange.com/a/2891712/476145 – Archer Nov 24 '18 at 04:12
  • Isn't Akash Roy's comment wrong? – Archer Nov 24 '18 at 04:12
  • The text I was using defined "independence" to be what that link calls "mutual independence" and didn't have a special word for a single occurrence of the intersection/product equation without it necessarily holding for proper subsets. Of course, it's all the same for a set of two events, so this issue of terminology only comes up with three or more events. – Ned Nov 24 '18 at 04:19
  • i am unable to understand when the simple multiplication wont work. Could you please see the example i have added to my question? – Archer Nov 24 '18 at 04:24
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    That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc – Ned Nov 24 '18 at 04:28
  • @Ned I actually had constructed a formula for independence of 3 events and the formula which Abcd has given was only valid for a special case. – Akash Roy Nov 24 '18 at 05:32
  • @Abcd, here is a non-elegant example. Let U = {1,2,3,...,27} with all 27 outcomes equally likely, let A = {1,2,3,...,8, 27}, B = { 9,10,11,...,16, 27}, C = { 17,18,19,...,24, 27}. Then each of A, B, C has probability 1/3, and the intersection-product equation holds for {A,B,C}, but not for any two of them. – Ned Nov 24 '18 at 10:52
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If I remember right, the idea here is this:

$P(A,B,C) = P(A|B,C)*P(B,C) = P(A|B,C)*P(B|C)*P(C)$

Chain rule for conditional probability

Makina
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If the 3 events are just pairwise independent then it is easy to find counterexamples to the product rule. For instance consider the experiment "toss two coins", and the following events:

A = getting heads on the first coin.

B = getting heads on the second coin.

C = getting the same outcome (heads or tails) on both coins.

Those events are pairwise independent, we have $P(A) = P(B) = P(C) = 1/2$, and we have $P(A\cap B) = P(A)P(B) =1/4$, $P(A\cap C) = P(A)P(C) = 1/4$, and $P(B\cap C) = P(B)P(C) = 1/4$, however $P(A\cap B\cap C) = 1/4 \neq P(A)P(B)P(C)$.

If what you mean is that besides being pairwise independent each event is also independent from the other two, i.e., $P(A|B\cap C) = P(A)$, etc., then the claim is true: $$ P(A\cap B\cap C) = P(A|B\cap C) P(B\cap C) = P(A) P(B) P(C) \,. $$

For the second part, I assume it depends on how the sample space is altered.

mlerma54
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