# Pairwise independent and mutually independent

This is the question problem i have in my tutorial sheet in school. It is about pairwise independent and mutually independent

Generally, i have listened to my lecturer explaining how those two are different but i still can not get out of the confusion and can’t really understand or how to differentiate between those two.

The problem below has very short answer. Just two calculation, but i do not know how they arrive to that answer. Specifically, for every pair of color, we have the probability is .25. But if it is pairwise between 3 colors(?), it is still .25 ( i don’t get this point) and that is different from mutually between 3 colors which is .125. And the problem is solved!

Events $A_1,...,A_n$ are (mutually) independent if for any $m$ indices $1 ≤ i_1 < i_2 < ··· < i_m ≤ n$, $P(\bigcap_jA_i) = P(A_i)$ .

Consider a regular tetrahedron (a polyhedron with 4 identical triangular face with one face painted green, one red, one blue and another one which is painted with all 3 colors. The tetrahedron is rolled and we note the face it lands on. Let $G, R,$ and $B$ be the events that the face has green, red and blue respectively. Show that these events are pairwise independent but not (mutually) independent.

Thank you so much :)

We evaluate ${\mathsf P(R)=0.5\\\mathsf P(B)=0.5\\\mathsf P(G)=0.5}\qquad{\mathsf P(R\cap B)=0.25\\\mathsf P(R\cap G)=0.25\\\mathsf P(B\cap G)=0.25}\qquad{\mathsf P(R\cap B\cap G)=0.25}$
The events in the collection will be pairwise independent if every pair of events are independent. Are they?$$\mathsf P(R\cap B)\overset?=\mathsf P(R)\mathsf P(B)\\\mathsf P(R\cap G)\overset?=\mathsf P(R)\mathsf P(G)\\\mathsf P(B\cap G)\overset?=\mathsf P(B)\mathsf P(G)$$
The events in the collection will be mutually independent if every subcollection is independent. Are they? $$\mathsf P(R\cap B)\overset?=\mathsf P(R)\mathsf P(B)\\\mathsf P(R\cap G)\overset?=\mathsf P(R)\mathsf P(G)\\\mathsf P(B\cap G)\overset?=\mathsf P(B)\mathsf P(G)\\\mathsf P(R\cap B\cap G)\overset?=\mathsf P(R)\mathsf P(B)\mathsf P(G)$$