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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.

Please help me explaining that question

Thank you so much :)

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We evaluate ${\mathsf P(R)=0.5\\\mathsf P(B)=0.5\\\mathsf P(G)=0.5}\qquad{\mathsf P(R\cap B)=0.25\\\mathsf P(R\cap G)=0.25\\\mathsf P(B\cap G)=0.25}\qquad{\mathsf P(R\cap B\cap G)=0.25}$

Reason: Each color shows on two of the four faces, but only one face has more than one colour-to be precise all three colours.

Now just apply the definitions.

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? $$\mathsf P(R\cap B)\overset?=\mathsf P(R)\mathsf P(B)\\\mathsf P(R\cap G)\overset?=\mathsf P(R)\mathsf P(G)\\\mathsf P(B\cap G)\overset?=\mathsf P(B)\mathsf P(G)\\\mathsf P(R\cap B\cap G)\overset?=\mathsf P(R)\mathsf P(B)\mathsf P(G)$$

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