Intuition/Real-life Examples: Pairwise Independence vs (Mutual) Independence Would someone please advance/discuss some real-life situations falsities $1, 2$?
I'd like to intuit why these are false. As a neophyte, since I still need to compute the probabilities for the examples in the two answers to apprehend them, I haven't naturalised these concepts yet.
Thus, I beg leave to ask about other real-life examples which require less or no computations.
I tried and would appreciate less numerical examples than http://notesofastatisticswatcher.wordpress.com/2012/01/02/pairwise-independence-does-not-imply-mutual-independence/ and http://econ.la.psu.edu/~hbierens/INDEPENDENCE.PDF, and  Examples 1.22 and 1.23 on P39 of Introduction to Pr by Bertsekas.
$1.$ Pairwise Independence $\require{cancel} \cancel{\implies}$ (Mutual) Independence.    
$2.$ Pairwise Independence $\require{cancel} \cancel{\Longleftarrow}$ (Mutual) Independence.    

P38 defines (Mutual) Independence : 
  For all $S \subseteq \{1, ..., n - 1, n\} $ and events $A_i$, $Pr(\cap_{i \in S} A_i) = \Pi_{i \in S} A_i.$


Head appears in the first toss.
B: Head appears in the second toss.
C: Head appears in the third toss.
D: A and B yield the same outcome.
Mutual independence:
Firstly we only consider the $A, B, C$ (just treat $D$ as nonexistent). It is obvious that they are mutual independent. And here are two perspectives of this statement.

*

*$$P(A\cap B\cap
C)=P(A)P(B)P(C)=\frac{1}{8}$$ $$P(A\cap B)=P(A)P(B)=\frac{1}{4}$$ $$P(A\cap C)=P(A)P(C)=\frac{1}{4}$$ $$P(B\cap C)=P(B)P(C)=\frac{1}{4}.$$


*We firstly calculate the joint probability: $$P(A\cap B\cap
C)=P(A)P(B)P(C)=\frac{1}{8}$$
And then get the marginal probability $P(B, C)$ by summing out A:
$$P(B, C) = P(A, B, C) + P(\neg A, B, C)=\frac{1}{4}$$
And we can verify the other two easily due to the symmetry of A, B and C.

Mutual independence implies pairwise independence because we can just marginalize out the variables not in each pair.

Pairwise independent but not mutual independent
Let's consider A, B and D(just treat C as nonexistent).
$$P(A\cap B\cap
D)=\frac{1}{4}\neq P(A)P(B)P(D)=\frac{1}{8}$$ $$P(A\cap B)=P(A)P(B)=\frac{1}{4}$$ $$P(A\cap D)=P(A)P(D)=\frac{1}{4}$$ $$P(B\cap D)=P(B)P(D)=\frac{1}{4}.$$
We can see that A, B and C are mutually independent(thereof pairwise independent), but A, B and D are only pairwise independent and don't satisfy the mutual independence.
Refrence:

*

*Pairsewise independence versus mutual independence by Isaac (Ed) Leonard.

*Mutually Independent Events by Albert R Meyer

*Probabilistic graphical model by Stefano Ermon
