Some elements dependent, but collectively independent? Is it possible that some collection could be independent, even if some of its elements were dependent?
Or is collection always independent iff its elements are independent?
Or perhaps this calls for notions such as "pair-wise independent, but not three-wise".
 A: Let $X_1, \dotsc, X_k$ be a finite collection of random variables, defined on the same probability space.  For simplicity, assume in addition they are all real-valued. Then these random variables are mutually independent if for all $x_1, \dotsc, x_k$ we have 
$$ \DeclareMathOperator{\P}{\mathbb{P}}  
   \P\{ X_1 \le x_1, \dotsc, X_k\le x_k\} = \prod_{i=1}^k \P\{X_i\le x_i\}.
$$
Now let a subcollection be given by $1\le i_1 < \dotsm < i_r \le k$ and the complementary $k-r$ indices be given by $j_1,\dotsc,j_{k-r}$. Then we have 
$$
  \P\{X_{i_1}\le x_{i_1}, \dotsc, X_{i_r}\le x_{i_r}\}= \\
\P\{ X_{i_1}\le x_{i_1},\dotsc,X_{i_r}\le x_{i_r}, X_{j_1}\le +\infty, \dotsc, X_{j_{k-r}}\le +\infty \}=\\
\prod_{i \in \{i_1,\dotsc,i_r\}}\P\{ X_i\le x_i \}   \prod_{j\in\{j_1, \dots,j_{k-r}  \}} \P\{ X_j\le +\infty \}= \\
\prod_{i \in \{i_1,\dotsc,i_r\}}\P\{ X_i\le x_i \} 
$$ which indeed means that the subcollection is mutually independent. This indeed means that the collection cannot be mutually independent if any of its subcollections are not. 
While these proof where written for real random variables, it is valid more generally. The proof in the general case consists basically of translating the above argument into measure theory.  
A: A standard example involves two dice: #1 #2
$A$ = Event that the sum on #1 and #2 is 7
$B$ = Event that #1 shows 1
$C$ = Event that #2 shows 6
$P(A) = P(B) = P(C) = \frac{6}{36} = \frac{1}{6}$
$P(A \cap B) = P(B \cap C) = P(A \cap C) = \frac{1}{36} = P(A)P(B) = P(B)P(C) = P(A)P(C)$
But $P(A \cap B \cap C) = \frac{1}{36} \neq P(A)P(B)P(C)  $
$\implies$ $A$, $B$, $C$ are pairwise independent but not mutually independent
A: 1 No 2 Yes 


*

*Definition: A family $\mathscr F$ of events is independent if, for every finite number of distinct events $A_1$, $A_2$, $\ldots$, $A_n$ in $\mathscr F$, $$P\left(\bigcap_{i=1}^nA_i\right) =\prod_{i=1}^nP(A_i)$$


So if $A_{10}$ and $A_{777}$ are not independent and are both in $\mathscr F$, we just choose the subcollection $\{A_{10}, A_{777}\} \subseteq \mathscr F$ to show that $\mathscr F$ is not an independent family of events.
3 Pairwise independence of events does not imply independence of events. In general $k$-wise independence of events, for some k, does not imply independence of events. However, independent of events implies $k$-wise independence for any $k$.
