Are A, B and C mutually independent? Chose a number uniformly at random on $[0,1]$. Consider the events
$A= [0,1/2] \quad B= [1/3,2/3], \quad C= [11/24,17/24]$
Are A, B, C independent (i.e. mutually independent)?
I realize that I need to show that $\mathbb{P}(A\cap B\cap C)=\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)$. But I don't understand what the events are in the first place. How are we choosing those two numbers on $[0,1]$?
 A: To show that two events are independent, you need only show that $P(A\cap B)=P(A)P(B)$, but to show that three events are independent, you need to not only show $P(A\cap B\cap C)=P(A)P(B)P(C)$ but also pairwise independence $P(A\cap B)=P(A)P(B),P(A\cap C)=P(A)P(C), \text { and }P(B\cap C)=P(B)P(C)$.
For this problem, A is the event that the random number falls into the interval given, and similarly for B and C. It is easiest to draw the intervals on a number line to see where they overlap.
$P(A)=\frac 1 2, P(B)=\frac 1 3, P(C)=\frac 1 4, P(A\cap B\cap C)=\frac 1 2 - \frac {11} {24}=P(A)P(B)P(C)$, so this condition holds.
Check the conditions for pairwise independence:
$P(A\cap B)=\frac 1 2 - \frac 1 3=P(A)P(B)$
$P(A\cap C)=\frac 1 2 - \frac {11} {24} \approx .042 \ne P(A)P(C)=\frac  1 8 =.125$. Hence the three events are not independent.

In general, to show that a set of events $A_1,...,A_k$ is independent, you must show that for every subset $A_{1_{j_l}},...,A_{i_{j_l}};j=2,...,k;i=1,...,j;l=1,...,{k \choose j}$, where j is the size of the subset, i runs from 1 to the subset size, and l runs from 1 to the number of subsets of that size, that $P(A_{1_{j_l}}\cap...\cap A_{i_{j_l}})=P(A_{1_{j_l}})...P(A_{i_{j_l}})$.
