Number of ordered pairs of $A,B$ in Probability 
Let $|X|$ denote the number of elements in a set $X,$ 
Let $S = \{1,2,3,4,5,6\}$ be a sample space,
where each element is equally likely to occur. 
If $A$ and $B$ are independent events associated with
  $S,$ 
Then the number of ordered pairs $(A,B)$ 
such that $1 \leq |B| < |A|,$ equals

what i try
Let number of elements in $A,B,A \cap B$ is $x,y,z$ respectively
conditions given as $1\leq y\leq x.$
For $2$ Independent events $A,B$
is $\displaystyle P(A \cap B)=P(A)\cdot P(B)\Rightarrow \frac{z}{6}=\frac{x}{6}\cdot \frac{y}{6}\Rightarrow z=xy/6$

The equation
$$z=\frac{xy}{6}$$
shows that the events $A$ and $B$ can be independent only if


*

*$A=S$, or

*$|A|=3,|B|=2$

*$|A|=4,|B|=3$
In the 2nd case $A$ and $B$ have 1 common element; in the 3rd case $A$ and $B$ have 2 common elements.
In the 1st case, we have $2^6-2$ solutions ($B$ can be anything except $S$ and $\varnothing$).
In the 2nd case, we can choose $A$ by $\binom{6}{3}$ ways and then $B$ by $3\cdot 3$ ways. 
In the 3rd case we can choose common elements by $\binom{6}{2}$ ways, then 2 remaining A elements by $\binom{4}{2}$ ways and the remaining B element by 2 ways.
Summing it all
$$64-2+\frac{6\cdot 5\cdot 4}{6}\cdot 3 \cdot 3 + \frac{6\cdot 5}{2}\cdot \frac{4\cdot 3}{2}\cdot 2= 422$$

PS: thanks @karthikeya kurella for noticing the 3rd case.
